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Marvin Ray Burns

Try to beat these MRB constant records!

Marvin Ray Burns
Posted 5 years ago
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As with any scientific paper, this post contains only reproducible results with all methods disclosed. These records represent the advancement of consumer level computers and clever programming over the past 20 years. I see others breaking these records, even after I die!

Map:

  • First we have these record number of digits of the MRB constant computations.
  • Then we have some hints for anyone serious about breaking my record.
  • Followed by speed records,
  • a program Richard Crandall wrote to check his code for computing record number of digits
  • and a conversation about whether Mathematica uses the same algorithm for computing MRB by a couple of different methods.
  • Then, for a few replies, we compute the MRB constant from Crandall's eta derivative formulas and see records there.
  • There are three replies about "NEW RECORD ATTEMPTS OF 4,000,000 DIGITS!" and the computation is now complete!!!!!.
  • We see where I am on a 5,000,000 digit calculation. (ACTIVE and IN PROGRESS!!!!!!!!!!!!)
  • Finally, I introduce the MRB supercomputer!!!!!! (faster than some computers with dual platinum Xeon processors)

POSTED BY: Marvin Ray Burns.

MKB constant calculations,

enter image description here , have been moved to their own discussion at Calculating the digits of the MKB constant.

The MRB constant is defined below. See http://mathworld.wolfram.com/MRBConstant.html.

$$\text{MRB}=\sum _{n=1}^{\infty } (-1)^n \left(n^{1/n}-1\right).$$

Here are some record computations. If you know of any others let me know.

  1. On or about Dec 31, 1998 I computed 1 digit of the (additive inverse of the) MRB constant with my TI-92's, by adding 1-sqrt(2)+3^(1/3)-4^(1/4) as far as I could That first digit, by the way, is just 0 Then by using the sum feature to compute $\sum _{n=1}^{1000 } (-1)^n \left(n^{1/n}\right),$ I computed the first decimal (.1).

  2. On Jan 11, 1999 I computed 4 decimals(.1878) of the MRB constant with the Inverse Symbolic Calculator, with the command evalf( 0.1879019633921476926565342538468+sum((-1)^n* (n^(1/n)-1),n=140001..150000)); were 0.1879019633921476926565342538468 was the running total of t=sum((-1)^n* (n^(1/n)-1),n=1..10000), then t= t+10001.. 20000, then t=t+20001..30000 up to t=t+130001..140000.

  3. In Jan of 1999 I computed 5 correct decimals (rounded to .18786)of the MRB constant using Mathcad 3.1 on a 50 MHz 80486 IBM 486 personal computer operating on Windows 95.

  4. Shortly afterwards I tried to compute 9 digits of the MRB constant using Mathcad 7 professional on the Pentium II mentioned below, by summing to 10,000,000, 20,000,000, and a many more, then linearly approximating the sum to a what a few billion terms would have given.

  5. On Jan 23, 1999 I computed 500 digits of the MRB constant with the online tool called Sigma. Remarkably the sum in 4. was correct to 6 of the 9 decimal places! See http://marvinrayburns.com/OriginalMRBPost.html if you can read the printed and scanned copy there.

  6. In September of 1999, I computed the first 5,000 digits of the MRB Constant on a 350 MHz Pentium II with 64 Mb of ram using the simple PARI commands \p 5000;sumalt(n=1,((-1)^n*(n^(1/n)-1))), after allocating enough memory.

  7. On June 10-11, 2003 over a period, of 10 hours, on a 450mh P3 with an available 512mb RAM, I computed 6,995 accurate digits of the MRB constant.

  8. Using a Sony Vaio P4 2.66 GHz laptop computer with 960 MB of available RAM, on 2:04 PM 3/25/2004, I finished computing 8000 digits of the MRB constant.

  9. On March 01, 2006 with a 3GH PD with 2GBRAM available, I computed the first 11,000 digits of the MRB Constant.

  10. On Nov 24, 2006 I computed 40, 000 digits of the MRB Constant in 33hours and 26min via my own program in written in Mathematica 5.2. The computation was run on a 32-bit Windows 3GH PD desktop computer using 3.25 GB of Ram.

  11. Finishing on July 29, 2007 at 11:57 PM EST, I computed 60,000 digits of MRB Constant. Computed in 50.51 hours on a 2.6 GH AMD Athlon with 64 bit Windows XP. Max memory used was 4.0 GB of RAM.

  12. Finishing on Aug 3 , 2007 at 12:40 AM EST, I computed 65,000 digits of MRB Constant. Computed in only 50.50 hours on a 2.66GH Core2Duo using 64 bit Windows XP. Max memory used was 5.0 GB of RAM.

  13. Finishing on Aug 12, 2007 at 8:00 PM EST, I computed 100,000 digits of MRB Constant. They were computed in 170 hours on a 2.66GH Core2Duo using 64 bit Windows XP. Max memory used was 11.3 GB of RAM. Median (typical) daily record of memory used was 8.5 GB of RAM.

  14. Finishing on Sep 23, 2007 at 11:00 AM EST, I computed 150,000 digits of MRB Constant. They were computed in 330 hours on a 2.66GH Core2Duo using 64 bit Windows XP. Max memory used was 22 GB of RAM. Median (typical) daily record of memory used was 17 GB of RAM.

  15. Finishing on March 16, 2008 at 3:00 PM EST, I computed 200,000 digits of MRB Constant using Mathematica 5.2. They were computed in 845 hours on a 2.66GH Core2Duo using 64 bit Windows XP. Max memory used was 47 GB of RAM. Median (typical) daily record of memory used was 28 GB of RAM.

  16. Washed away by Hurricane Ike -- on September 13, 2008 sometime between 2:00PM - 8:00PM EST an almost complete computation of 300,000 digits of the MRB Constant was destroyed. Computed for a long 4015. Hours (23.899 weeks or 1.4454*10^7 seconds) on a 2.66GH Core2Duo using 64 bit Windows XP. Max memory used was 91 GB of RAM. The Mathematica 6.0 code used follows:

    Block[{$MaxExtraPrecision = 300000 + 8, a, b = -1, c = -1 - d, 
 d = (3 + Sqrt[8])^n, n = 131 Ceiling[300000/100], s = 0}, a[0] = 1; 
 d = (d + 1/d)/2; For[m = 1, m < n, a[m] = (1 + m)^(1/(1 + m)); m++]; 
 For[k = 0, k < n, c = b - c; 
  b = b (k + n) (k - n)/((k + 1/2) (k + 1)); s = s + c*a[k]; k++]; 
 N[1/2 - s/d, 300000]]

  17. On September 18, 2008 a computation of 225,000 digits of MRB Constant was started with a 2.66GH Core2Duo using 64 bit Windows XP. It was completed in 1072 hours. Memory usage is recorded in the attachment pt 225000.xls, near the bottom of this post.

  18. 250,000 digits was attempted but failed to be completed to a serious internal error which restarted the machine. The error occurred sometime on December 24, 2008 between 9:00 AM and 9:00 PM. The computation began on November 16, 2008 at 10:03 PM EST. Like the 300,000 digit computation this one was almost complete when it failed. The Max memory used was 60.5 GB.

  19. On Jan 29, 2009, 1:26:19 pm (UTC-0500) EST, I finished computing 250,000 digits of the MRB constant. with a multiple step Mathematica command running on a dedicated 64bit XP using 4Gb DDR2 Ram on board and 36 GB virtual. The computation took only 333.102 hours. The digits are at http://marvinrayburns.com/250KMRB.txt . The computation is completely documented in the attached 250000.pd at bottom of this post.

  20. On Sun 28 Mar 2010 21:44:50 (UTC-0500) EST, I started a computation of 300000 digits of the MRB constant using an i7 with 8.0 GB of DDR3 Ram on board, but it failed due to hardware problems.

  21. I computed 299,998 Digits of the MRB constant. The computation began Fri 13 Aug 2010 10:16:20 pm EDT and ended 2.23199*10^6 seconds later | Wednesday, September 8, 2010. I used Mathematica 6.0 for Microsoft Windows (64-bit) (June 19, 2007) That is an average of 7.44 seconds per digit.. I used my Dell Studio XPS 8100 i7 860 @ 2.80 GH 2.80 GH with 8GB physical DDR3 RAM. Windows 7 reserved an additional 48.929 GB virtual Ram.

  22. I computed exactly 300,000 digits to the right of the decimal point of the MRB constant from Sat 8 Oct 2011 23:50:40 to Sat 5 Nov 2011 19:53:42 (2.405*10^6 seconds later). This run was 0.5766 seconds per digit slower than the 299,998 digit computation even though it used 16GB physical DDR3 RAM on the same machine. The working precision and accuracy goal combination were maximized for exactly 300,000 digits, and the result was automatically saved as a file instead of just being displayed on the front end. Windows reserved a total of 63 GB of working memory of which at 52 GB were recorded being used. The 300,000 digits came from the Mathematica 7.0 command

    Quit; DateString[]
digits = 300000; str = OpenWrite[]; SetOptions[str, 
PageWidth -> 1000]; time = SessionTime[]; Write[str, 
NSum[(-1)^n*(n^(1/n) - 1), {n, \[Infinity]}, 
WorkingPrecision -> digits + 3, AccuracyGoal -> digits, 
Method -> "AlternatingSigns"]]; timeused = 
SessionTime[] - time; here = Close[str]
DateString[]

  23. 314159 digits of the constant took 3 tries do to hardware failure. Finishing on September 18, 2012 I computed 314159 digits, taking 59 GB of RAM. The digits are came from the Mathematica 8.0.4 code

    DateString[]
NSum[(-1)^n*(n^(1/n) - 1), {n, \[Infinity]}, 
WorkingPrecision -> 314169, Method -> "AlternatingSigns"] // Timing
DateString[]

  24. Sam Noble of Apple computed 1,000,000 digits of the MRB constant in 18 days 9 hours 11 minutes 34.253417 seconds.

  25. Finishing on Dec 11, 2012 Richard Crandall, an Apple scientist, computed 1,048,576 digits in a lighting fast 76.4 hours computation time (from the timing command). That's on a 2.93 Ghz 8-core Nehalem. In Aug of 2018 I computed 1,004,993 digits of the MRB constant in 53.5 hours with 10 DDR4 RAM (of up to 3000 MHz) supported processor cores overclocked up to 4.7 GHz! Search this post for "53.5" for documentation. Sept 21, 2018: I computed 1,004,993 digits of the MRB constant in 50.37 hours of absolute time (35.4 hours computation time) with 18 (DDR3 and DDR4) processor cores! Search this post for "50.37 hours" for documentation.** Finally, now, May 11, 2019, I computed over 1,004,993 digits, using 28 kernels on 18 DDR4 RAM (of up to 3200MHz) supported cores overclocked up to 5.1 GHz in 45,5 hours of absolute time and only 32.5 hours of computation time! Search 'Documented in the attached ":3 fastest computers together 3.nb." ' for the post that has the attached documenting notebook.

  26. Previously, I computed a little over 1,200,000 digits of the MRB constant in 11 days, 21 hours, 17 minutes, and 41 seconds (finishing on on March 31 2013). I used a six core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz 3.20 GHz.

  27. On May 17, 2013 I finished a 2,000,000 or more digit computation of the MRB constant, using only around 10GB of RAM. It took 37 days 5 hours 6 minutes 47.1870579 seconds. I used my six core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz 3.20 GHz.

  28. A previous world record computation of the MRB constant was finished on Sun 21 Sep 2014 18:35:06. It took 1 month 27 days 2 hours 45 minutes 15 seconds.The processor time from the 3,000,000+ digit computation was 22 days. I computed the 3,014,991 digits of the MRB constant with Mathematica 10.0. I Used my new version of Richard Crandall's code in the attached 3M.nb, optimized for my platform and large computations. I also used a six core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz 3.20 GHz with 64 GB of RAM of which only 16 GB was used. Can you beat it (in more number of digits, less memory used, or less time taken)? This confirms that my previous "2,000,000 or more digit computation" was actually accurate to 2,009,993 digits. they were used to check the first several digits of this computation. See attached 3M.nb for the full code and digits.

  29. Finished on Wed 16 Jan 2019 19:55:20, I computed over 4 million digits of the MRB constant. It took 4 years of continuous tries. This successful run took 65.13 days computation time, with a processor time of 25.17 days, on a 3.7 GH overclocked up to 4.7 GH on all cores Intel 6 core computer with 3000 MHz RAM. According to this computation, the previous record, 3,000,000+ digit computation, was accurate to 3,014,871 decimals, as this computation used my own algorithm for computing n^(1/n) as found at chapter 3 in the paper at

https://www.sciencedirect.com/science/article/pii/0898122189900242

and the 3 million+ computation used Crandall's algorithm. Both algorithms outperform Newton's method per calculation and iteration.

See attached notebook.

M R Burns' algorithm:

x = SetPrecision[x, pr];
y = x^n; z = (n - y)/y;
t = 2 n - 1; t2 = t^2;
x = 
x*(1 + SetPrecision[4.5, pr] (n - 1)/t2 + (n + 1) z/(2 n t) - 
SetPrecision[13.5, pr] n (n - 1) 1/(3 n t2 + t^3 z));
(*N[Exp[Log[n]/n],pr]*)

Example:

 ClearSystemCache[]; n = 123456789;
(*n is the n in n^(1/n)*)
x = N[n^(1/n),100];
 (*x starts out as a relatively small precision approximation to n^(1/n)*)
 pc = Precision[x]; pr = 10000000;
  (*pr is the desired presicion of your n^(1/n)*)
  Print[t0 = Timing[While[pc < pr, pc = Min[4 pc, pr];
     x = SetPrecision[x, pc];
     y = x^n; z = (n - y)/y;
     t = 2 n - 1; t2 = t^2;
     x = x*(1 + SetPrecision[4.5, pc] (n - 1)/t2 + (n + 1) z/(2 n t)
     - SetPrecision[13.5, pc] n (n - 1)/(3 n t2 + t^3 z))];
     (*You get a much faster version of N[n^(1/n),pr]*)
    N[n - x^n, 10]](*The error*)];
    ClearSystemCache[]; n = 123456789; Print[t1 = Timing[N[n - N[n^(1/n), pr]^n, 10]]]

Gives

  {25.5469,0.*10^-9999984}

  {101.359,0.*10^-9999984}

R Crandall's algorithm:

While[pc < pr, pc = Min[3 pc, pr];
x = SetPrecision[x, pc];
y = x^n - n;
x = x (1 - 2 y/((n + 1) y + 2 n n));];
(*N[Exp[Log[n]/ n],pr]*)

Example:

    ClearSystemCache[]; n = 123456789;
   (*n is the n in n^(1/n)*)
    x = N[n^(1/n)];
   (*x starts out as a machine precision approximation to n^(1/n)*)
   pc = Precision[x]; pr = 10000000;
     (*pr is the desired presicion of your n^(1/n)*)
     Print[t0 = Timing[While[pc < pr, pc = Min[3 pc, pr];
        x = SetPrecision[x, pc];
        y = x^n - n;
        x = x (1 - 2 y/((n + 1) y + 2 n n));];
       (*N[Exp[Log[n]/n],pr]*)
       N[n - x^n, 10]](* The error*)]; Print[
     t1 = Timing[N[n - N[n^(1/n), pr]^n, 10]]]

Gives

  {32.1406,0.*10^-9999984}

  {104.516,0.*10^-9999984}

More information available upon request.

Here is my mini cluster of the fastest 3 computers mentioned below: The one to the left is my custom built extreme edition 6 core and later with a 8 core Xeon processor. The one in the center is my fast little 4 core Asus with 2400 MHz RAM. Then the one on the right is my fastest -- a Digital Storm 6 core overclocked to 4.7 GHz on all cores and with 3000 MHz RAM.

enter image description here

Attachments:
250000.pdf
3M.nb
58 hour million.nb
kernel priority ...nb
pf 225000.xls
POSTED BY: Marvin Ray Burns
Answer
41 Replies
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The following might help anyone serious about breaking my record.

Richard Crandall wrote about couple of new formulae for computing the MRB constant faster on pp 28 and 29 here. a true jewel of an expression Interestingly, as pointed out by J. Borwein,

By the way, I wrote Crandall about formula (44) -- it seems it should have a minus sign in front of it -- and he wrote back: enter image description here
POSTED BY: Marvin Ray Burns
Answer

The following email Crandall sent me before he died might be useful for anyone checking their results: enter image description here
POSTED BY: Marvin Ray Burns
Answer

Perhaps some of these speed records will be easier to beat.

enter image description here

The ultimate speed record to beat, though, is Ricahrd Crandall's (Chief cryptographer for Apple, 2012) 1,048,576 digits in a blistering fast 76.4 hours, while the Apple computation group(also in 2012) computed 1,000,000 digits of the MRB constant in a long 18 days 9 hours 11 minutes 34.253417 seconds, so progress vs time is not as linear as my results plotted above.
POSTED BY: Marvin Ray Burns
Answer

I found the following program Crandall wrote to check his code. (It gives the MRB constant to the desired level of precision.) It's hard to follow! But here it is:

t1 = 1/2 (2 EulerGamma Log[2] - Log[2]^2); t2 = 
 1/24 (\[Pi]^2 Log[2]^2 - 
    2 \[Pi]^2 Log[
      2] (EulerGamma + Log[2] - 12 Log[Glaisher] + Log[\[Pi]]) - 
    6 (Zeta^\[Prime]\[Prime])[2]);
FLAG = 1;

prec = 200;
$MaxPrecision = Infinity;
expM[pr_] := Module[{a, d, s, k, b, c}, n = Floor[1.3 pr]; Print[n];
   d = N[(3 + Sqrt[8])^n, prec]; d = 1/2 (d + 1/d);

   {b, c, s} = {-1, -d, 0};
   T0 = T00 = SessionTime[];
   Do[c = b - c;
     s = s + 
      c*N[E^(Log[k + 1]/(k + 1)) - 1 - 
         FLAG * (Log[k + 1]/(k + 1) + 1/2! Log[k + 1]^2/(k + 1)^2), 
        pr]; 
    (* s=s+c*N[(k+1)^(1/(k+1))-1,pr]; *)
    b = (k + n) (k - n) b/((k + 1) (k + 1/2));
    If[Mod[k, 1000] == 0, Print[{k, SessionTime[] - T0}];
     T0 = SessionTime[]], {k, 0, n - 1}];
   N[-s/d, pr]];

(*Begin test to given precision.*)
(*t1=Timing[MRBtrue=NSum[(-1)^n*(n^(1/n)-1),{n,1,Infinity},\
WorkingPrecision\[Rule]prec,Method\[Rule]"AlternatingSigns"];];*)

t3 = Timing[MRBtest = expM[prec];]
SessionTime[] - T00
MRBtest +  FLAG * (t1 + t2)
POSTED BY: Marvin Ray Burns
Answer

A missing post, that is commented on bellow, follows.

enter image description here

c1 = NSum[(-1)^n*(n^(1/n) - 1), {n, 1, Infinity}, 
Method -> "AlternatingSigns", WorkingPrecision -> 200, 
PrecisionGoal -> 200];

c2 = NSum[(2 k)^(1/(2 k)) - (2 k - 1)^(1/(2 k - 1)), {k, 1, Infinity},
WorkingPrecision -> 600, PrecisionGoal -> 600, NSumTerms -> 2^22];
c1 - c2

gives -6.6444030376101061743388604*10^-173 While, if you try it naively,

NSum[(-1)^n*(n^(1/n) - 1), {n, 1, \[Infinity]}, 
  Method -> "AlternatingSigns", WorkingPrecision -> 100, 
  PrecisionGoal -> 100] - 
 N[Sum[(2 k)^(1/(2 k)) - (2 k - 1)^(1/(2 k - 1)), {k, 
    1, \[Infinity]}], 100]

you get 8.65025879128817600921262272409653679780852599347660397129681915033728\ 91*10^-28, with only 28 digits of accuracy!

Attachments:
5 million digits.nb
POSTED BY: Marvin Ray Burns
Answer

It is hard to be certain that c1 and c2 are correct to 77 digits even though they agree to that extent. I'm not saying that they are incorrect and presumably you have verified this. Just claiming that whatever methods NSum may be using to accelerate convergence, there is really no guarantee that they apply to this particular computation. So c1 aand c2 could agree to that many places because they are computed in a similar manner without all digits actually being correct.
POSTED BY: Daniel Lichtblau
Answer

For c1 I used a method for alternating series to get a very precise sum approximation: That is from the "AlternatingSigns" option.
POSTED BY: Marvin Ray Burns
Answer

C1, the approximation to enter image description hereand a special case of enter image description here, is computed correctly by Henri Cohen, Fernando Rodriguez Villegas, and Don Zagier,s Convergence Acceleration of Alternating Series as found next.

enter image description hereenter image description here

Below it is computed by that method to at least 1000 decimals.

  (* Newer loop with Newton interior. *)
 prec = 1000;(*Number of required decimals.*)
 expM[pr_] := Module[{a, d, s, k, b, c}, n = Floor[1.32 pr];
   Print["Iterations required: ", n];
   d = N[(3 + Sqrt[8])^n, pr + 10];
   d = Round[1/2 (d + 1/d)];
   {b, c, s} = {-1, -d, 0};
   T0 = SessionTime[];
   Do[c = b - c;
    x = N[E^(Log[k + 1]/(k + 1)), iprec = Ceiling[prec/128]];
    pc = iprec;
    Do[nprec = Min[2 pc, pr];
     x = SetPrecision[x, nprec];(*Adjust precision*)
     x = N[x - x/(k + 1) + 1/x^k, nprec];
     pc *= 2;
     If[nprec >= pr, Break[]], {ct, 1, 19}];
    s += c*(x - 1);
    b *= 2 (k + n) (k - n)/((k + 1) (2 k + 1));
    If[Mod[k, 1000] == 0, 
     Print["Iterations: ", k, "    Cumulative time (sec): ", 
       SessionTime[] - T0];], {k, 0, n - 1}];
   N[-s/d, pr]];

 MRBtest2 = expM[prec]

During evaluation of In[17]:= Iterations required: 1320

During evaluation of In[17]:= Iterations: 0 Cumulative time (sec): 0.*10^-8

During evaluation of In[17]:= Iterations: 1000 Cumulative time (sec): 0.1872004

Out[18]= 0. 1878596424620671202485179340542732300559030949001387861720046840894772 3156466021370329665443310749690384234585625801906123137009475922663043 8929348896184120837336626081613602738126379373435283212552763962171489 3217020762820621715167154084126804483635416719985197680252759893899391 4457983505561350964852107120784442309586812949768852694956420425558648 3670441042527952471060666092633974834103115781678641668915460034222258 8380025455396892947114212218910509832871227730802003644521539053639505 5332203470627551159812828039510219264914673176293516190659816018664245 8249506972033819929584209355151625143993576007645932912814517090824249 1588320416906640933443591480670556469280678700702811500938060693813938 5953360657987405562062348704329360737819564603104763950664893061360645 5280675151935082808373767192968663981030949496374962773830498463245634 7931157530028921252329181619562697369707486576547607117801719578736830 0965902260668753656305516567361288150201438756136686552210674305370591 0397357561914891

                 c1 = NSum[(-1)^n*(n^(1/n) - 1), {n, 1, \[Infinity]}, 
                   Method -> "AlternatingSigns", WorkingPrecision -> 100, 
                   PrecisionGoal -> 100]

0.18785964246206712024851793405427323005590309490013878617200468408947 723156466021370329665443310750
POSTED BY: Marvin Ray Burns
Answer

Above where Cohen said,

enter image description here

the coefficients of the numerators are furthered below. I know the pattern for the first and last coefficient of each line, but can you figure out a simple pattern for the rest of them? My hat's off to you if you can!

2
16, 8
98, 80, 32
576, 544, 384, 128
3362, 3312, 2912, 1792, 512
19600, 19528, 18688, 15104, 8192, 2048
114242, 114144, 112576, 103168, 76288, 36864, 8192
665856, 665728, 663040, 641536, 557056, 376832, 163840, 32768
3880898, 3880736, 3876416, 3832064, 3603968, 2945024, 1826816, 720896, 131072
POSTED BY: Marvin Ray Burns
Answer

One more record, Richard Crandall was the first one to compute 319,000 digits of the MRB constant in one day*, as early as Nov 24, 2012. Here is a portion of his email:

enter image description here

*He may have been counting CPU time alone (from the Timing[] comand which is an under estimate of the actual time taken).. My personal records are in actual time taken.
POSTED BY: Marvin Ray Burns
Answer

Tue 24 Apr 2018

Here are some timings from version 11.3 and my first custom built computer with it's two separate processors:

enter image description here[41]

Sat 28 Apr 2018

Chapter 3 in the paper at https://www.sciencedirect.com/science/article/pii/0898122189900242 , gives a 4th degree convergence rate algorithm for the nth root of x, which I used to add a new interior to my version of Crandall's MRB constant computation program, making it a hybrid. As the paper describes the algorithm, it is best used for getting many digits from a sufficiently precise initial term. (I used Crandall's 3rd degree convergence interior to get the sufficiently precise term.) This algorithm better optimizes my program for large calculations a little according to the following:

Here is a table of the results so far, which also show that RAM speed is the most important factor in a computer for computing the MRB constant!:

enter image description here

Here are a timing table with All MRB method, Crandall's method and a hybrid of the two on this 4 core ASUS Intel Core i7 7700 Desktop computer K31CD-DS71, 32 GB ram DDR4, 2TB SATA in Windows 10 Pro:

table

Here are the 3 programs referenced in the above table:

All MRB:

 Print["Start time is " "Start time is ", ds = DateString[], "."];
 prec = 10000;
 (**Number of required decimals.*.*)ClearSystemCache[];
 T0 = SessionTime[];
 expM[pre_] := 
   Module[{a, d, s, k, bb, c, end, iprec, xvals, x, pc, cores = 16(*=4*
     number of physical cores*), tsize = 2^7, chunksize, start = 1, ll,
      ctab, pr = Floor[1.0002 pre]}, chunksize = cores*tsize;
    n = Floor[1.32 pr];
    end = Ceiling[n/chunksize];
    Print["Iterations required: ", n];
    Print["Will give ", end, 
     " time estimates, each more accurate than the previous."];
    Print["Will stop at ", end*chunksize, 
     " iterations to ensure precsion of around ", pr, 
     " decimal places."]; d = ChebyshevT[n, 3];
    {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
    iprec = pr/2^6;
    Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l;
         x = N[E^(Log[ll]/(ll)), iprec];
         pc = iprec;


         While[pc < pr, pc = Min[4 pc, pr];
          x = SetPrecision[x, pc];
          xll = x^ll; z = (ll - xll)/xll;
          t = 2 ll - 1; t2 = t^2;
          x =    x*(1 + SetPrecision[4.5, pc] (ll - 1)/
                t2 + (ll + 1) z/(2 ll t) - 
              SetPrecision[13.5, 
                2 pc] ll (ll - 1)/ (3 ll t2 + t^3 z))];(**N[Exp[Log[
         ll]/ll],pr]**)


         x, {l, 0, tsize - 1}], {j, 0, cores - 1}, 
        Method -> 
         "EvaluationsPerKernel" -> 
          16(*a power of 2 commensurate with available RAM*)]];
     ctab = ParallelTable[Table[c = b - c;
        ll = start + l - 2;
        b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
        c, {l, chunksize}], 
       Method -> 
        "EvaluationsPerKernel" -> 
         16(*a power of 2 commensurate with processor strength*)];
     s += ctab.(xvals - 1);
     start += chunksize;
     st = SessionTime[] - T0; kc = k*chunksize;
     ti = (st)/(kc + 10^-10)*(n)/(3600)/(24);
     If[kc > 1, 
      Print[kc, " iterations done in ", N[st - stt, 4], " seconds.", 
       " Should take ", N[ti, 4], " days or ", ti*3600*24, 
       "s, finish ", DatePlus[ds, ti], "."], 
      Print["Denominator computed in  " , stt = st, "s."]];, {k, 0, 
      end - 1}];
    N[-s/d, pr]];
 t2 = Timing[MRBtest2 = expM[prec];]; Print["Finished on ", 
  DateString[], ". Processor and total time were ", 
  t2[[1]], " and ", st, " s respectively."];
 (*Print[*)MRBtest2(*]*)(*Remove (**) or enter MRBtest2 to print \
 output*); Print["Enter MRBtest2 to print ", 
  Floor[Precision[
    MRBtest2]], " digits"]; Print["If you saved m3M, the difference \
 between this and 3,014,991 known digits is ", N[MRBtest2 - m3M, 10]]

Crandall

Print["Start time is ", ds = DateString[], "."];
prec = 100000;
(**Number of required decimals.*.*)ClearSystemCache[];
T0 = SessionTime[];
expM[pre_] := 
  Module[{a, d, s, k, bb, c, end, iprec, xvals, x, pc, cores = 32(*=4*
    number of physical cores*), tsize = 2^7, chunksize, start = 1, ll,
     ctab, pr = Floor[1.005 pre]}, chunksize = cores*tsize;
   n = Floor[1.32 pr];
   end = Ceiling[n/chunksize];
   Print["Iterations required: ", n];
   Print["Will give ", end, 
    " time estimates, each more accurate than the previous."];
   Print["Will stop at ", end*chunksize, 
    " iterations to ensure precsion of around ", pr, 
    " decimal places."]; d = ChebyshevT[n, 3];
   {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
   iprec = Ceiling[pr/27];
   Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l;
        x = N[E^(Log[ll]/(ll)), iprec];
        pc = iprec;
        While[pc < pr, pc = Min[3 pc, pr];
         x = SetPrecision[x, pc];
         y = x^ll - ll;
         x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];(**N[Exp[Log[ll]/
        ll],pr]**)x, {l, 0, tsize - 1}], {j, 0, cores - 1}, 
       Method -> 
        "EvaluationsPerKernel" -> 
         64(*a power of 2 commensurate with processor strength*)]];
    ctab = ParallelTable[Table[c = b - c;
       ll = start + l - 2;
       b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
       c, {l, chunksize}], 
      Method -> 
       "EvaluationsPerKernel" -> 
        16(*a power of 2 commensurate with processor strength*)];
    s += ctab.(xvals - 1);
    start += chunksize;
    st = SessionTime[] - T0; kc = k*chunksize;
    ti = (st)/(kc + 10^-4)*(n)/(3600)/(24);
    If[kc > 1, 
     Print[kc, " iterations done in ", N[st, 3], " seconds.", 
      " Should take ", N[ti, 2], " days or ", N[ti*24*3600, 2], 
      "s, finish ", DatePlus[ds, ti], "."]];, {k, 0, end - 1}];
   N[-s/d, pr]];
t2 = Timing[MRBtest2 = expM[prec];]; Print["Finished on ", 
 DateString[], ". Processor time was ", 
 t2[[1]], " s. Actual time was ", SessionTime[] - T0, "."];
(*Print[*)MRBtest2(*]*)(*Remove (**) or enter MRBtest2 to print \
output*); Print["Enter MRBtest2 to print ", 
 Floor[Precision[
   MRBtest2]], " digits"]; Print["If you saved m3M, the difference \
between this and 3,014,991 known digits is ", N[MRBtest2 - m3M, 10]]

Hybrid

Print["Start time is ", ds = DateString[], "."];
prec = 300000;
(**Number of required decimals.*.*)ClearSystemCache[];
T0 = SessionTime[];
expM[pre_] := 
  Module[{a, d, s, k, bb, c, end, iprec, xvals, x, pc, cores = 16(*=4*
    number of physical cores*), tsize = 2^7, chunksize, start = 1, ll,
     ctab, pr = Floor[1.005 pre]}, chunksize = cores*tsize;
   n = Floor[1.32 pr];
   end = Ceiling[n/chunksize];
   Print["Iterations required: ", n];
   Print["Will give ", end, 
    " time estimates, each more accurate than the previous."];
   Print["Will stop at ", end*chunksize, 
    " iterations to ensure precsion of around ", pr, 
    " decimal places."]; d = ChebyshevT[n, 3];
   {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
   iprec = Ceiling[pr/27];
   Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l;
        x = N[E^(Log[ll]/(ll)), iprec];
        pc = iprec;
        While[pc < pr/4, pc = Min[3 pc, pr/4];
         x = SetPrecision[x, pc];
         y = x^ll - ll;
         x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];(**N[Exp[Log[ll]/
        ll],pr/4]**)x = SetPrecision[x, pr];
        xll = x^ll; z = (ll - xll)/xll;
        t = 2 ll - 1; t2 = t^2;
        x = 
         x*(1 + SetPrecision[4.5, pr] (ll - 1)/
              t2 + (ll + 1) z/(2 ll t) - 
            SetPrecision[13.5, pr] ll (ll - 1) 1/(3 ll t2 + t^3 z));(**
        N[Exp[Log[ll]/ll],pr]**)x, {l, 0, tsize - 1}], {j, 0, 
        cores - 1}, 
       Method -> 
        "EvaluationsPerKernel" -> 
         32(*a power of 2 commensurate with available RAM*)]];
    ctab = ParallelTable[Table[c = b - c;
       ll = start + l - 2;
       b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
       c, {l, chunksize}], 
      Method -> 
       "EvaluationsPerKernel" -> 
        16(*a power of 2 commensurate with processor strength*)];
    s += ctab.(xvals - 1);
    start += chunksize;
    st = SessionTime[] - T0; kc = k*chunksize;
    ti = (st)/(kc + 10^-4)*(n)/(3600)/(24);
    If[kc > 1, 
     Print[kc, " iterations done in ", N[st, 4], " seconds.", 
      " Should take ", N[ti, 4], " days or ", N[ti*24*3600, 4], 
      "s, finish ", DatePlus[ds, ti], "."]];, {k, 0, end - 1}];
   N[-s/d, pr]];
t2 = Timing[MRBtest2 = expM[prec];]; Print["Finished on ", 
 DateString[], ". Proccessor time was ", t2[[1]], " s."];
(*Print[*)MRBtest2(*]*)(*Remove (**) or enter MRBtest2 to print \
output*); Print["Enter MRBtest2 to print ", 
 Floor[Precision[
   MRBtest2]], " digits"]; Print["If you saved m3M, the difference \
between this and 3,014,991 known digits is ", N[MRBtest2 - m3M, 10]]

Aug 8 2018

I've been having great success with the Wolfram Lightweight Grid and "threadpriority," like in the following:

 In[23]:= Needs["SubKernels`LocalKernels`"]
 Block[{$mathkernel = $mathkernel <> " -threadpriority=2"}, 
  LaunchKernels[]]

 Out[24]= {"KernelObject"[43, "burns"], "KernelObject"[44, "burns"], 
  "KernelObject"[45, "burns"], "KernelObject"[46, "burns"], 
  "KernelObject"[47, "burns"], "KernelObject"[48, "burns"], 
  "KernelObject"[49, "burns"], "KernelObject"[50, "burns"], 
  "KernelObject"[51, "local"], "KernelObject"[52, "local"], 
  "KernelObject"[53, "local"], "KernelObject"[54, "local"], 
  "KernelObject"[55, "local"], "KernelObject"[56, "local"]}

.

I just now computed 1,004,754 digits of the MRB constant in 58 hours of absolute time!

I think that worked because my computer with the local kernels is the faster one.

See attached "58 hour million."

Aug 9, 2018, I just now computed 1,004,993 digits of the MRB constant in 53.5 hours of absolute time! Summarized below.

Print["Start time is ", ds = DateString[], "."];
prec = 1000000;
(**Number of required decimals.*.*)ClearSystemCache[];
T0 = SessionTime[];
expM[pre_] := 
  Module[{a, d, s, k, bb, c, end, iprec, xvals, x, pc, cores = 16(*=4*
    number of physical cores*), tsize = 2^7, chunksize, start = 1, ll,
     ctab, pr = Floor[1.005 pre]}, chunksize = cores*tsize;
   n = Floor[1.32 pr];
   end = Ceiling[n/chunksize];
   Print["Iterations required: ", n];
   Print["Will give ", end, 
    " time estimates, each more accurate than the previous."];
   Print["Will stop at ", end*chunksize, 
    " iterations to ensure precsion of around ", pr, 
    " decimal places."]; d = ChebyshevT[n, 3];
   {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
   iprec = Ceiling[pr/6912];
   Do[xvals = Flatten[Parallelize[Table[Table[ll = start + j*tsize + l;
         x = N[E^(Log[ll]/(ll)), iprec];


         pc = iprec;
         While[pc < pr/1024, pc = Min[3 pc, pr/1024];
          x = SetPrecision[x, pc];
          y = x^ll - ll;
          x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];
         (**N[Exp[Log[ll]/ll],pr/1024]**)


         x = SetPrecision[x, pr/256];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/256] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/256] ll (ll - 1) 1/(3 ll t2 + t^3 z));(*N[Exp[Log[
         ll]/ll],pr/256]*)

         x = SetPrecision[x, pr/64];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/64] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/64] ll (ll - 1) 1/(3 ll t2 + t^3 z));(**N[Exp[Log[
         ll]/ll],pr/64]**)

         x = SetPrecision[x, pr/16];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/16] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/16] ll (ll - 1) 1/(3 ll t2 + t^3 z));(**N[Exp[Log[
         ll]/ll],pr/16]**)

         x = SetPrecision[x, pr/4];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/4] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/4] ll (ll - 1) 1/(3 ll t2 + t^3 z));(**N[Exp[Log[
         ll]/ll],pr/4]**)


         x = SetPrecision[x, pr];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr] ll (ll - 1) 1/(3 ll t2 + t^3 z));(*N[Exp[Log[ll]/
         ll],pr]*)

         x, {l, 0, tsize - 1}], {j, 0, cores - 1}]]];
    ctab = ParallelTable[Table[c = b - c;
       ll = start + l - 2;
       b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
       c, {l, chunksize}], Method -> "Automatic"];
    s += ctab.(xvals - 1);
    start += chunksize;
    st = SessionTime[] - T0; kc = k*chunksize;
    ti = (st)/(kc + 10^-4)*(n)/(3600)/(24);
    Print[kc, " iterations done in ", N[st, 4], " seconds.", 
     " Should take ", N[ti, 4], " days or ", N[ti*24*3600, 4], 
     "s, finish ", DatePlus[ds, ti], "."], {k, 0, end - 1}];
   N[-s/d, pr]];
t2 = Timing[MRBtest2 = expM[prec];]; Print["Finished on ", 
 DateString[], ". Proccessor time was ", 
 t2[[1]], " s."]; Print["Actual time was ", st];
(*Print[*)MRBtest2(*]*)(*Remove (**) or enter MRBtest2 to print \
output*); Print["Enter MRBtest2 to print ", 
 Floor[Precision[
   MRBtest2]], " digits"]; Print["If you saved m3M, the difference \
between this and 3,014,991 known digits is ", N[MRBtest2 - m3M, 10]]

Start time is Wed 8 Aug 2018 14:09:15.

Iterations required: 1326598

Will give 648 time estimates, each more accurate than the previous.

Will stop at 1327104 iterations to ensure precsion of around 1004999 decimal places.

0 iterations done in 259.9 seconds. Should take 3.991*10^7 days or 3.448*10^12s, finish Fri 24 Dec 111294 02:34:55.

2048 iterations done in 523.6 seconds. Should take 3.926 days or 3.392*10^5s, finish Sun 12 Aug 2018 12:22:00.

4096 iterations done in 790.2 seconds. Should take 2.962 days or 2.559*10^5s, finish Sat 11 Aug 2018 13:14:46.

6144 iterations done in 1058. seconds. Should take 2.645 days or 2.285*10^5s, finish Sat 11 Aug 2018 05:38:01.

8192 iterations done in 1329. seconds. Should take 2.490 days or 2.152*10^5s, finish Sat 11 Aug 2018 01:55:27.

10240 iterations done in 1602. seconds. Should take 2.402 days or 2.075*10^5s, finish Fri 10 Aug 2018 23:47:32.

12288 iterations done in 1875. seconds. Should take 2.343 days or 2.024*10^5s, finish Fri 10 Aug 2018 22:22:49.

14336 iterations done in 2151. seconds. Should take 2.304 days or 1.990*10^5s, finish Fri 10 Aug 2018 21:26:34.

16384 iterations done in 2424. seconds. Should take 2.272 days or 1.963*10^5s, finish Fri 10 Aug 2018 20:41:04.

18432 iterations done in 2700. seconds. Should take 2.250 days or 1.944*10^5s, finish Fri 10 Aug 2018 20:08:36.

20480 iterations done in 2977. seconds. Should take 2.232 days or 1.928*10^5s, finish Fri 10 Aug 2018 19:42:40.

22528 iterations done in 3256. seconds. Should take 2.219 days or 1.917*10^5s, finish Fri 10 Aug 2018 19:24:31.

24576 iterations done in 3533. seconds. Should take 2.207 days or 1.907*10^5s, finish Fri 10 Aug 2018 19:07:43.

26624 iterations done in 3811. seconds. Should take 2.198 days or 1.899*10^5s, finish Fri 10 Aug 2018 18:53:53.
 ...


1320960 iterations done in 1.921*10^5 seconds. Should take 2.232 days or 1.929*10^5s, finish Fri 10 Aug 2018 19:43:46.

1323008 iterations done in 1.923*10^5 seconds. Should take 2.232 days or 1.929*10^5s, finish Fri 10 Aug 2018 19:43:28.

1325056 iterations done in 1.926*10^5 seconds. Should take 2.232 days or 1.928*10^5s, finish Fri 10 Aug 2018 19:43:08.

Finished on Fri 10 Aug 2018 19:39:26. Proccessor time was 122579. s.

Actual time was 192609.7247443

Enter MRBtest2 to print 1004992 digits

If you saved m3M, the difference between this and 3,014,991 known digits is 0.*10^-1004993

Sept 21, 2018, I just now computed 1,004,993 digits of the MRB constant in 50.37 hours of absolute time (35.4 hours processor time)! See attached "50 hour million."

Here is a table of my speed progress in computing the MRB constant since 2012:

top bottom

See attached "kernel priority 2 computers.nb" and "3 fastest computers together.nb" for some documentation of the timings in the last 2 columns, respectively. See "3million 2 computers" for the estimated time of 24 days for computing 3,000,000 digits.

01/08/2019

Here is an update of 100k digits (in seconds): enter image description here Notice, I broke the 1,000 second mark!!!!!

Attachments:
3 fastest comput...nb
3million 2 computers.nb
50 hour million.nb
58 hour million.nb
kernel priority ...nb
POSTED BY: Marvin Ray Burns
Answer

What Richard Crandall and maybe others did to come up with that method is really good and somewhat mysterious. I still don't really understand the inner workings, and I had shown him how to parallelize it. So the best I can say is that it's really hard to compete against magic. (I don't want to discourage others, I'm just explaining why I myself would be reluctant to tackle this. Someone less familiar might actually have a better chance of breaking new ground.)

In a way this should be good news. Should it ever become "easy" to compute, the MRB number would lose what is perhaps its biggest point of interest. It just happens to be on that cusp of tantalizingly "close" to easily computable (perhaps as sums of zeta function and derivatives thereof), yet still hard enough that it takes a sophisticated scheme to get more than a few dozen digits.
POSTED BY: Daniel Lichtblau
Answer

Daniel Lichtblau,
Which one do you not understand? how he derived the eta' formula:enter image description here from enter image description here?

Or is it how and why he added the extra loop to Cohen's method:`

ll = start + l - 2; b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1)); ?`

I've been asking for a proof of the eta formula derivation. I think it could become the foundation to a great paper.

If the extra loop changes Cohen's method for enter image description here to a method for enter image description hereI think the code in that loop could help prove the derivation.
POSTED BY: Marvin Ray Burns
Answer

I can't say I understand either. My guess is the Eta stuff comes from summing (-1)^k*(Log[k]/k)^n over k, as those are the terms that appear in the double sum you get from expanding k^(1/k)-1 in powers of Log[k]/k (use k^(1/k)=Exp[Log[k]/k] and the power series for Exp). Even if it does come from this the details remain elusive..
POSTED BY: Daniel Lichtblau
Answer

Daniel Lichtblau and others, Richard Crandall did intend to explian his work on the MRB constant and his program to compute it. When I wrote him with a possible small improvement to his program he said, "It's worth observing when we write it up." See screenshot: enter image description here
POSTED BY: Marvin Ray Burns
Answer

Crandall is not using his eta formulas directly!!!!!!! He computes Sum[(-1)^k*(k^(1/k) - 1), {k, 1, Infinity}] directly!

Going back to Crandall's code:

(*Fastest (at RC's end) as of 30 Nov 2012.*)prec = 500000;(*Number of \
required decimals.*)ClearSystemCache[];
T0 = SessionTime[];
expM[pre_] := 
  Module[{a, d, s, k, bb, c, n, end, iprec, xvals, x, pc, cores = 4, 
    tsize = 2^7, chunksize, start = 1, ll, ctab, 
    pr = Floor[1.02 pre]}, chunksize = cores*tsize;
   n = Floor[1.32 pr];
   end = Ceiling[n/chunksize];
   Print["Iterations required: ", n];
   Print["end ", end];
   Print[end*chunksize];
   d = N[(3 + Sqrt[8])^n, pr + 10];
   d = Round[1/2 (d + 1/d)];
   {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
   iprec = Ceiling[pr/27];
   Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l;
        x = N[E^(Log[ll]/(ll)), iprec];
        pc = iprec;
        While[pc < pr, pc = Min[3 pc, pr];
         x = SetPrecision[x, pc];
         y = x^ll - ll;
         x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];(*N[Exp[Log[ll]/ll],
        pr]*)x, {l, 0, tsize - 1}], {j, 0, cores - 1}, 
       Method -> "EvaluationsPerKernel" -> 1]];
    ctab = Table[c = b - c;
      ll = start + l - 2;
      b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
      c, {l, chunksize}];
    s += ctab.(xvals - 1);
    start += chunksize;
    Print["done iter ", k*chunksize, " ", SessionTime[] - T0];, {k, 0,
      end - 1}];
   N[-s/d, pr]];

t2 = Timing[MRBtest2 = expM[prec];];
MRBtest2 - MRBtest3

x = N[E^(Log[ll]/(ll)), iprec]; Gives k^(1/k) to only 1 decimal place; they are either 1.0, 1.1, 1.2, 1.3 or 1.4 (usually 1.1 or 1.0).. On the other hand,

While[pc < pr, pc = Min[3 pc, pr];
 x = SetPrecision[x, pc];
 y = x^ll - ll;
 x = x (1 - 2 y/((ll + 1) y + 2 ll ll));],

takes the short precision x and gives it the necessary precision and accuracy for k^(1/k) (k Is ll there.) It actually computes k^(1/k). Then he remarks, "(N[Exp[Log[ll]/ll], pr])."

After finding a fast way to compute k^(1/k) to necessary precision he uses Cohen's algorithm 1 (See a screenshot in a previous post.) to accelerate convergence of Sum[(-1)^k*(k^(1/k) - 1), {k, 1, Infinity}]. That is his secret!!

As I mentioned in a previous post the "MRBtest2 - MRBtest3" is for checking with a known-to-be accurate approximation to the MRB constant, MRBtest3

I'm just excited that I figured it out! as you can tell.
POSTED BY: Marvin Ray Burns
Answer

Nice work. Worth a bit of excitement, I' d say.
POSTED BY: Daniel Lichtblau
Answer

Richard Crandall might of had some help in developing his method. He wrote one time:

"Marvin I am working on a highly efficient method for your constant, and I've been in touch with other mathematics scholars.

Please be patient...

rec

Sent from my iPhone."
POSTED BY: Marvin Ray Burns
Answer

The result of k^(1/k) from the iteration in Crandall's code grows like the following example of 123^(1/123)..

Timing[ll = 123; x = 1 + 1/10; 
 Table[Block[{$MaxExtraPrecision = Infinity}, 
   x = SetPrecision[x, 20 + a^6]; y = x^ll - ll; 
   x = x (1 - 2 y/((ll + 1) y + 2 ll ll)); 
   N[ll^(1/ll) - x, a + 20]], {a, 1, 13}]]

That gives the following timing and accuracies per iteration.

  {8.829657, {-0.04239426367665370783, -0.02519017014329854597401, \
  -0.0097174389047505410455275, -0.000933400633967433388167272, \
  -9.456045144810702667437515*10^-7, \
  -9.8570302819631224677862339*10^-16, \
  -1.11649387558952636447488240*10^-42, \
  -1.622508976729293063326569833*10^-123, \
  -4.9794273187141078980478269830*10^-366, \
  -1.43931637321260733551982880202*10^-1093, \
  -3.476056847637409626605825363818*10^-3276, \
  -4.8964202322309104109089811118496*10^-9824, \
  -1.36852937628135955635109182098985*10^-29467}}

Here is a more extreme example from my big computer where we get about 64,000,000 digits of accuracy of 123^(1/123) in only 20 iterations!

In[47]:= Timing[ll = 123; x = 1 + 1/10;
 Table[Block[{$MaxExtraPrecision = Infinity}, 
   x = SetPrecision[x, 20 + a^6]; y = x^ll - ll;
   x = x (1 - 2 y/((ll + 1) y + 2 ll ll));
   N[ll^(1/ll) - x, a + 20]], {a, 1, 20}]]

Out[47]= {210.180147, {-0.04239426367665370783, \
-0.02519017014329854597401, -0.0097174389047505410455275, \
-0.000933400633967433388167272, -9.456045144810702667437515*10^-7, \
-9.8570302819631224677862339*10^-16, \
-1.11649387558952636447488240*10^-42, \
-1.622508976729293063326569833*10^-123, \
-4.9794273187141078980478269830*10^-366, \
-1.43931637321260733551982880202*10^-1093, \
-3.476056847637409626605825363818*10^-3276, \
-4.8964202322309104109089811118496*10^-9824, \
-1.36852937628135955635109182098985*10^-29467, \
-2.987998984519890677800153644718129*10^-88398, \
-3.1099929855547029560727524646233719*10^-265190, \
-3.50668133180579791324881778853196092*10^-795566, \
-5.026977910888395698424738167604170974*10^-2386694, \
-1.4809452514436132712292881458571004047*10^-7160077, \
-3.78647491015226246115034702337662410645*10^-21480228, 
  0.*10^-64000020}}

Furthermore this process can compute 6,379,358,787 digits of the Sqrt[2], from an initial guess of machine precision, in 18 iterations:

In[90]:= Timing[ll = 2; x = 2.^(1/2);
 Table[Block[{$MaxExtraPrecision = Infinity}, 
   x = SetPrecision[x, 20 + a^7.5]; y = x^ll - ll;
   x = x (1 - 2 y/((ll + 1) y + 2 ll ll));
   N[ll^(1/ll) - x, a + 20]], {a, 1, 18}]]

Out[90]= {1913.882668, {0.*10^-21, -1.800462644283455490473*10^-148, \
-7.2956225729359623776786*10^-445, \
-4.85397000370369097837099*10^-1334, \
-1.429556345255675471801415*10^-4001, \
-3.6518576944392011682818847*10^-12004, \
-6.08767627470915271802076570*10^-36012, \
-2.820100999650988078782491560*10^-108035, \
-2.8035222068968058313783728204*10^-324105, \
-2.75436831997714080172674040062*10^-972315, \
-2.612017346393607427831858236332*10^-2916945, \
-2.2276049962448401256762035120120*10^-8750835, \
-1.38173437723345455709717486529180*10^-26252505, \
-3.297491628267321222864512567028780*10^-78757516, \
-4.4818892212256882776648121052875660*10^-236272548, \
-1.12536490316593766234907240580868063*10^-708817643, 
  0.*10^-1073741813, 0.*10^-1073741813}}

The above pattern is 3 times the previous number of accurate digits per iteration.. `708817643 * 3^2= 6379358787`

I believe it will only take 191+16 = 207 iterations to compute a google digits of the Sqrt[2]:

In[55]:= Log[10.^100/708817643]/Log[3]

Out[55]= 191.04
POSTED BY: Marvin Ray Burns
Answer

Jan 2015

How about computing the MRB constant from Crandall's eta derivative formulas? They are mentioned in a previous post but here they are again:

enter image description here

I computed and checked 500 digits, using the first eta derivative formula in 38.6 seconds. How well can you do? Can you improve my program? (It is a 51.4% improvement of one of Crandall's programs.) I would like a little competition in some of these records! (That formula takes just 225 summands, compared to 10^501 summands using -1^(1/1)+2^(1/2)-3^(1/3)+.... See http://arxiv.org/pdf/0912.3844v3.pdf for more summation requirements for other summation methods.) 

In[37]:= mm = 
  0.187859642462067120248517934054273230055903094900138786172004684089\
4772315646602137032966544331074969038423458562580190612313700947592266\
3043892934889618412083733662608161360273812637937343528321255276396217\
1489321702076282062171516715408412680448363541671998519768025275989389\
9391445798350556135096485210712078444230958681294976885269495642042555\
8648367044104252795247106066609263397483410311578167864166891546003422\
2258838002545539689294711421221891050983287122773080200364452153905363\
9505533220347062755115981282803951021926491467317629351619065981601866\
4245824950697203381992958420935515162514399357600764593291281451709082\
4249158832041690664093344359148067055646928067870070281150093806069381\
3938595336065798740556206234870432936073781956460310476395066489306136\
0645528067515193508280837376719296866398103094949637496277383049846324\
5634793115753002892125232918161956269736970748657654760711780171957873\
6830096590226066875365630551656736128815020143875613668655221067430537\
0591039735756191489093690777983203551193362404637253494105428363699717\
0244185516548372793588220081344809610588020306478196195969537562878348\
1233497638586301014072725292301472333336250918584024803704048881967676\
7601198581116791693527968520441600270861372286889451015102919988536905\
7286592870868754254925337943953475897035633134403826388879866561959807\
3351473990256577813317226107612797585272274277730898577492230597096257\
2562718836755752978879253616876739403543214513627725492293131262764357\
3214462161877863771542054231282234462953965329033221714798202807598422\
1065564890048536858707083268874877377635047689160983185536281667159108\
4121934201643860002585084265564350069548328301205461932`1661.\
273833491444;

In[30]:= Timing[
 etaMM[m_, pr_] := 
  Module[{a, d, s, k, b, c}, a[j_] := Log[j + 1]^m/(j + 1)^m;
   n = Floor[1.32 pr];
   d = Cos[n ArcCos[3]];
   {b, c, s} = {-1, -d, 0};
   Do[c = b - c;
    s = s + c a[k];
    b = (k + n) (k - n) b/((k + 1) (k + 1/2)), {k, 0, n - 1}];
   N[s/d, pr] (-1)^m];
 eta[s_] := (1 - 2^(1 - s)) Zeta[s];
 eta1 = Limit[D[eta[s], s], s -> 1];
 MRBtrue = mm;
 prec = 500;
 MRBtest = 
  eta1 - Sum[(-1)^m etaMM[m, prec]/m!, {m, 2, Floor[.45 prec]}];
 MRBtest - MRBtrue]

Out[30]= {36.831836, 0.*10^-502}

Here is a short table of computation times with that program:

Digits      Seconds

500        36.831836
1000       717.308198
1500       2989.759165
2000       3752.354453

I just now retweaked the program. It is now

Timing[etaMM[m_, pr_] := 
  Module[{a, d, s, k, b, c}, 
   a[j_] := N[(-PolyLog[1, -j]/(j + 1))^m, pr];
   n = Floor[1.32 pr];
   d = Cos[n ArcCos[3]];
   {b, c, s} = {-1, -d, 0};
   Do[c = b - c;
    s = s + c a[k];
    b = N[(k + n) (k - n) b/((k + 1) (k + 1/2)), pr], {k, 0, n - 1}];
   Return[N[s/d, pr] (-1)^m]];
 eta[s_] := (1 - 2^(1 - s)) Zeta[s];
 eta1 = Limit[D[eta[s], s], s -> 1];
 MRBtrue = mm;
 prec = 1500;
 MRBtest = 
  eta1 - Sum[(-1)^m etaMM[m, prec]/Gamma[m + 1], {m, 2, 
     Floor[.45 prec]}, Method -> "Procedural"];
 MRBtest - MRBtrue]

Feb 2015

Here are my best eta derivative records:

Digits        Seconds
 500          9.874863
 1000        62.587601
 1500        219.41540
 2000       1008.842867
 2500       2659.208646
 3000       5552.902395
 3500       10233.821601

That is using V10.0.2.0 Kernel. Here is a sample

Timing[etaMM[m_, pr_] := 
          Module[{a, d, s, k, b, c}, 
           a[j_] := N[(-PolyLog[1, -j]/(j + 1))^m, pr];
           n = Floor[1.32 pr];
           d = Cos[n ArcCos[3]];
           {b, c, s} = {-1, -d, 0};
           Do[c = b - c;
            s = s + c a[k];
            b = N[(k + n) (k - n) b/((k + 1) (k + 1/2)), pr], {k, 0, n - 1}];
           Return[N[s/d, pr] (-1)^m]];
         eta[s_] := (1 - 2^(1 - s)) Zeta[s];
         eta1 = Limit[D[eta[s], s], s -> 1];
         MRBtrue = mm;
         prec = 500;
         MRBtest = 
          eta1 - Sum[(-1)^m etaMM[m, prec]/Gamma[m + 1], {m, 2, 
             Floor[.45 prec]}];
        ]
         N::meprec: Internal precision limit $MaxExtraPrecision = 50. reached while evaluating 
             -Cos[660 ArcCos[3]].

         N::meprec: Internal precision limit $MaxExtraPrecision = 50. reached while evaluating 
             -Cos[660 ArcCos[3]].

         N::meprec: Internal precision limit $MaxExtraPrecision = 50. reached while evaluating 
             -Cos[660 ArcCos[3]].

         General::stop: Further output of N::meprec will be suppressed during this calculation.

         Out[1]= {9.874863, Null}

Aug 2016

enter image description here

V 11 has a significant improvement in my new most recently mentioned fastest program for calculating digits of the MRB constant via the eta formula, Here are some timings:

Digits           seconds

1500                42.6386632

2000             127.3101969

3000             530.4442911

4000           1860.1966540

5000           3875.6978162

6000           8596.9347275



 10,000        53667.6315476

From an previous message that starts with "How about computing the MRB constant from Crandall's eta derivative formulas?" here are my first two sets of records to compare with the just mentioned ones. You can see that I increased time efficiency by 10 to 29 to even 72 fold for select computations! In the tests used in that "previous message," 4000 or more digit computations produced a seemingly indefinitely long hang-on.

Digits      Seconds

500        36.831836
1000       717.308198
1500       2989.759165
2000       3752.354453


Digits        Seconds
 500          9.874863
 1000        62.587601
 1500        219.41540
 2000       1008.842867
 2500       2659.208646
 3000       5552.902395
 3500       10233.821601

Comparing first of the just mentioned 2000 digit computations with the "significant improvement" one we get the following.

3752/127 ~=29.

And from the slowest to the fastest 1500 digit run we get

2989/42 ~=72,
POSTED BY: Marvin Ray Burns
Answer

Talking about the eta formula for the MRB constant

Here is another great improvement in calculating digits of MRB though this first Crandall eta formula found in the psu.edu paper Unified algorithms for polylogarithm, L-series, and zeta variants

enter image description here

From an even earlier message that starts with "How about computing the MRB constant from Crandall's eta derivative formulas?" here are my first two sets of records to compare with the just mentioned ones. You can see that I increased time efficiency by 10 to 29 to even 72 fold for select computations! In the tests used in that "previous message," 4000 or more digit computations produced a seemingly indefinitely long hang-on.

Until mentioned otherwise, the following tables are results from my 3.4 GHz Xeon with 64 GB 1666 MHz RAM.

Digits      Seconds

500        36.831836
1000       717.308198
1500       2989.759165
2000       3752.354453


Digits        Seconds
 500          9.874863
 1000        62.587601
 1500        219.41540
 2000       1008.842867
 2500       2659.208646
 3000       5552.902395
 3500       10233.821601

Comparing first of the just mentioned 2000 digit computations with the "significant improvement" one, we get the following.

3752/127 ~=29.

And from the slowest to the fastest 1500 digit run we get

2989/42 ~=72,

Here is an example calculating 1500 digits in less than 23 seconds. First calculate known accurate digits:

mTrue = NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 2000];

Then use this program:

prec = 1500;
to = SessionTime[];
etaMM[m_, pr_] := 
  Module[{a, s, k, b, c}, a[j_] := (Log[j + 1]/(j + 1))^m;
   {b, c, s} = {-1, -d, 0};
   Do[c = b - c;
    s = s + c a[k];
    b = (k + n) (k - n) b/((k + 1) (k + 1/2)), {k, 0, n - 1}];
   Return[N[s/d, pr] (-1)^m]];
eta1 = N[EulerGamma Log[2] - Log[2]^2/2, prec]; n = 
 Floor[132/100 prec]; d = N[ChebyshevT[n, 3], prec];
MRBtest = 
  eta1 - ParallelSum[(Cos[Pi m]) etaMM[m, prec]/
      N[Gamma[m + 1], prec], {m, 2, Floor[.365 prec]}];
Print[mTrue - MRBtest]; SessionTime[] - to

The larger prec (precission) you want to choose, the smaller number you can replace .365 with in
ParallelSum[(Cos[Pi m]) etaMM[m, prec]/N[Gamma[m + 1], prec], {m, 2, Floor[.365 prec]}. For prec=5000, change .365 to .31 (approximately). This advantage for smaller upper limits being needed in the sum is due to the Gamma (!) in the denominator. On my big computer 4000 digits now gives a timing of around 1800 seconds (1/2 an hour). See message above the previous occurrence of 8/13/2017 for previous records and records before those.

V 11 has a significant improvement in my new most recently mentioned fastest program for calculating digits of the MRB constant via the eta formula, Here are some timings:

Digits           seconds

1500                42.6386632

2000             127.3101969

3000             530.4442911

4000           1860.1966540

5000           3875.6978162

6000           8596.9347275

10,000        53667.6315476

UPDATE 5/30/2018

I made some programming improvements to my code for the eta formula computation of the MRB constant.

Here is the new code:

In[88]:= mTrue = 
  NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 5000];

In[82]:= prec = 1500;
to = SessionTime[];
etaMM[m_, pr_] := 
  Block[{a, s, k, b, c}, 
   a[j_] := (SetPrecision[Log[j + 1], prec]/(j + 1))^m;
   {b, c, s} = {-1, -d, 0};
   Do[c = b - c;
    s = s + c a[k];
    b = (k + n) (k - n) b/((k + 1) (k + 1/2)), {k, 0, n - 1}];
   Return[N[s/d, pr] (-1)^m]];
eta1 = N[EulerGamma Log[2] - Log[2]^2/2, prec]; n = 
 Floor[132/100 prec]; d = N[ChebyshevT[n, 3], prec];
MRBtest = 
  eta1 - ParallelSum[(Cos[Pi m]) etaMM[m, prec]/
      N[Gamma[m + 1], prec], {m, 2, Floor[.365 prec]}];
Print[mTrue - MRBtest]; SessionTime[] - to

During evaluation of In[82]:= 0.*10^-1500

Out[87]= 15.4875433

Here is a comparison of my eta formula speed records:

enter image description here

Here is the code for 10,000 digits:

prec = 10000;
to = SessionTime[];
etaMM[m_, pr_] := 
  Block[{a, s, k, b, c}, 
   a[j_] := (SetPrecision[Log[j + 1], prec]/(j + 1))^m;
   {b, c, s} = {-1, -d, 0};
   Do[c = b - c;
    s = s + c a[k];
    b = (k + n) (k - n) b/((k + 1) (k + 1/2)), {k, 0, n - 1}];
   Return[N[s/d, pr] (-1)^m]];
eta1 = N[EulerGamma Log[2] - Log[2]^2/2, prec]; n = 
 Floor[132/100 prec]; d = N[ChebyshevT[n, 3], prec];
MRBtest = 
  eta1 - ParallelSum[(Cos[Pi m]) etaMM[m, prec]/
      N[Gamma[m + 1], prec], {m, 2, Floor[.295 prec]}];
Print[mTrue - MRBtest]; SessionTime[] - to

0.*10^-9999

20621.99102

I tried a little something I got from the Mathematica documentation center. It speed the calculations up just a little more. My newest sample code as of 5/31/2018:

prec = 1500;
to = SessionTime[];
etaMM[m_, pr_] := 
  Block[{a, s, k, b, c}, 
   a[j_] := (SetPrecision[Log[j + 1], prec]/(j + 1))^m;
   {b, c, s} = {-1, -d, 0};
   Do[c = b - c;
    s = s + c a[k];
    b = (k + n) (k - n) b/((k + 1) (k + 1/2)), {k, 0, n - 1}];
   Return[N[s/d, pr] (-1)^m]];
eta1 = N[EulerGamma Log[2] - Log[2]^2/2, prec]; n = 
 Floor[132/100 prec]; d = N[ChebyshevT[n, 3], prec];
MRBtest = 
  eta1 - ParallelSum[(Cos[Pi m]) etaMM[m, prec]/
      N[Gamma[m + 1], prec], {m, 2, Floor[.365 prec]}, 
    Method -> "CoarsestGrained"];
Print[mTrue - MRBtest]; SessionTime[] - to

That leads to the following table.

enter image description here

See attached "eta may 31 2018.nb" for the work.

enter image description here

May 31 2018 | 9:00PM EST

I tried this form using the ParallelCombine command:

prec = 1500;
to = SessionTime[];
etaMM[m_, pr_] := 
  Block[{a, s, k, b, c}, 
   a[j_] := (SetPrecision[Log[j + 1], prec]/(j + 1))^m;
   {b, c, s} = {-1, -d, 0};
   Do[c = b - c;
    s = s + c a[k];
    b = (k + n) (k - n) b/((k + 1) (k + 1/2)), {k, 0, n - 1}];
   Return[N[s/d, pr] (-1)^m]];
eta1 = N[EulerGamma Log[2] - Log[2]^2/2, prec]; n = 
 Floor[132/100 prec]; d = N[ChebyshevT[n, 3], prec];
MRBtest = eta1 -
   Total[ParallelCombine[((Cos[Pi #]) etaMM[#, prec]/
         N[Gamma[# + 1], prec]) &, Range[2, Floor[.365 prec]], 
     Method -> "CoarsestGrained"]];
Print[mTrue - MRBtest];

SessionTime[] - to

, which lead to this remarkable set of results from my fast 4 core machine with 2400 MHz DDR4 RAM:

2a

The code and computations from Crandall's original code and my first improvement of it can be found by searching for the post above that begins with

"How about computing the MRB constant from Crandall's eta derivative formulas?"

See eta may 31 2018.nb attachment for more work.

UPDATE 2:51PM, June 1, 2018

**Last night at about 9:30PM EST, I started a 20,000 digit calculation of the digits of the MRB constant via eta derivatives using the "Combine" method.

UPDATE

The 20,000 digits finished at Fri 1 Jun 2018 21:11:49 -- it took 83728.3439962 sec -- see attached 20 k eta style.nb .

That timing of 83728.3439962 sec for 20,000 digits fits perfectly with a pattern compared of previous timings. approx 5000 digits took 548sec. 10000 digits took 6806sec.

6806sec(6806sec/548sec) = 84528.532* sec.

That surprises me because my choice in the multiple of prec, as in Floor[.365 prec] for prec=20000 was only a guess. (I used 0.285.) which means the program took 5700 iterations. (It got 3.5 digits per iteration.) My plan was to always use the minimum number of iterations to get a desired number of digits.

As I wrote above:

The larger prec (precission) you want to choose, the smaller number you can replace .365 with in ParallelSum[(Cos[Pi m]) etaMM[m, prec]/N[Gamma[m + 1], prec], {m, 2, Floor[.365 prec]}. For prec=5000, change .365 to .31 (approximately). This advantage for smaller upper limits being needed in the sum is due to the Gamma (!) in the denominator. On my big computer 4000 digits now gives a timing of around 1800 seconds (1/2 an hour). See message above the previous occurrence of 8/13/2017 for previous records and records before those.

WARNING

I posted some about 12 kernels:

I hoped to produce some extra success by lying to Kernel Configuration as follows:

A big Lie!

Which did lead to the following, great results:

enter image description here

Here is a sample:

prec = 10000;
to = SessionTime[];
etaMM[m_, pr_] := 
  Block[{a, s, k, b, c}, 
   a[j_] := (SetPrecision[Log[j + 1], prec]/(j + 1))^m;
   {b, c, s} = {-1, -d, 0};
   Do[c = b - c;
    s = s + c a[k];
    b = (k + n) (k - n) b/((k + 1) (k + 1/2)), {k, 0, n - 1}];
   Return[N[s/d, pr] (-1)^m]];
eta1 = N[EulerGamma Log[2] - Log[2]^2/2, prec]; n = 
 Floor[132/100 prec]; d = N[ChebyshevT[n, 3], prec];
MRBtest = 
  eta1 - Total[
    ParallelCombine[((Cos[Pi #]) etaMM[#, prec]/
         N[Gamma[# + 1], prec]) &, Range[2, Floor[.295 prec]], 
     Method -> "CoarsestGrained"]];
Print[mTrue - MRBtest];

SessionTime[] - to

During evaluation of In[52]:= 0.*10^-10000

Out[58]= 3413.3092082

See attached 16 socalled cores.nb.

There is an example of the results for using more and more kernels on my 8 duel core Zeon: -- in attached "more and more kernels effect.nb ."

I would be absolutely delighted if anyone with a larger kernel limit license could run it and show us what your results are!!!

Here is a sample of using more and more kernels on my 8 duel core Zeon:

 Quit[];
mTrue =  0.18785964246206712024851793405427323005590309490013878617200468408947\
7231564660213703296654433107496903842345856258019061231370094759226630\
4389293488961841208373366260816136027381263793734352832125527639621714\
8932170207628206217151671540841268044836354167199851976802527598938993\
9144579835055613509648521071207844423095868129497688526949564204255586\
4836704410425279524710606660926339748341031157816786416689154600342222\
5883800254553968929471142122189105098328712277308020036445215390536395\
0553322034706275511598128280395102192649146731762935161906598160186642\
4582495069720338199295842093551516251439935760076459329128145170908242\
4915883204169066409334435914806705564692806787007028115009380606938139\
3859533606579874055620623487043293607378195646031047639506648930613606\
4552806751519350828083737671929686639810309494963749627738304984632456\
3479311575300289212523291816195626973697074865765476071178017195787368\
3009659022606687536563055165673612881502014387561366865522106743053705\
9103973575619148909369077798320355119336240463725349410542836369971702\
4418551654837279358822008134480961058802030647819619596953756287834812\
3349763858630101407272529230147233333625091858402480370404888196767676\
0119858111679169352796852044160027086137228688945101510291998853690572\
8659287086875425492533794395347589703563313440382638887986656195980733\
5147399025657781331722610761279758527227427773089857749223059709625725\
6271883675575297887925361687673940354321451362772549229313126276435732\
1446216187786377154205423128223446295396532903322171479820280759842210\
6556489004853685870708326887487737763504768916098318553628166715910841\
2193420164386000258508426556435006954832830120546193205155935040023508\
3512613359217408970073297842771289673651619602250771173880842623256978\
8546537869046222708567487474709306935732666859085616282375386551243297\
5647464914619179575869342996208149878536663170197264534260468378010759\
0551486787190395783150604524441907570445113820585333984692194828794764\
8657593178595816527492977822095977440911371434216929624593175324537340\
1299593995004917912983680848547143925846704238528608320053664510586678\
1511964596760791964317343076715344983004971288694016566004270621110790\
5316472150455632994388400521115239016877311545696102836920503689610880\
6031603660382896533239383524154510137534165673472607464891120088099838\
1520466954150263770355732835929966306427173051589721163519991611359546\
7031540872528724399819787250274679738863889705686743537785798105855619\
2492185716949135673462704077491448799682065482817465880642236348160780\
9507770579393134958298066028252721284916888092303252902700599177550596\
1583591999319086939303973661164651485821997292533710676873868623504791\
5879737968269847878082223410618789674667450680064404065553875213281494\
9807002098581322206201090112659034497174108010632475647128346095492843\
7006514745021822612041564393030885982642625682812609249113673396723593\
3714534216902560140050169469983875907342920361729301531400405936246406\
78140077947561307736973240992352946479458077816460769624086460;
Quiet[Table[LaunchKernels[1]; prec = 3000;
  to = SessionTime[];
  etaMM[m_, pr_] := 
   Block[{a, s, k, b, c}, 
    a[j_] := (SetPrecision[Log[j + 1], prec]/(j + 1))^m;
    {b, c, s} = {-1, -d, 0};
    Do[c = b - c;
     s = s + c a[k];
     b = (k + n) (k - n) b/((k + 1) (k + 1/2)), {k, 0, n - 1}];
    Return[N[s/d, pr] (-1)^m]];
  eta1 = N[EulerGamma Log[2] - Log[2]^2/2, prec]; 
  n = Floor[132/100 prec]; d = N[ChebyshevT[n, 3], prec];
  MRBtest = 
   eta1 - Total[
     ParallelCombine[((Cos[Pi #]) etaMM[#, prec]/
          N[Gamma[# + 1], prec]) &, Range[2, Floor[.365 prec]], 
      Method -> "CoarsestGrained"]];
  Print["Error was ", mTrue - MRBtest, ", using"];

  Print[xx, "  kernels, and gave a time of ", SessionTime[] - to, 
   " s."], {xx, 1, 16}]]

  Error was 0.*10^-3000, using

1  kernels, and gave a time of 477.9303673 s.

Error was 0.*10^-3000, using

2  kernels, and gave a time of 257.3167709 s.

Error was 0.*10^-3000, using

3  kernels, and gave a time of 181.8228406 s.

Error was 0.*10^-3000, using

4  kernels, and gave a time of 147.4913411 s.

Error was 0.*10^-3000, using

5  kernels, and gave a time of 118.8902094 s.

Error was 0.*10^-3000, using

6  kernels, and gave a time of 103.5240060 s.

Error was 0.*10^-3000, using

7  kernels, and gave a time of 92.3546592 s.

Error was 0.*10^-3000, using

8  kernels, and gave a time of 85.5143638 s.

Error was 0.*10^-3000, using

9  kernels, and gave a time of 75.6869920 s.

Error was 0.*10^-3000, using

10  kernels, and gave a time of 72.1320176 s.

Error was 0.*10^-3000, using

11  kernels, and gave a time of 68.4291460 s.

Error was 0.*10^-3000, using

12  kernels, and gave a time of 62.7524163 s.

Error was 0.*10^-3000, using

13  kernels, and gave a time of 63.8671114 s.

Error was 0.*10^-3000, using

14  kernels, and gave a time of 61.3992399 s.

Error was 0.*10^-3000, using

15  kernels, and gave a time of 61.0743278 s.

Error was 0.*10^-3000, using

16  kernels, and gave a time of 60.1936369 s.

I just updated the attachment, "more and more kernels effect.nb ."

UPDATE July 29, 2018

Here are some new stats:

enter image description here

12/31/2018

Ending this year, I subscribed to a cloud service and used a 48 core Xeon Platinum server to compare with my old 8 core Xeon that I sold, and got the following results for MRB constant computations using only Crandall's eta formula.

enter image description here

Here is the code used on that Intel Xeon Platinum 48 core computer, for 10,000 digits in less than 1426 seconds:

prec = 10000;
to = SessionTime[];
etaMM[m_, pr_] := 
  Block[{a, s, k, b, c}, 
   a[j_] := (SetPrecision[Log[j + 1], prec]/(j + 1))^m;
   {b, c, s} = {-1, -d, 0};
   Do[c = b - c;
    s = s + c a[k];
    b = (k + n) (k - n) b/((k + 1) (k + 1/2)), {k, 0, n - 1}];
   Return[N[s/d, pr] (-1)^m]];
eta1 = N[EulerGamma Log[2] - Log[2]^2/2, prec]; n = 
 Floor[132/100 prec]; d = N[ChebyshevT[n, 3], prec];
MRBtest = 
  eta1 - Total[
    ParallelCombine[((Cos[Pi #]) etaMM[#, prec]/
         N[Gamma[# + 1], prec]) &, Range[2, Floor[.298 prec]], 
     Method -> "CoarsestGrained"]];
Export["accuracy.dat",mTrue - MRBtest];

Export["time.dat",SessionTime[] - to]

3/27/2019

I ran an improved version of the above code (Cradall first eta formula) for 10,000 digits of MRB using 48 Mma kernels on 1 node (24 cores) of IU's Carbonate supercomputer.

configuration: enter image description here

code:

Iprec = 10000;
to = SessionTime[];
etaMM[m_, pr_] := 
  Module[{a, s, k, b, c}, 
   a[j_] := (SetPrecision[Log[j + 1], prec]/(j + 1))^m;
   {b, c, s} = {-1, -d, 0};
   Do[c = b - c;
    s = s + c a[k];
    b = (k + n) (k - n) b/((k + 1) (k + 1/2)), {k, 0, n - 1}];
   Return[N[s/d, pr] (-1)^m]];
eta1 = N[EulerGamma Log[2] - Log[2]^2/2, prec]; n = 
 Floor[132/100 prec]; d = N[ChebyshevT[n, 3], prec];
MRBtest = 
  eta1 - Total[
    ParallelCombine[((Cos[Pi #]) etaMM[#, prec]/
         N[Gamma[# + 1], prec]) &, Range[2, Floor[.295 prec]], 
     Method -> "CoarsestGrained"]];
Print[N[MRBtest2 - MRBtest, 10]];

SessionTime[] - to

output:

 0.*10^-10000

 1333.552401

That 1333 seconds is a new speed record for 10,000 digits via eta derivatives!

Attachments:
16 socalled cores.nb
20 k eta style.nb
eta all cores as...nb
eta may 31 2018.nb
more and more ke...nb
POSTED BY: Marvin Ray Burns
Answer

02/12/2019

Using my 2 nodes of the MRB constant supercomputer (3.7 GH overclocked up to 4.7 GH, Intel 6core, 3000MH RAM,and 4 cores from my 3.6 GH, 2400MH RAM) I computed 34,517 digits of the MRB constant using Crandall's first eta formula:

prec = 35000;
to = SessionTime[];
etaMM[m_, pr_] := 
  Block[{a, s, k, b, c}, 
   a[j_] := (SetPrecision[Log[j + 1], prec]/(j + 1))^m;
   {b, c, s} = {-1, -d, 0};
   Do[c = b - c;
    s = s + c a[k];
    b = (k + n) (k - n) b/((k + 1) (k + 1/2)), {k, 0, n - 1}];
   Return[N[s/d, pr] (-1)^m]];
eta1 = N[EulerGamma Log[2] - Log[2]^2/2, prec]; n = 
 Floor[132/100 prec]; d = N[ChebyshevT[n, 3], prec];
MRBtest = 
  eta1 - Total[
    ParallelCombine[((Cos[Pi #]) etaMM[#, prec]/
         N[Gamma[# + 1], prec]) &, Range[2, Floor[.250 prec]], 
     Method -> "CoarsestGrained"]];
Print[N[MRBtest2 - MRBtest,10]];

SessionTime[] - to

giving -2.166803252*10^-34517 for a difference and 208659.2864422 seconds or 2.415 days for a timing.

Where MRBtest2 is 36000 digits computed through acceleration methods of n^(1/n)

3/28/2019

Here is an updated table of speed eta formula records: eta records 12 31 18

04/03/2019

Using my 2 nodes of the MRB constant supercomputer (3.7 GH overclocked up to 4.7 GH, Intel 6core, 3000MH RAM,and 4 cores from my 3.6 GH, 2400MH RAM) I computed 50,000 digits of the MRB constant using Crandall's first eta formula in 5.79 days.

 prec = 50000;
to = SessionTime[];
etaMM[m_, pr_] := 
  Module[{a, s, k, b, c}, 
   a[j_] := 
    SetPrecision[SetPrecision[Log[j + 1]/(j + 1), prec]^m, prec];
   {b, c, s} = {-1, -d, 0};
   Do[c = b - c;
    s = s + c a[k];
    b = (k + n) (k - n) b/((k + 1) (k + 1/2)), {k, 0, n - 1}];
   Return[N[s/d, pr] (-1)^m]];
eta1 = N[EulerGamma Log[2] - Log[2]^2/2, prec]; n = 
 Floor[132/100 prec]; d = N[ChebyshevT[n, 3], prec];
MRBtest = 
  eta1 - Total[
    ParallelCombine[((Cos[Pi #]) etaMM[#, prec]/
         N[Gamma[# + 1], prec]) &, Range[2, Floor[.245 prec]], 
     Method -> "CoarsestGrained"]];
Print[N[MRBtest2 - MRBtest, 10]];

SessionTime[] - to

 (* 0.*10^-50000

  500808.4835750*)
POSTED BY: Marvin Ray Burns
Answer

4/22/2019

Let $$M=\sum _{n=1}^{\infty } \frac{(-1)^{n+1} \eta ^n(n)}{n!}=\sum _{n=1}^{\infty } (-1)^n \left(n^{1/n}-1\right).$$ Then using what I learned about the absolute convergence of $\sum _{n=1}^{\infty } \frac{(-1)^{n+1} \eta ^n(n)}{n!}$ from https://math.stackexchange.com/questions/1673886/is-there-a-more-rigorous-way-to-show-these-two-sums-are-exactly-equal, combined with an identity from Richard Crandall: enter image description here, Also using what Mathematica says:

$$\sum _{n=1}^1 \frac{\underset{m\to 1}{\text{lim}} \eta ^n(m)}{n!}=\gamma (2 \log )-\frac{2 \log ^2}{2},$$

I figured out that

$$\sum _{n=2}^{\infty } \frac{(-1)^{n+1} \eta ^n(n)}{n!}=\sum _{n=1}^{\infty } (-1)^n \left(n^{1/n}-\frac{\log (n)}{n}-1\right).$$

So I made the following major breakthrough in computing MRB from Candall's first eta formula. See attached 100 k eta 4 22 2019. Also shown below.

eta 18 to19 n 2.JPG

The time grows 10,000 times slower than the previous method!

I broke a new record, 100,000 digits: Processor and total time were 806.5 and 2606.7281972 s respectively.. See attached 2nd 100 k eta 4 22 2019.

Here is the work from 100,000 digits. enter image description here

Print["Start time is ", ds = DateString[], "."];
prec = 100000;
(**Number of required decimals.*.*)ClearSystemCache[];
T0 = SessionTime[];
expM[pre_] := 
  Module[{a, d, s, k, bb, c, end, iprec, xvals, x, pc, cores = 16(*=4*
    number of physical cores*), tsize = 2^7, chunksize, start = 1, ll,
     ctab, pr = Floor[1.005 pre]}, chunksize = cores*tsize;
   n = Floor[1.32 pr];
   end = Ceiling[n/chunksize];
   Print["Iterations required: ", n];
   Print["Will give ", end, 
    " time estimates, each more accurate than the previous."];
   Print["Will stop at ", end*chunksize, 
    " iterations to ensure precsion of around ", pr, 
    " decimal places."]; d = ChebyshevT[n, 3];
   {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
   iprec = Ceiling[pr/27];
   Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l;
        x = N[E^(Log[ll]/(ll)), iprec];
        pc = iprec;
        While[pc < pr/4, pc = Min[3 pc, pr/4];
         x = SetPrecision[x, pc];
         y = x^ll - ll;
         x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];(**N[Exp[Log[ll]/
        ll],pr/4]**)x = SetPrecision[x, pr];
        xll = x^ll; z = (ll - xll)/xll;
        t = 2 ll - 1; t2 = t^2;
        x = 
         x*(1 + SetPrecision[4.5, pr] (ll - 1)/
              t2 + (ll + 1) z/(2 ll t) - 
            SetPrecision[13.5, pr] ll (ll - 1) 1/(3 ll t2 + t^3 z));(**
        N[Exp[Log[ll]/ll],pr]**)x, {l, 0, tsize - 1}], {j, 0, 
        cores - 1}, Method -> "EvaluationsPerKernel" -> 32]];
    ctab = ParallelTable[Table[c = b - c;
       ll = start + l - 2;
       b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
       c, {l, chunksize}], Method -> "EvaluationsPerKernel" -> 16];
    s += ctab.(xvals - 1);
    start += chunksize;
    st = SessionTime[] - T0; kc = k*chunksize;
    ti = (st)/(kc + 10^-4)*(n)/(3600)/(24);
    If[kc > 1, 
     Print[kc, " iterations done in ", N[st, 4], " seconds.", 
      " Should take ", N[ti, 4], " days or ", N[ti*24*3600, 4], 
      "s, finish ", DatePlus[ds, ti], "."]];, {k, 0, end - 1}];
   N[-s/d, pr]];
t2 = Timing[MRB = expM[prec];]; Print["Finished on ", 
 DateString[], ". Proccessor time was ", t2[[1]], " s."];
Print["Enter MRBtest2 to print ", Floor[Precision[MRBtest2]], 
  " digits"];


 (Start time is )^2Tue 23 Apr 2019 06:49:31.

 Iterations required: 132026

 Will give 65 time estimates, each more accurate than the previous.

 Will stop at 133120 iterations to ensure precsion of around 100020 decimal places.

 Denominator computed in  17.2324041s.

...

129024 iterations done in 1011. seconds. Should take 0.01203 days or 1040.s, finish Mon 22 Apr 
2019 12:59:16.

131072 iterations done in 1026. seconds. Should take 0.01202 days or 1038.s, finish Mon 22 Apr 
2019 12:59:15.

Finished on Mon 22 Apr 2019 12:59:03. Processor time was 786.797 s.

enter image description here

 Print["Start time is " "Start time is ", ds = DateString[], "."];
 prec = 100000;
 (**Number of required decimals.*.*)ClearSystemCache[];
 T0 = SessionTime[];
 expM[pre_] := 
   Module[{lg, a, d, s, k, bb, c, end, iprec, xvals, x, pc, cores = 16(*=
     4*number of physical cores*), tsize = 2^7, chunksize, start = 1, 
     ll, ctab, pr = Floor[1.0002 pre]}, chunksize = cores*tsize;
    n = Floor[1.32 pr];
    end = Ceiling[n/chunksize];
    Print["Iterations required: ", n];
    Print["Will give ", end, 
     " time estimates, each more accurate than the previous."];
    Print["Will stop at ", end*chunksize, 
     " iterations to ensure precsion of around ", pr, 
     " decimal places."]; d = ChebyshevT[n, 3];
    {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
    iprec = pr/2^6;
    Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l;
         lg = Log[ll]/(ll); x = N[E^(lg), iprec];
         pc = iprec;
         While[pc < pr, pc = Min[4 pc, pr];
          x = SetPrecision[x, pc];
          xll = x^ll; z = (ll - xll)/xll;
          t = 2 ll - 1; t2 = t^2;
          x = 
           x*(1 + SetPrecision[4.5, pc] (ll - 1)/
                t2 + (ll + 1) z/(2 ll t) - 
              SetPrecision[13.5, 2 pc] ll (ll - 1)/(3 ll t2 + t^3 z))];
          x - lg, {l, 0, tsize - 1}], {j, 0, cores - 1}, 
        Method -> "EvaluationsPerKernel" -> 16]];
     ctab = ParallelTable[Table[c = b - c;
        ll = start + l - 2;
        b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
        c, {l, chunksize}], Method -> "EvaluationsPerKernel" -> 16];
     s += ctab.(xvals - 1);
     start += chunksize;
     st = SessionTime[] - T0; kc = k*chunksize;
     ti = (st)/(kc + 10^-10)*(n)/(3600)/(24);
     If[kc > 1, 
      Print[kc, " iterations done in ", N[st - stt, 4], " seconds.", 
       " Should take ", N[ti, 4], " days or ", ti*3600*24, 
       "s, finish ", DatePlus[ds, ti], "."], 
      Print["Denominator computed in  ", stt = st, "s."]];, {k, 0, 
      end - 1}];
    N[-s/d, pr]];
 t2 = Timing[MRBeta2toinf = expM[prec];]; Print["Finished on ", 
  DateString[], ". Processor and total time were ", 
  t2[[1]], " and ", st, " s respectively."];

Start time is  Tue 23 Apr 2019 06:49:31.

Iterations required: 132026

Will give 65 time estimates, each more accurate than the previous.

Will stop at 133120 iterations to ensure precision of around 100020 decimal places.

Denominator computed in  17.2324041s.

...

131072 iterations done in 2589. seconds. Should take 0.03039 days or 2625.7011182s, finish Tue 23 Apr 2019 07:33:16.

Finished on Tue 23 Apr 2019 07:32:58. Processor and total time were 806.5 and 2606.7281972 s respectively.

enter image description here

 MRBeta1 = EulerGamma Log[2] - 1/2 Log[2]^2

 EulerGamma Log[2] - Log[2]^2/2

enter image description here

   N[MRBeta2toinf + MRBeta1 - MRB, 10]

   1.307089967*10^-99742
Attachments:
2nd 100 k eta 4 ...nb
50 k eta 4 22 2019.nb
POSTED BY: Marvin Ray Burns
Answer

As mentioned a few times above Richard Crandall made the following observations about the MRB constant: enter image description here

His last comment is that we might find some algorithm advantages by focusing on the fixed argument of 0 for the eta derivatives. He gives Chapter 3 as a possible source to find the algorithm advantages. (Here is another link to his paper: http://marvinrayburns.com/UniversalTOC25.pdf .)

Here is what he was looking for.

c[0] = 0; 
c[1] = Log[2]; Table[
c[n] = -2*Derivative[n - 1][Zeta][0], {n, 2, 22}]; 
l = CoefficientList[Normal[Series[Exp[(-x)*E^x], {x, 0, 21}]], 
x]; N[-Sum[
l[[n + 1]]*((-1)^
n*(Sum[(-1)^x*(c[x + 1]*Binomial[n, x]*c[1]^(n - x)), {x, 1, 
n - 1}] + c[1]^n + 
c[0]) + 1/2 c[n + 1]), {n, 1, 17}]]

Mathematica still has trouble with higher zeta derivatives of 0, however. I'm going to look for a procedure for computing them better.

[Edit]

I computed the zeta derivatives of 0 by 

   zeta0[p_] := (-1)^p Sum[StieltjesGamma[p + k]/k!, {k, 0, Infinity}] - 
     p!

and got the same results Mathematica got for the higher zeta derivatives of 0.

I also checked the CoefficientList of 

CoefficientList[Normal[Series[Exp[(-x)*E^x], {x, 0, 21}]]

by the following, and haven't found where Mathematica is having trouble!

c[0] = 0; c[1] = Log[2]; Table[c[n] = -2*Derivative[n - 1][Zeta][0], 
{n, 2, 
18}]; N[-Sum[(Sum[(-1)^d*d^(n - d)*Binomial[n, d], {d, 1, n}]*
((-1)^
n*(Sum[(-1)^x*(c[x + 1]*Binomial[n, x]*c[1]^(n - x)), 
{x, 1, n - 1}] + c[1]^n + c[0]) + (1/2)*
c[n + 1]))/n!, {n, 1, 17}]]

I even checked the binomials by their formula. Also the StieltjesGamma and Derivative[n ][Zeta][0]] results appear smooth for discrete operations:

 Table[N[Derivative[n ][Zeta][0]], {n, 10, 25}]

and

 N[Table[StieltjesGamma[x], {x, 1, 30}]]

.

[Edit]

I found the problem as to why Mathematica doesn't increase its accuracy in the above formula after 17 iterations: Mathematica will give a numeric solution to many zeta derivatives of 0, only to machine precision!

For machine precision:

  Table[N[Derivative[n ][Zeta][0]], {n, 1, 30}]

gives the machine sized numeric to many derivatives:

 {-0.918939, -2.00636, -6.00471, -23.9971, -120., -720.001, \
-5040., -40320., -362880., -3.6288*10^6, -3.99168*10^7, \
-4.79002*10^8, -6.22702*10^9, -8.71783*10^10, -1.30767*10^12, \
-2.09228*10^13, -3.55687*10^14, -6.40237*10^15, -1.21645*10^17, \
-2.4329*10^18, -5.10909*10^19, -1.124*10^21, -2.5852*10^22, \
-6.20448*10^23, -1.55112*10^25, -4.03291*10^26, -1.08889*10^28, \
-3.04888*10^29, -8.84176*10^30, -2.65253*10^32}.

Then

   Table[N[Derivative[n][Zeta][0], 10], {n, 1, 30}]

gives the high precision numeric for only the first 3 derivatives:

{-0.91893853320467274178032973640561763986`10., \
-2.0063564559085848512101000267299604382`10., 
   -6.0047111668622544478`10., Derivative[4][Zeta][0], 
 Derivative[5][Zeta][0], Derivative[6][Zeta][0], 
   Derivative[7][Zeta][0], Derivative[8][Zeta][0], 
 Derivative[9][Zeta][0], Derivative[10][Zeta][0], 
   Derivative[11][Zeta][0], Derivative[12][Zeta][0], 
 Derivative[13][Zeta][0], Derivative[14][Zeta][0], 
   Derivative[15][Zeta][0], Derivative[16][Zeta][0], 
 Derivative[17][Zeta][0], Derivative[18][Zeta][0], 
   Derivative[19][Zeta][0], Derivative[20][Zeta][0], 
 Derivative[21][Zeta][0], Derivative[22][Zeta][0], 
   Derivative[23][Zeta][0], Derivative[24][Zeta][0], 
 Derivative[25][Zeta][0], Derivative[26][Zeta][0], 
   Derivative[27][Zeta][0], Derivative[28][Zeta][0], 
 Derivative[29][Zeta][0], Derivative[30][Zeta][0]}.

With some confidence I can say

formula

I fixed it!!!

I forced Mathematica to give up the needed digits of the zeta derivatives.

My new program does not really compete with the previously mentioned record-breakers. But it is a fairly efficient use of Crandall's formula (44). First it calculates many digits on MRB the old fashion way to use as a check. So here is my new program:

m = NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 120];

Block[{$MaxExtraPrecision = 1000}, s = 100(*number of terms*);
 c[1] = Log[2];
 c[n_] := N[1 - 2*Derivative[n - 1][Zeta][0], Floor[3/2 s]] - 1;
 mtest = (-Sum[(Sum[(-1)^d*d^(n - d)*Binomial[n, d], {d, 1, 
          n}]*((-1)^
            n*(Sum[(-1)^x*(c[x + 1]*Binomial[n, x]*c[1]^(n - x)), {x, 
               1, n - 1}] + c[1]^n) + (1/2)*c[n + 1]))/n!, {n, 1, 
      s}])]; Print[s, " terms give ", mtest, ", and the error is ", 
 m - mtest]

100 terms give 0.18785964486341229249700337222, and the error is -2.40134517224848543816*10^-9

Changing s to 200 shows all correct digits:

200 terms give 0.187859642465, and the error is -3.*10^-12
POSTED BY: Marvin Ray Burns
Answer

I figured out how to rapidly compute AND CHECK a computation of the MRB constant! (The timing given is in processor time [for computing and checking] only. T0 can be used with another SessionTime[] call at the end to figure out all time expired during running of the program.) I used both of Crandall's methods for computing it and used for a check, the nontrivial identityenter image description here ,where gamma is the Euler constant and M is the MRB constant.

Below is my first version of the code with results. If nothing else, I thought, the code pits Crandall's 2 methods against each other to show if one is wrong they both are wrong. (So it is not a real proof.) But these are two totally different methods! (the first of which has been proven by Henry Cohen to be theoretically correct here). For a second check mm is a known approximation to the constant; over 3 million non checked (as of now) digits are found in the attached file 3M.nb. (You will have to change the Format/Style to Input to use the digits.)

In[15]:= (*MRB constant computation with verification! The constant's \
decimal approximation is saved as MRBtest*)prec = 5000;(*Number of \
required decimals.*)ClearSystemCache[];
T0 = SessionTime[];
expM[pre_] := 
 Module[{a, d, s, k, bb, c, n, end, iprec, xvals, x, pc, cores = 4, 
   tsize = 2^7, chunksize, start = 1, ll, ctab, pr = Floor[1.02 pre]},
   chunksize = cores*tsize;
  n = Floor[1.32 pr];
  end = Ceiling[n/chunksize];
  d = N[(3 + Sqrt[8])^n, pr + 10];
  d = Round[1/2 (d + 1/d)];
  {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
  iprec = Ceiling[pr/27];
  Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l;
        x = N[E^(Log[ll]/(ll)), iprec];
        pc = iprec;
        While[pc < pr, pc = Min[3 pc, pr];
         x = SetPrecision[x, pc];
         y = x^ll - ll;
         x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];(*N[Exp[Log[ll]/ll],
        pr]*)x, {l, 0, tsize - 1}], {j, 0, cores - 1}, 
       Method -> "EvaluationsPerKernel" -> 1]];
    ctab = Table[c = b - c;
      ll = start + l - 2;
      b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
      c, {l, chunksize}];
    s += ctab.(xvals - 1);
    start += chunksize;, {k, 0, end - 1}];
  etaMs = N[-s/d - (EulerGamma Log[2] - Log[2]^2/2), pr]]; t2 = 
 Timing[MRBtest2 = expM[prec];];
Print["The MRB constant was computed and checked to ", prec, " digits \
in ", t1 = t2[[1]] + Timing[eta[s_] := (1 - 2^(1 - s)) Zeta[s];
     eta1 = Limit[D[eta[s], s], s -> 1];
     MRBtrue = mm;
     MRBtest = eta1 + etaMs;
     check = MRBtest - MRBtrue][[1]], " seconds"]; check

During evaluation of In[15]:= The MRB constant was computed and checked to 5000 digits in 2.12161 seconds

Out[18]= 0.*10^-5000

In[19]:= MRBtest - mm

Out[19]= 0.*10^-5000
Attachments:
3M.nb
POSTED BY: Marvin Ray Burns
Answer

The identity in question is straightforward. Write n^(1/n) as Exp[Log[n]/n], take a series expansion at 0, and subtract the first term from all summands. That means subtracting off Log[n]/n in each summand. This gives your left hand side. We know it must be M - the sum of the terms we subtracted off. Now add all of them up, accounting for signs.

Expand[Sum[(-1)^n*Log[n]/n, {n, 1, Infinity}]]

(* Out[74]= EulerGamma Log[2] - Log[2]^2/2 *)

So we recover the right hand side.

I have not understood whether this identity helps with Crandall's iteration. One advantage it confers, a good one in general, is that it converts a conditionally convergent alternating series into one that is absolutely convergent. From a numerical computation point of view this is always good.
POSTED BY: Daniel Lichtblau
Answer

Daniel Lichtblau and others, I just deciphered an Identity Crandall used for checking computations of the MRB constant just before he died. It is used in a previous post about checking, where I said it was hard to follow. The MRB constant is B here. B=`enter image description here In input form that is

   B= Sum[(-1)^(k + 1)*(-1 + (1 + k)^(1/(1 + k)) - Log[1 + k]/(1 + k) - 
         Log[1 + k]^2/(2*(1 + k)^2)), {k, 0, Infinity}] + 
     1/24 (\[Pi]^2 Log[2]^2 - 
        2 \[Pi]^2 Log[
          2] (EulerGamma + Log[2] - 12 Log[Glaisher] + Log[\[Pi]]) - 
        6 (Zeta^\[Prime]\[Prime])[2]) + 
     1/2 (2 EulerGamma Log[2] - Log[2]^2)

For 3000 digit numeric approximation, it is

B=NSum[((-1)^(
    k + 1) (-1 + (1 + k)^(1/(1 + k)) - Log[1 + k]/(1 + k) - 
      Log[1 + k]^2/(2 (1 + k)^2))), {k, 0, Infinity}, 
  Method -> "AlternatingSigns", WorkingPrecision -> 3000] + 
 1/24 (\[Pi]^2 Log[2]^2 - 
    2 \[Pi]^2 Log[
      2] (EulerGamma + Log[2] - 12 Log[Glaisher] + Log[\[Pi]]) - 
    6 (Zeta^\[Prime]\[Prime])[2]) + 
 1/2 (2 EulerGamma Log[2] - Log[2]^2)

It is anylitaclly straight forward too because

Sum[(-1)^(k + 1)*Log[1 + k]^2/(2 (1 + k)^2), {k, 0, Infinity}]

gives

1/24 (-\[Pi]^2 (Log[2]^2 + EulerGamma Log[4] - 
      24 Log[2] Log[Glaisher] + Log[4] Log[\[Pi]]) - 
   6 (Zeta^\[Prime]\[Prime])[2])

That is enter image description here I wonder why he chose it?
POSTED BY: Marvin Ray Burns
Answer

The new sum is this.

Sum[(-1)^(k + 1)*(-1 + (1 + k)^(1/(1 + k)) - Log[1 + k]/(1 + k) - 
         Log[1 + k]^2/(2*(1 + k)^2)), {k, 0, Infinity}]

That appears to be the same as for MRB except now we subtract two terms from the series expansion at the origin of k^(1/k). For each k these terms are Log[k]/k + 1/2*(Log[k]/k)^2. Accounting for the signs (-1)^k and summing, as I did earlier for just that first term, we get something recognizable.

Sum[(-1)^(k)*(Log[k]/(k) + Log[k]^2/(2*k^2)), {k, 1, Infinity}]

(* Out[21]= 1/24 (24 EulerGamma Log[2] - 2 EulerGamma \[Pi]^2 Log[2] - 
   12 Log[2]^2 - \[Pi]^2 Log[2]^2 + 24 \[Pi]^2 Log[2] Log[Glaisher] - 
   2 \[Pi]^2 Log[2] Log[\[Pi]] - 6 (Zeta^\[Prime]\[Prime])[2]) *)

So what does this buy us? For one thing, we get even better convergence from brute force summation, because now our largest terms are O((logk/k)^3) and alternating (which means if we sum in pairs it's actually O~(1/k^4) with O~ denoting the "soft-oh" wherein one drops polylogarithmic factors).

How helpful is this? Certainly it cannot hurt. But even with 1/k^4 size terms, it takes a long time to get even 40 digits, let alone thousands. So there is more going on in that Crandall approach.
POSTED BY: Daniel Lichtblau
Answer

If you copy and paste the code for Zeta''[2] you get a non evaluable input!

I get the following results comparing two methods:

In[655]:= 
Timing[MRBtrue = 
   NSum[(-1)^n*(n^(1/n) - 1), {n, 1, Infinity}, 
    WorkingPrecision -> prec, Method -> "AlternatingSigns"]][[1]]

Out[655]= 8.049652

In[656]:= 
Timing[B = 
   NSum[((-1)^(k + 1) (-1 + (1 + k)^(1/(1 + k)) - Log[1 + k]/(1 + k) -
          Log[1 + k]^2/(2 (1 + k)^2))), {k, 0, Infinity}, 
     Method -> "AlternatingSigns", WorkingPrecision -> 3000] + 
    1/24 (\[Pi]^2 Log[2]^2 - 
       2 \[Pi]^2 Log[
         2] (EulerGamma + Log[2] - 12 Log[Glaisher] + Log[\[Pi]]) - 
       6 (Zeta''[2])) + 1/2 (2 EulerGamma Log[2] - Log[2]^2)][[1]]

Out[656]= 12.277279
POSTED BY: Marvin Ray Burns
Answer

Daniel Lichtblau,

Notice Log[k]/(k) and Log[k]^2/(2k^2) are of the form Log[k]^n/(nk^n). We can take that summing of Log[k]/(k) + Log[k]^2/(2*k^2)+... to its end and write B=enter image description here or compute

NSum[(-1)^(k)*(k^(1/k) - 1 + 
     Sum[Log[k]^n/(n*k^n), {n, 1, Infinity}]), {k, 1, Infinity}, 
  WorkingPrecision -> 1000, Method -> "AlternatingSigns"] - 
 NSum[(-1)^(1 + k) Log[(k - Log[k])/k], {k, 1, Infinity}, 
  WorkingPrecision -> 1000, Method -> "AlternatingSigns"]

Simplified,.B=enter image description here,

What does that do to our "better convergence from brute force summation?"

P.S. I only used 2 terms in my sample data, so I might not have the correct pattern for Log[k]/(k) + Log[k]^2/(2*k^2)+... .

The reason that I suspect that I've got the wrong pattern is, in the following code the summands Log[k]/(k) and Log[k]^2/(2k^2) give 0 where the Log[k]^n/(nk^n).for n>2 does not. (etaMM[m, n] Gives a numeric approximation for the mth derivative of m.)

   etaMM[m_, pr_] := 
  Module[{a, d, s, k, b, c}, a[j_] := Log[j + 1]^m/(j + 1)^m;
   n = Floor[1.32 pr];
   d = Cos[n ArcCos[3]];
   {b, c, s} = {-1, -d, 0};
   Do[c = b - c;
    s = s + c a[k];
    b = (k + n) (k - n) b/((k + 1) (k + 1/2)), {k, 0, n - 1}];
   Return[N[s/d, pr] (-1)^m]]; Table[-Sum[(-1)^m*etaMM[m, 30]/m!, {m, 
      1, a}] - 
   Sum[(-1)^(k)*(Sum[Log[k]^n/(n*k^n), {n, 1, a}]), {k, 1, 
     Infinity}], {a, 1, 15}]
POSTED BY: Marvin Ray Burns
Answer

I could really use your opinions here! If your getting tired of my posts let me know.

Power towers are defined below. See http://mathworld.wolfram.com/PowerTower.html .

enter image description here ;;;enter image description here

How about breaking my record for computing the infinite power tower of the MRB constant. Here's how I have to go about computing n digits of it: I compute a decimal expansion of the MRB constant, which I save as m. Then I compute l = Log[m]; N[-ProductLog[-l]/l, n]. My record is 3 million digits, in which I used my 3,014,991 digit computation of the MRB constant (mentioned in the first post of this thread) for m.

It would be EXTREMELY helpful if anyone could find a way to compute the exact power tower without first computing the MRB constant! Such a revelation would be synonymous with finding a closed for solution to the MRB constant! I've been trying to find the exact power tower, lately, with no success. Even a high precision approximation to the power tower would generate a nearly equally precise approximation to the MRB constant. (Let ll be the infinite power tower of m, then ll^(1/ll)=m)

Attached are 3 million digits of the power tower.
POSTED BY: Marvin Ray Burns
Answer

The next active post starts out with

"NEW RECORD ATTEMPTS OF 4,000,000 DIGITS!"
POSTED BY: Marvin Ray Burns
Answer

The MRB constant is now a Scholar subject:

https://scholar.google.com/scholar?q=%22MRB+constant%22
POSTED BY: Marvin Ray Burns
Answer

MKB constant calculations have been moved to their own discussion at http://community.wolfram.com/groups/-/m/t/1323951?ppauth=W3TxvEwH .

enter image description here

Attachments are still here, though.

Attachments:
10000MKB.pdf
10KMKB.nb
faster10KMKB.nb
fastest10KMKB.nb
POSTED BY: Marvin Ray Burns
Answer

Try to beat these MRB constant records!
POSTED BY: Marvin Ray Burns
Answer
l van Veen
l van Veen
Posted 1 year ago

nice system!
POSTED BY: l van Veen
Answer

NEW RECORD ATTEMPTS OF 4,000,000 DIGITS!

.

.

First effort, Sep 04, 2015

At 10:00 PM Friday Sept 04, 2015 I started the 4,000,000 run of the MRB constant proper. Does anyone want to work on it with me?

Here is the code I am using:

(*Fastest (at MRB's end) as of 24 dEC 2014.*)
Block[{$MaxExtraPrecision = 50}, 
 prec = 4000000;(*Number of required decimals.*)ClearSystemCache[];
 T0 = SessionTime[];
 expM[pre_] := 
  Module[{a, d, s, k, bb, c, n, end, iprec, xvals, x, pc, cores = 16, 
    tsize = 2^7, chunksize, start = 1, ll, ctab, 
    pr = Floor[1.005 pre]}, chunksize = cores*tsize;
   n = Floor[1.32 pr];
   end = Ceiling[n/chunksize];
   Print["Iterations required: ", n];
   Print["end ", end];
   Print[end*chunksize]; d = Cos[n ArcCos[3]];
   {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
   iprec = Ceiling[pr/27];
   Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l;
        h = Log[ll]/ll; x = N[Exp[h], iprec];
        pc = iprec;
        While[pc < pr, pc = Min[3 pc, pr];
         x = SetPrecision[x, pc];
         y = x^ll - ll;
         x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];(*N[Exp[Log[ll]/ll],
        pr]*)x, {l, 0, tsize - 1}], {j, 0, cores - 1}, 
       Method -> "EvaluationsPerKernel" -> 4]];
    ctab = ParallelTable[Table[c = b - c;
       ll = start + l - 2;
       b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
       c, {l, chunksize}], Method -> "EvaluationsPerKernel" -> 2];
    s += ctab.(xvals - 1);
    start += chunksize;
    Print["done iter ", k*chunksize, " ", SessionTime[] - T0];, {k, 0,
      end - 1}];
   N[-s/d, pr]];
 t2 = Timing[MRBtest2 = expM[prec];]; 
 Print["Difference from 3014991 known digits is ", MRBtest2 - m3M];
 MRBtest2]

3014991 digits are saved as m3M.

I will post several updates as edits here as I go.

EDIT: As of Sat, Sept 5, 2015, the program is indicating that the run will take 3.3 months.

EDIT: Having to run the program for over 3 months, there is a chance that I would suffer a power outage longer than the 15 minutes or so that my battery backup will run my computer! Of course I'll keep you posted if that happens. (I've tried calculations that took longer than that before!)

...Updates were installed closing Mathematica..

Second effort, Sept 10, 2015

EDIT I didn't know that upgrading to Windows 10 caused my Windows update option to read, automatically install updates, so my computer restarted. I started the 4,000,000 digit run again at 8:34 PM, Thursday Sept 10, 2015. This time I made sure the deffer updates option was chosen!!

EDIT As of Sept 19,2015, the program says it will take 94 more days to complete the 4 million digit run.

EDIT As of Sept 23,2015, 9:20 PM, the program says it will take 90.9563 more days to complete the 4 million digit run.

EDIT As of Sept 27,2015, 9:20 PM, the program says it will take 86.9283 more days to complete the 4 million digit run

...Updates were installed closing Mathematica..

Third effort, Sept 28,2015

Well I started the 4 million digit search again. This time I installed the Windows 10 Fall Creators Update to try to avoid the automatic restarts! Per my habit above I will post updates in this message. The code I'm using is the same mentioned twice above that starts with

The Windows 10 Fall Creators Update restarted my machine on Thursday, Nov 16, 2017 at 5:33 pm!

Fourth effort, Dec 19, 2017

I applied some new update policies via LogMeIn, for windows to download but not apply updates. That should work!

The new run looks like

  1 : 15 : 58 pm EST | Sunday, December 17, 2017
   ...  done iter 96256 165140.1164436.

Then

  (165140.1164436) sec/(96256 iterations) (5306880 iterations)/(3600sec/hour)/(24 hour/day)

gives 105.378 day.

Entering "December 17, 2017+106 days" into Wolfram Alpha tells us the computation should be done around Monday, April 2, 2018.

My computer went to sleep sometime between 10:15:51 pm EST | Thursday, December 28, 2017 and 9:22:00 am EST | Sunday, December 31, 2017 .

Fifth effort, Jan 09, 2018

The output starts out with

"Sun 7 Jan 2018 15:14:50"

Iterations required: 5306398

 end 3455

 5306880

 done iter 0     2374.6810043.

The latest result is done iter 92160 153642.6180409.

Entering 

 (153642.6180409) sec/(92160 iterations) (5306880iterations)/(3600 sec/hour)/(24 hour/day)

gives

102.399 day or 3.4 months for the computation to run.

Here is the latest output: done iter 3048960 5.1636602440434*10^6.

N[3048960/5306880*100] says the computation is 57.453% done.

... My computer failed on 3/11/2018 at 8:30 am. I will break the program up, where it does a little at a time and saves a file with what progress has been made!

Sixth effort, Sat 14 Apr 2018

I optimized my program to calculating 4019999 digits with my new system in Mathematica v. 11.3 (not kernel version).

Posted below. On my new Xeon system, this optimized version gives V11.3 (not kernel version) a slight advantage over 11.2 (not kernel version)!

Print["Start time is ", ds = DateString[], "."];
prec = 4000000;
(**Number of required decimals.*.*)ClearSystemCache[];
T0 = SessionTime[];
expM[pre_] := 
  Module[{a, d, s, k, bb, c, end, iprec, xvals, x, pc, cores = 32(*=4*number of physical cores*), 
    tsize = 2^7, chunksize, start = 1, ll, ctab, 
    pr = Floor[1.005 pre]}, chunksize = cores*tsize;
   n = Floor[1.32 pr];
   end = Ceiling[n/chunksize];
   Print["Iterations required: ", n];
   Print["Will give ", end, 
    " time estimates, each more accurate than the previous."];
   Print["Will stop at ", end*chunksize, 
    " iterations to ensure precsion of around ", pr, 
    " decimal places."]; d = ChebyshevT[n, 3];
   {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
   iprec = Ceiling[pr/27];
   Do[xvals = Flatten[ParallelTable[Table[ll = start + j*tsize + l;
        x = N[E^(Log[ll]/(ll)), iprec];
        pc = iprec;
        While[pc < pr, pc = Min[3 pc, pr];
         x = SetPrecision[x, pc];
         y = x^ll - ll;
         x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];(**N[Exp[Log[ll]/
        ll],pr]**)x, {l, 0, tsize - 1}], {j, 0, cores - 1}, 
       Method -> "EvaluationsPerKernel" -> 16(* a power of 2 commensurate with processor strength*)]];
    ctab = ParallelTable[Table[c = b - c;
       ll = start + l - 2;
       b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
       c, {l, chunksize}], Method -> "EvaluationsPerKernel" -> 16(* a power of 2 commensurate with processor strength*)];
    s += ctab.(xvals - 1);
    start += chunksize;
    st = SessionTime[] - T0; kc = k*chunksize; 
    ti = (st)/(kc + 10^-4)*(n)/(3600)/(24);
    If[kc > 1, 
     Print[kc, " iterations done in ", N[st, 3], " seconds.", 
      " Should take ", N[ti, 2], " days or ", N[ti*24*3600, 2], 
      "s, finish ", DatePlus[ds, ti], "."]];, {k, 0, end - 1}];
   N[-s/d, pr]];
t2 = Timing[MRBtest2 = expM[prec];]; Print["Finished on ", 
 DateString[], ". Proccessor time was ", t2[[1]], " s."];
(*Print[*)MRBtest2(*]*)(*Remove (**) or enter MRBtest2 to print \
output*); Print["Enter MRBtest2 to print ", 
 Floor[Precision[
   MRBtest2]], " digits"]; Print["If you saved m3M, the difference \
between this and 3,014,991 known digits is ", N[MRBtest2 - m3M, 10]]

I'm going to play around with it an see what my patience and my computer's endurance amounts to.

Here is the result:

Start time is Sat 14 Apr 2018 14:00:33.

Iterations required: 5306398

Will give 1296 time estimates, each more accurate than the previous.

Will stop at 5308416 iterations to ensure precsion of around 4019999 decimal places.

4096 iterations done in 1.1810^4 seconds. Should take 1.810^2 days or 1.5*10^7s, finish Tue 9 Oct 2018 01:49:36.

...

2789376 iterations done in 4.21*10^6 seconds. Should take 93. days or 8.0*10^6s, finish Mon 16 Jul 2018 07:21:24.

That is 43.92 days from now (9:15:15 am EDT | Saturday, June 2, 2018).

...

My computer crashed!

Seventh! effort 27 Aug 2018

I started another 4 million digit run with my 3.7 GH overclocked up to 4.7 GH, Intel 6core and 6 kernels at 3000MH RAM, w/ 4 cores and 8 kernels of 3.6 GH at 2400MH RAM with optimized use of kernel priority.

It is ultra fast!!!!!! Instead of around 100 days it says it will be done in less than 50 days.

Here are the results thus far:

Print["Start time is ", ds = DateString[], "."];
prec = 4000000;
(**Number of required decimals.*.*)ClearSystemCache[];
T0 = SessionTime[];
expM[pre_] := 
  Module[{a, d, s, k, bb, c, end, iprec, xvals, x, pc, cores = 16(*=4*
    number of physical cores*), tsize = 2^7, chunksize, start = 1, ll,
     ctab, pr = Floor[1.005 pre]}, chunksize = cores*tsize;
   n = Floor[1.32 pr];
   end = Ceiling[n/chunksize];
   Print["Iterations required: ", n];
   Print["Will give ", end, 
    " time estimates, each more accurate than the previous."];
   Print["Will stop at ", end*chunksize, 
    " iterations to ensure precsion of around ", pr, 
    " decimal places."]; d = ChebyshevT[n, 3];
   {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
   iprec = Ceiling[pr/24768];
   Do[xvals = Flatten[Parallelize[Table[Table[ll = start + j*tsize + l;
         x = N[E^(Log[ll]/(ll)), iprec];



         pc = iprec;
         While[pc < pr/4096, pc = Min[3 pc, pr/4096];
          x = SetPrecision[x, pc];
          y = x^ll - ll;
          x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];
         (**N[Exp[Log[ll]/ll],pr/4096]**)

         x = SetPrecision[x, pr/1024];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/1024] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/1024] ll (ll - 1) 1/(3 ll t2 + t^3 z));(*N[Exp[Log[
         ll]/ll],pr/1024]*)

         x = SetPrecision[x, pr/256];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/256] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/256] ll (ll - 1) 1/(3 ll t2 + t^3 z));(*N[Exp[Log[
         ll]/ll],pr/256]*)

         x = SetPrecision[x, pr/64];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/64] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/64] ll (ll - 1) 1/(3 ll t2 + t^3 z));(**N[Exp[Log[
         ll]/ll],pr/64]**)

         x = SetPrecision[x, pr/16];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/16] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/16] ll (ll - 1) 1/(3 ll t2 + t^3 z));(**N[Exp[Log[
         ll]/ll],pr/16]**)

         x = SetPrecision[x, pr/4];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/4] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/4] ll (ll - 1) 1/(3 ll t2 + t^3 z));(**N[Exp[Log[
         ll]/ll],pr/4]**)


         x = SetPrecision[x, pr];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr] ll (ll - 1) 1/(3 ll t2 + t^3 z));(*N[Exp[Log[ll]/
         ll],pr]*)

         x, {l, 0, tsize - 1}], {j, 0, cores - 1}]]];
    ctab = ParallelTable[Table[c = b - c;
       ll = start + l - 2;
       b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
       c, {l, chunksize}], Method -> "Automatic"];
    s += ctab.(xvals - 1);
    start += chunksize;
    st = SessionTime[] - T0; kc = k*chunksize;
    ti = (st)/(kc + 10^-4)*(n)/(3600)/(24);
    Print[kc, " iterations done in ", N[st, 4], " seconds.", 
     " Should take ", N[ti, 4], " days or ", N[ti*24*3600, 4], 
     "s, finish ", DatePlus[ds, ti], "."], {k, 0, end - 1}];
   N[-s/d, pr]];
t2 = Timing[MRBtest2 = expM[prec];]; Print["Finished on ", 
 DateString[], ". Proccessor time was ", 
 t2[[1]], " s."]; Print["Actual time was ", st];
(*Print[*)MRBtest2(*]*)(*Remove (**) or enter MRBtest2 to print \
output*); Print["Enter MRBtest2 to print ", 
 Floor[Precision[
   MRBtest2]], " digits"]; Print["If you saved m3M, the difference \
between this and 3,014,991 known digits is ", N[MRBtest2 - m3M, 10]]

Start time is Mon 27 Aug 2018 17:22:11.

Iterations required: 5306398

Will give 2592 time estimates, each more accurate than the previous.

Will stop at 5308416 iterations to ensure precsion of around 4019999 decimal places.

0 iterations done in 1408. seconds. Should take 8.646*10^8 days or 7.470*10^13s, finish Sun 25 Jul 2369306 03:19:20.

2048 iterations done in 2831. seconds. Should take 84.90 days or 7.335*10^6s, finish Tue 20 Nov 2018 14:58:22.

4096 iterations done in 4261. seconds. Should take 63.90 days or 5.521*10^6s, finish Tue 30 Oct 2018 14:52:54.

6144 iterations done in 5752. seconds. Should take 57.49 days or 4.967*10^6s, finish Wed 24 Oct 2018 05:12:32.

8192 iterations done in 7237. seconds. Should take 54.26 days or 4.688*10^6s, finish Sat 20 Oct 2018 23:35:31.
...

210944 iterations done in 1.513*10^5 seconds. Should take 44.06 days or 3.807*10^6s, finish Wed 10 Oct 2018 18:53:41.

212992 iterations done in 1.528*10^5 seconds. Should take 44.06 days or 3.807*10^6s, finish Wed 10 Oct 2018 18:49:24.

215040 iterations done in 1.543*10^5 seconds. Should take 44.06 days or 3.807*10^6s, finish Wed 10 Oct 2018 18:48:16.

217088 iterations done in 1.558*10^5 seconds. Should take 44.07 days or 3.808*10^6s, finish Wed 10 Oct 2018 19:04:29.

219136 iterations done in 1.573*10^5 seconds. Should take 44.07 days or 3.808*10^6s, finish Wed 10 Oct 2018 19:07:25.


...



548864 iterations done in 3.954*10^5 seconds. Should take 44.24 days or 3.822*10^6s, finish Wed 10 Oct 2018 23:06:24.

550912 iterations done in 3.968*10^5 seconds. Should take 44.24 days or 3.822*10^6s, finish Wed 10 Oct 2018 23:08:05.

552960 iterations done in 3.983*10^5 seconds. Should take 44.24 days or 3.822*10^6s, finish Wed 10 Oct 2018 23:09:03.

...

782336 iterations done in 5.655*10^5 seconds. Should take 44.40 days or 3.836*10^6s, finish Thu 11 Oct 2018 02:52:31.

784384 iterations done in 5.671*10^5 seconds. Should take 44.40 days or 3.836*10^6s, finish Thu 11 Oct 2018 03:00:20.

786432 iterations done in 5.685*10^5 seconds. Should take 44.40 days or 3.836*10^6s, finish Thu 11 Oct 2018 02:58:41.

...
907264 iterations done in 6.568*10^5 seconds. Should take 44.46 days or 3.841*10^6s, finish Thu 11 Oct 2018 04:26:21.

909312 iterations done in 6.583*10^5 seconds. Should take 44.46 days or 3.842*10^6s, finish Thu 11 Oct 2018 04:29:41.

911360 iterations done in 6.598*10^5 seconds. Should take 44.47 days or 3.842*10^6s, finish Thu 11 Oct 2018 04:33:53.


...

1028096 iterations done in 7.454*10^5 seconds. Should take 44.53 days or 3.848*10^6s, finish Thu 11 Oct 2018 06:07:13.

1030144 iterations done in 7.470*10^5 seconds. Should take 44.53 days or 3.848*10^6s, finish Thu 11 Oct 2018 06:11:43.

1032192 iterations done in 7.485*10^5 seconds. Should take 44.54 days or 3.848*10^6s, finish Thu 11 Oct 2018 06:12:48.

A long-term power outage stopped this computation!

I so dearly want to see more digits of the MRB constant: Please, anyone, please help here!!
POSTED BY: Marvin Ray Burns
Answer

Would anyone let me know how this comes out on your computer?

Here is a link to my Nov 16, 2018 update of calculating 4 million digits of MRB: HERE.

Here is a link to my Nov 18, 2018 update of calculating 4 million digits of MRB: HERE.

Here is a link to my Nov 19, 2018 update of calculating 4 million digits of MRB: HERE,

Here is a link to my Nov 20, 2018 update of calculating 4 million digits of MRB: HERE

Here is a link to my Nov 21, 2018 update of calculating 4 million digits of MRB: HERE

Here is a link to my Nov 22, 2018 update of calculating 4 million digits of MRB: HERE.

Here is a link to my Nov 24, 2018 update of calculating 4 million digits of MRB: HERE.

Here is a link to my Nov 27, 2018 update of calculating 4 million digits of MRB: HERE.

Here is a link to my Dec 05, 2018 update of calculating 4 million digits of MRB: HERE.

Here is a link to my Dec 10, 2018 update of calculating 4 million digits of MRB: HERE.

Last update of 2018

5:37:20 pm EST | Monday, December 31, 2018 (49.04 days from start time), 75.2315 percent completed . 3,024,300 digits computed.3993600 iterations done in 4.23810^6 seconds. Should take 65.18 days or 5.63110^6s, finish Wed 16 Jan 2019 20:57:47. (16.64 days from now).

Probably the last update

12:47:20 am EST | Monday, January 14, 2019 (62.3495 days from start time), 95.679 percent completed . 3,846,295 digits computed. 5079040 iterations done in 5.38710^6 seconds. Should take 65.15 days or 5.62910^6s, finish Wed 16 Jan 2019 20:13:07. (2.81 days from now).
POSTED BY: Marvin Ray Burns
Answer

Finished on Wed 16 Jan 2019 19:55:20, I computed over 4 million digits of the MRB constant!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!..... It took 65.13 days with a processor time of 25.17 days.On a 3.7 GH overclocked up to 4.7 GH on all cores Intel 6 core computer with 3000 MHz RAM.

See attached notebook.

Watch my reaction here.

Attachments:
4 million 11 2018.nb
POSTED BY: Marvin Ray Burns
Answer

Attempts at a 5,000,000 digit calculation

First effort: Tue 22 May 2018

I started a 5,000,000 digits calculation on my 4 core ASUS Intel Core i7 7700 Desktop computer K31CD-DS71, 32 GB DDR4 ram running at 2400 MHz in Windows 10 Pro using a version of my Burns-Crandall hybrid code supercharged with 4 iterations of my 4th order convergence subroutine.

See April 2018 5 million digits try.nb for exact code and results of the incomplete computation.

Here are some of the results:

Start time is Tue 22 May 2018 01:02:37.
Iterations required: 6632998
Will give 3239 time estimates, each more accurate than the previous.
Will stop at 6633472 iterations to ensure precision of around 5024999 decimal places.
0 iterations done in 3301. seconds. Should take 2.534*10^9 days or 2.189*10^14s, finish Mon 11 Nov 6939709 17:32:23.
2048 iterations done in 6503. seconds. Should take 243.8 days or 2.106*10^7s, finish Sun 20 Jan 2019 19:11:08.
4096 iterations done in 9721. seconds. Should take 182.2 days or 1.574*10^7s, finish Tue 20 Nov 2018 05:57:47.
6144 iterations done in 1.295*10^4 seconds. Should take 161.8 days or 1.398*10^7s, finish Tue 30 Oct 2018 21:15:01.
8192 iterations done in 1.623*10^4 seconds. Should take 152.1 days or 1.314*10^7s, finish Sun 21 Oct 2018 03:44:45.
10240 iterations done in 1.964*10^4 seconds. Should take 147.3 days or 1.272*10^7s, finish Tue 16 Oct 2018 07:34:30.
12288 iterations done in 2.312*10^4 seconds. Should take 144.4 days or 1.248*10^7s, finish Sat 13 Oct 2018 11:43:37.
14336 iterations done in 2.667*10^4 seconds. Should take 142.8 days or 1.234*10^7s, finish Thu 11 Oct 2018 20:49:04.
16384 iterations done in 2.998*10^4 seconds. Should take 140.5 days or 1.214*10^7s, finish Tue 9 Oct 2018 12:13:43.
18432 iterations done in 3.326*10^4 seconds. Should take 138.5 days or 1.197*10^7s, finish Sun 7 Oct 2018 13:48:39.

I felt this was going too slow so I got more computing power!!!

Second effort: Sun 20 Jan 2019 21:06:58

I've made a lot better progress with the following!!!

6 kernels, 3.7 GH overclocked up to 4.7 GH, Intel 6core 3000MH RAM, w/ 8 kernels of 3.6 GH 4 core, 2400MH RAM (with optimized use of kernel priority)

Needs["SubKernels`LocalKernels`"]
Block[{$mathkernel = $mathkernel <> " -threadpriority=2"}, 
 LaunchKernels[]](*optimized use of kernel priority for master computer being the fastest one*)

Out[3]= {"KernelObject"[1, "burns"], "KernelObject"[2, "burns"], "KernelObject"[3, "burns"], "KernelObject"[4, "burns"], "KernelObject"[5, "burns"], "KernelObject"[6, "burns"], "KernelObject"[7, "burns"], "KernelObject"[8, "burns"], "KernelObject"[9, "local"], "KernelObject"[10, "local"], "KernelObject"[11, "local"], "KernelObject"[12, "local"], "KernelObject"[13, "local"], "KernelObject"[14, "local"]}

Print["Start time is ", ds = DateString[], "."];
prec = 5000000;
(**Number of required decimals.*.*)ClearSystemCache[];
T0 = SessionTime[];
expM[pre_] := 
  Module[{a, d, s, k, bb, c, end, iprec, xvals, x, pc, cores = 16(*=4*
    number of physical cores*), tsize = 2^7, chunksize, start = 1, ll,
     ctab, pr = Floor[1.005 pre]}, chunksize = cores*tsize;
   n = Floor[1.32 pr];
   end = Ceiling[n/chunksize];
   Print["Iterations required: ", n];
   Print["Will give ", end, 
    " time estimates, each more accurate than the previous."];
   Print["Will stop at ", end*chunksize, 
    " iterations to ensure precsion of around ", pr, 
    " decimal places."]; d = ChebyshevT[n, 3];
   {b, c, s} = {SetPrecision[-1, 1.1*n], -d, 0};
   iprec = Ceiling[pr/24768];
   Do[xvals = Flatten[Parallelize[Table[Table[ll = start + j*tsize + l;
         x = N[E^(Log[ll]/(ll)), iprec];
         pc = iprec;
         While[pc < pr/4096, pc = Min[3 pc, pr/4096];
          x = SetPrecision[x, pc];
          y = x^ll - ll;
          x = x (1 - 2 y/((ll + 1) y + 2 ll ll));];
         (**N[Exp[Log[ll]/ll],pr/4096]**)x = SetPrecision[x, pr/1024];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/1024] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/1024] ll (ll - 1) 1/(3 ll t2 + t^3 z));(*N[Exp[Log[
         ll]/ll],pr/1024]*)x = SetPrecision[x, pr/256];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/256] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/256] ll (ll - 1) 1/(3 ll t2 + t^3 z));(*N[Exp[Log[
         ll]/ll],pr/256]*)x = SetPrecision[x, pr/64];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/64] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/64] ll (ll - 1) 1/(3 ll t2 + t^3 z));(**N[Exp[Log[
         ll]/ll],pr/64]**)x = SetPrecision[x, pr/16];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/16] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/16] ll (ll - 1) 1/(3 ll t2 + t^3 z));(**N[Exp[Log[
         ll]/ll],pr/16]**)x = SetPrecision[x, pr/4];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/4] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/4] ll (ll - 1) 1/(3 ll t2 + t^3 z));(**N[Exp[Log[
         ll]/ll],pr/4]**)x = SetPrecision[x, pr];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr] ll (ll - 1) 1/(3 ll t2 + t^3 z));(*N[Exp[Log[ll]/
         ll],pr]*)x, {l, 0, tsize - 1}], {j, 0, cores - 1}]]];
    ctab = ParallelTable[Table[c = b - c;
       ll = start + l - 2;
       b *= 2 (ll + n) (ll - n)/((ll + 1) (2 ll + 1));
       c, {l, chunksize}], Method -> "Automatic"];
    s += ctab.(xvals - 1);
    start += chunksize;
    st = SessionTime[] - T0; kc = k*chunksize;
    ti = (st)/(kc + 10^-4)*(n)/(3600)/(24);
    Print[kc, " iterations done in ", N[st, 4], " seconds.", 
     " Should take ", N[ti, 4], " days or ", N[ti*24*3600, 4], 
     "s. Finish ", DatePlus[ds, ti], "."]; 
    Print[Text[
      Style[{IntegerPart[kc/(end*chunksize)*100], "% done."}, 
       FontSize -> Large]]], {k, 0, end - 1}];
   N[-s/d, pr]];
t2 = Timing[MRBtest2 = expM[prec];]; Print["Finished on ", 
 DateString[], ". Proccessor time was ", 
 t2[[1]], " s."]; Print["Actual time was ", st];
(*Print[*)MRBtest2(*]*)(*Remove (**) or enter MRBtest2 to print \
output*); Print["Enter MRBtest2 to print ", 
 Floor[Precision[
   MRBtest2]], " digits"]; Print["If you saved m3M, the difference \
between this and 4,000,000 known digits is ", N[MRBtest2 - m4M, 10]]

  Start time is Sun 20 Jan 2019 21:06:58.

Iterations required: 6632998

Will give 3239 time estimates, each more accurate than the previous.

Will stop at 6633472 iterations to ensure precsion of around 5024999 decimal places.

0 iterations done in 1893. seconds. Should take 1.454*10^9 days or 1.256*10^14s. Finish Sun 16 Apr 3981600 03:48:25.

{0,% done.}

2048 iterations done in 3801. seconds. Should take 142.5 days or 1.231*10^7s. Finish Wed 12 Jun 2019 08:24:54.

{0,% done.}

4096 iterations done in 5774. seconds. Should take 108.2 days or 9.351*10^6s. Finish Thu 9 May 2019 02:36:43.

{0,% done.}

6144 iterations done in 7744. seconds. Should take 96.76 days or 8.360*10^6s. Finish Sat 27 Apr 2019 15:27:12.

{0,% done.}

8192 iterations done in 9655. seconds. Should take 90.48 days or 7.818*10^6s. Finish Sun 21 Apr 2019 08:42:56.

{0,% done.}

10240 iterations done in 1.160*10^4 seconds. Should take 86.95 days or 7.512*10^6s. Finish Wed 17 Apr 2019 19:49:38.

{0,% done.}

12288 iterations done in 1.351*10^4 seconds. Should take 84.43 days or 7.295*10^6s. Finish Mon 15 Apr 2019 07:29:22.

{0,% done.}

14336 iterations done in 1.544*10^4 seconds. Should take 82.67 days or 7.143*10^6s. Finish Sat 13 Apr 2019 13:14:41.

{0,% done.}

16384 iterations done in 1.737*10^4 seconds. Should take 81.39 days or 7.032*10^6s. Finish Fri 12 Apr 2019 06:34:15.

{0,% done.}

18432 iterations done in 1.931*10^4 seconds. Should take 80.42 days or 6.948*10^6s. Finish Thu 11 Apr 2019 07:06:55.

{0,% done.}

![enter image description here][1]

See 5 million 01 2019.nb for full progress. The cluster is processing 1.024 5,000,000 digit iterations of n^(1/n) per second. It is completing 1% every 18 hours, 3 minutes and 36 seconds.

Well, the power went out here in INDY, again.

My 6 year old battery backup is becoming worthless on handling 2 computers for very long. I will try to build some better household infrastructure. Plus, there is still the possibility of breaking my program up so it does smaller portions of the computation at a time. But, I haven't figured how to do that yet!

I will let you know when I'm ready to try again. It took over 8 tries for 4 million digits, so I'm not heartbroken that this second try of 5 million digit didn't work out.

Third effort: 5/3/2019

Using the MRB constant supercomputer described in the next post, we now have much better progress, as shown in 5 million 05 2019.nb:

   Start time is Sun 12 May 2019 10:23:26.

Iterations required: 6632998

Will give 3239 time estimates, each more accurate than the previous.

Will stop at 6633472 iterations to ensure precsion of around 5024999 decimal places.

0 iterations done in 1482. seconds. Should take 1.138*10^9 days or 9.829*10^13s. Finish Thu 23 Jan 3116775 18:57:34.

{0,% done.}

2048 iterations done in 2986. seconds. Should take 111.9 days or 9.672*10^6s. Finish Sun 1 Sep 2019 08:57:12.

{0,% done.}

4096 iterations done in 4503. seconds. Should take 84.39 days or 7.292*10^6s. Finish Sun 4 Aug 2019 19:49:39.

{0,% done.}

6144 iterations done in 6036. seconds. Should take 75.42 days or 6.516*10^6s. Finish Fri 26 Jul 2019 20:21:33.

{0,% done.}

8192 iterations done in 7562. seconds. Should take 70.87 days or 6.123*10^6s. Finish Mon 22 Jul 2019 07:13:15.

{0,% done.}

10240 iterations done in 9106. seconds. Should take 68.27 days or 5.898*10^6s. Finish Fri 19 Jul 2019 16:47:37.

{0,% done.}

12288 iterations done in 1.064*10^4 seconds. Should take 66.46 days or 5.743*10^6s. Finish Wed 17 Jul 2019 21:32:38.

{0,% done.}

14336 iterations done in 1.218*10^4 seconds. Should take 65.24 days or 5.636*10^6s. Finish Tue 16 Jul 2019 16:03:55.

{0,% done.}

16384 iterations done in 1.372*10^4 seconds. Should take 64.31 days or 5.556*10^6s. Finish Mon 15 Jul 2019 17:46:42.

{0,% done.}

18432 iterations done in 1.527*10^4 seconds. Should take 63.62 days or 5.496*10^6s. Finish Mon 15 Jul 2019 01:09:08.

{0,% done.}

Wish me luck!

Here is the latest output:

131072 iterations done in 1.026*10^5 seconds. Should take 60.11 days or 5.193*10^6s. Finish Thu 11 Jul 2019 13:00:35.

{1,% done.}

133120 iterations done in 1.042*10^5 seconds. Should take 60.10 days or 5.193*10^6s. Finish Thu 11 Jul 2019 12:53:13.

{2,% done.}

135168 iterations done in 1.058*10^5 seconds. Should take 60.09 days or 5.192*10^6s. Finish Thu 11 Jul 2019 12:35:29.

{2,% done.}

137216 iterations done in 1.074*10^5 seconds. Should take 60.09 days or 5.192*10^6s. Finish Thu 11 Jul 2019 12:29:39.

{2,% done.}

139264 iterations done in 1.090*10^5 seconds. Should take 60.08 days or 5.191*10^6s. Finish Thu 11 Jul 2019 12:15:25.

{2,% done.}.

Followed by

196608 iterations done in 1.543*10^5 seconds. Should take 60.27 days or 5.207*10^6s. Finish Thu 11 Jul 2019 16:48:37.

{2,% done.}

198656 iterations done in 1.559*10^5 seconds. Should take 60.26 days or 5.207*10^6s. Finish Thu 11 Jul 2019 16:40:38.

{2,% done.}

200704 iterations done in 1.575*10^5 seconds. Should take 60.25 days or 5.206*10^6s. Finish Thu 11 Jul 2019 16:26:21.

{3,% done.}

Followed by

264192 iterations done in 2.075*10^5 seconds. Should take 60.30 days or 5.210*10^6s. Finish Thu 11 Jul 2019 17:34:10.

{3,% done.}

266240 iterations done in 2.091*10^5 seconds. Should take 60.29 days or 5.209*10^6s. Finish Thu 11 Jul 2019 17:24:24.

{4,% done.}

Well, for an unknown reason Mathematica 11.3 asked whether I wanted to close the program and had not displayed any new information since my last checking it 18 hours ago.

Fourth effort: Thu 16 May 2019 15:39:40

I restarted the 5 million digit computation. For an unknown reason it is a little slower this time, but what's important is that it gets completed! See attached B 5 million 05 2019.nb.

0 iterations done in 1520. seconds. Should take 1.167*10^9 days or 1.008*10^14s. Finish Sun 13 Dec 3197801 13:09:30.

{0,% done.}

2048 iterations done in 3103. seconds. Should take 116.3 days or 1.005*10^7s. Finish Mon 9 Sep 2019 23:32:26.

{0,% done.}

4096 iterations done in 4697. seconds. Should take 88.04 days or 7.607*10^6s. Finish Mon 12 Aug 2019 16:42:17.

{0,% done.}

6144 iterations done in 6298. seconds. Should take 78.69 days or 6.799*10^6s. Finish Sat 3 Aug 2019 08:14:25.

{0,% done.}

8192 iterations done in 7899. seconds. Should take 74.02 days or 6.396*10^6s. Finish Mon 29 Jul 2019 16:13:37.

{0,% done.}

10240 iterations done in 9465. seconds. Should take 70.96 days or 6.131*10^6s. Finish Fri 26 Jul 2019 14:47:42.

{0,% done.}

12288 iterations done in 1.102*10^4 seconds. Should take 68.87 days or 5.950*10^6s. Finish Wed 24 Jul 2019 12:33:43.

{0,% done.}

14336 iterations done in 1.260*10^4 seconds. Should take 67.48 days or 5.830*10^6s. Finish Tue 23 Jul 2019 03:13:52.

{0,% done.}

16384 iterations done in 1.418*10^4 seconds. Should take 66.44 days or 5.740*10^6s. Finish Mon 22 Jul 2019 02:08:54.

{0,% done.}

6:56:20 pm EDT | Sunday, May 19, 2019:

{4,% done.}

331776 iterations done in 2.710*10^5 seconds. Should take 62.71 days or 5.418*10^6s. Finish Thu 18 Jul 2019 08:46:13.
{5,% done.}

9:53:00 am EDT | Monday, May 20, 2019:

 397312 iterations done in 3.248*10^5 seconds. Should take 62.76 days or 5.423*10^6s. Finish Thu 18 Jul 2019 09:56:53.

 {5,% done.}

 399360 iterations done in 3.265*10^5 seconds. Should take 62.77 days or 5.423*10^6s. Finish Thu 18 Jul 2019 10:02:57.

 {6,% done.}

1:55:40 am EDT | Tuesday, May 21, 2019:

462848 iterations done in 3.792*10^5 seconds. Should take 62.90 days or 5.435*10^6s. Finish Thu 18 Jul 2019 13:17:05.

{6,% done.}

464896 iterations done in 3.809*10^5 seconds. Should take 62.90 days or 5.435*10^6s. Finish Thu 18 Jul 2019 13:15:16.

{7,% done.}

It is completing 1% every 15 hours, 35 minutes and 0 seconds.

5:04:40 pm EDT | Tuesday, May 21, 2019;

530432 iterations done in 4.354*10^5 seconds. Should take 63.01 days or 5.444*10^6s. Finish Thu 18 Jul 2019 16:00:32.

{7,% done.}

532480 iterations done in 4.371*10^5 seconds. Should take 63.02 days or 5.445*10^6s. Finish Thu 18 Jul 2019 16:02:22.

{8,% done.}

8:04:40 am EDT | Wednesday, May 22, 2019

595968 iterations done in 4.894*10^5 seconds. Should take 63.05 days or 5.447*10^6s. Finish Thu 18 Jul 2019 16:49:21.

{8,% done.}

598016 iterations done in 4.911*10^5 seconds. Should take 63.04 days or 5.447*10^6s. Finish Thu 18 Jul 2019 16:42:25.

{9,% done.}

It is completing 1% every 15 hours, 9 minutes and 27 seconds.

Should reach 10% at 11:14:07 pm EDT | Wednesday, May 22, 2019.

11:14:40 pm EDT | Wednesday, May 22, 2019:

{9,% done.}

663552 iterations done in 5.457*10^5 seconds. Should take 63.14 days or 5.455*10^6s. Finish Thu 18 Jul 2019 18:57:53.

{10,% done.}

It is completing 1% every 15 hours, 9 minutes and 30 seconds.

Should reach 11% at 2:23:37 pm EDT | Thursday, May 23, 2019.

2:59:40 pm EDT | Thursday, May 23, 2019:

{10,% done.}

731136 iterations done in 6.024*10^5 seconds. Should take 63.25 days or 5.465*10^6s. Finish Thu 18 Jul 2019 21:45:37.

{11,% done.}

It is completing 1% every 15 hours, 12 minutes and 44 seconds.

Should reach 12% at 6:12:24 am EDT | Friday, May 24, 2019.

6:21:20 am EDT | Friday, May 24, 2019

{11,% done.}

796672 iterations done in 6.577*10^5 seconds. Should take 63.38 days or 5.476*10^6s. Finish Fri 19 Jul 2019 00:45:30.

{12,% done.}

It is completing 1% every 15 hours, 13 minutes and 28 seconds.

Should reach 13% at 9:34:48 pm EDT | Friday, May 24, 2019.

10:16:20 pm EDT | Friday, May 24, 2019

{12,% done.}

864256 iterations done in 7.150*10^5 seconds. Should take 63.51 days or 5.487*10^6s. Finish Fri 19 Jul 2019 03:53:13.

{13,% done.}

It is completing 1% every 15 hours, 55 minutes and 0 seconds.

Should reach 14% at 2:11:20 pm EDT | Saturday, May 25, 2019.

{13,% done.}

929792 iterations done in 7.701*10^5 seconds. Should take 63.58 days or 5.494*10^6s. Finish Fri 19 Jul 2019 05:38:05.

{14,% done.}

It is completing 1% every 15 hours, 18 minutes and 20 seconds.

Should reach 15% at 5:29:40 am EDT | Sunday, May 26, 2019.

4:59:00 am EDT | Sunday, May 26, 2019

{14,% done.}

995328 iterations done in 8.256*10^5 seconds. Should take 63.68 days or 5.502*10^6s. Finish Fri 19 Jul 2019 08:01:57.

{15,% done.}

It is completing 1% every 15 hours, 17 minutes and 20 seconds.

Should reach 16% at 8:16:19 pm EDT | Sunday, May 26, 2019.

8:57:20 pm EDT | Saturday, May 26, 2019

{15,% done.}

1062912 iterations done in 8.831*10^5 seconds. Should take 63.79 days or 5.511*10^6s. Finish Fri 19 Jul 2019 10:32:05.

{16,% done.}

It is completing 1% every 15 hours 19 minutes 53.75 seconds.

Should reach 17% at 12:17:13 pm EDT | Monday, May 27, 2019.

12:25:40 pm EDT | Monday, May 27, 2019

{16,% done.}

1128448 iterations done in 9.388*10^5 seconds. Should take 63.87 days or 5.518*10^6s. Finish Fri 19 Jul 2019 12:28:56.

{17,% done.}

It is completing 1% every 15 hours 28 minutes 19.92 seconds.

Should reach 18% at 3:54:00 am EDT | Tuesday, May 28, 2019.

4:29:00 am EDT | Tuesday, May 28, 2019

{17,% done.}

1196032 iterations done in 9.966*10^5 seconds. Should take 63.97 days or 5.527*10^6s. Finish Fri 19 Jul 2019 14:52:46.

{18,% done.}

It is completing 1% every 16 hours 3 minutes 20 seconds.

Should reach 19% at 8:32:20 pm EDT | Tuesday, May 28, 2019.

8:09:40 pm EDT | Tuesday, May 28, 2019

{18,% done.}

1261568 iterations done in 1.053*10^6 seconds. Should take 64.07 days or 5.536*10^6s. Finish Fri 19 Jul 2019 17:23:49.

{19,% done.}

It is completing 1% every 15 hours 40 minutes 0 seconds.

Should reach 20% at 11:49:40 am EDT | Sunday, May 29, 2019.

11:42:20 am EDT | Wednesday, May 29, 2019

{19,% done.}

1327104 iterations done in 1.109*10^6 seconds. Should take 64.17 days or 5.544*10^6s. Finish Fri 19 Jul 2019 19:42:37.

{20,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 21% at 3:15:40 am EDT | Thursday, May 30, 2019.

4:05:40 am EDT | Thursday, May 30, 2019

{20,% done.}

1394688 iterations done in 1.168*10^6 seconds. Should take 64.27 days or 5.553*10^6s. Finish Fri 19 Jul 2019 22:14:03.

{21,% done.}

It is completing 1% every 16 hours 23 minutes 20 seconds.

Should reach 22% at 8:29:00 pm EDT | Thursday, May 30, 2019.

7:39:00 pm EDT | Thursday, May 30, 2019

{21,% done.}

1460224 iterations done in 1.224*10^6 seconds. Should take 64.37 days or 5.562*10^6s. Finish Sat 20 Jul 2019 00:37:43.

{22,% done.}

It is completing 1% every 16 hours 6 minutes 40 seconds.

Should reach 23% at 11:12:20 am EDT | Friday, May 31, 2019.

11:45:40 am EDT | Friday, May 31, 2019

{22,% done.}

1525760 iterations done in 1.282*10^6 seconds. Should take 64.48 days or 5.571*10^6s. Finish Sat 20 Jul 2019 03:12:08.

{23,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 24% at 3:52:20 am EDT | Saturday, June 1, 2019.

4:09:00 am EDT | Saturday, June 1, 2019

{23,% done.}

1593344 iterations done in 1.341*10^6 seconds. Should take 64.61 days or 5.582*10^6s. Finish Sat 20 Jul 2019 06:17:52.

{24,% done.}

It is completing 1% every 16 hours 23 minutes 20.04 seconds.

Should reach 25% at 8:32:20 pm EDT | Saturday, June 1, 2019.

8:16:20 pm EDT | Saturday, June 1, 2019.

{24,% done.}

1658880 iterations done in 1.399*10^6 seconds. Should take 64.74 days or 5.594*10^6s. Finish Sat 20 Jul 2019 09:28:57.

{25,% done.}

It is completing 1% every 16 hours 11 minutes 40 seconds.

Should reach 26% at 12:39:00 pm EDT | Sunday, June 2, 2019.

12:39:40 pm EDT | Sunday, June 2, 2019

{25,% done.}

1726464 iterations done in 1.458*10^6 seconds. Should take 64.85 days or 5.603*10^6s. Finish Sat 20 Jul 2019 12:08:42.

{26,% done.}

It is completing 1% every 16 hours 23 minutes 20.04 seconds.

Should reach 27% at 5:03:00 am EDT | Monday, June 3, 2019.

3:56:20 am EDT | Monday, June 3, 2019

{26,% done.}

1792000 iterations done in 1.513*10^6 seconds. Should take 64.82 days or 5.600*10^6s. Finish Sat 20 Jul 2019 11:19:41.

{27,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 28% at 7:13:00 pm EDT | Monday, June 3, 2019.

7:13:00 pm EDT | Monday, June 3, 2019

{27,% done.}

1857536 iterations done in 1.568*10^6 seconds. Should take 64.79 days or 5.598*10^6s. Finish Sat 20 Jul 2019 10:36:19.

{28,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 29% at 10:29:40 am EDT | Tuesday, June 4, 2019.

10:46:20 am EDT | Tuesday, June 4, 2019

{28,% done.}

1925120 iterations done in 1.624*10^6 seconds. Should take 64.76 days or 5.595*10^6s. Finish Sat 20 Jul 2019 09:55:36.

{29,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 30% at 2:19:40 am EDT | Wednesday, June 5, 2019.

2:03:00 am EDT | Wednesday, June 5, 2019

{29,% done.}

1990656 iterations done in 1.679*10^6 seconds. Should take 64.74 days or 5.594*10^6s. Finish Sat 20 Jul 2019 09:26:48.

{30,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 31% at 5:19:40 pm EDT | Wednesday, June 5, 2019.

5:53:00 pm EDT | Wednesday, June 5, 2019

{30,% done.}

2058240 iterations done in 1.736*10^6 seconds. Should take 64.73 days or 5.593*10^6s. Finish Sat 20 Jul 2019 09:17:16.

{31,% done.}

It is completing 1% every 15 hours 50 minutes 0 seconds.

Should reach 32% at 9:43:00 am EDT | Thursday, June 6, 2019.

9:09:40 am EDT | Thursday, June 6, 2019

{31,% done.}

2123776 iterations done in 1.791*10^6 seconds. Should take 64.73 days or 5.592*10^6s. Finish Sat 20 Jul 2019 09:07:00.

{32,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 33% at 12:26:20 am EDT | Friday, June 7, 2019.

12:09:40 am EDT | Friday, June 7, 2019

{32,% done.}

2189312 iterations done in 1.845*10^6 seconds. Should take 64.69 days or 5.589*10^6s. Finish Sat 20 Jul 2019 08:10:40.

{33,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 34% at 3:09:40 pm EDT | Friday, June 7, 2019.

3:43:00 pm EDT | Friday, June 7, 2019

{33,% done.}

2256896 iterations done in 1.901*10^6 seconds. Should take 64.66 days or 5.586*10^6s. Finish Sat 20 Jul 2019 07:24:55.

{34,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 35% at 7:16:20 am EDT | Saturday, June 8, 2019.

6:43:00 am EDT | Saturday, June 8, 2019.

{34,% done.}

2322432 iterations done in 1.955*10^6 seconds. Should take 64.63 days or 5.584*10^6s. Finish Sat 20 Jul 2019 06:39:41.

{35,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 36% at 10:16:20 pm EDT | Saturday, June 8, 2019.

10:16:20 pm EDT | Saturday, June 8, 2019

{35,% done.}

2390016 iterations done in 2.011*10^6 seconds. Should take 64.59 days or 5.581*10^6s. Finish Sat 20 Jul 2019 05:55:37.

{36,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 37% at 1:49:40 pm EDT | Sunday, June 9, 2019

1:16:20 pm EDT | Sunday, June 9, 2019

{36,% done.}

2455552 iterations done in 2.065*10^6 seconds. Should take 64.56 days or 5.578*10^6s. Finish Sat 20 Jul 2019 05:09:04.

{37,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 38% at 4:16:20 am EDT | Monday, June 10, 2019.

4:33:00 am EDT | Monday, June 10, 2019

{37,% done.}

2521088 iterations done in 2.120*10^6 seconds. Should take 64.54 days or 5.576*10^6s. Finish Sat 20 Jul 2019 04:41:06.

{38,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 39% at 7:49:40 pm EDT | Monday, June 10, 2019.

8:06:20 pm EDT | Monday, June 10, 2019

{38,% done.}

2588672 iterations done in 2.176*10^6 seconds. Should take 64.53 days or 5.576*10^6s. Finish Sat 20 Jul 2019 04:26:31.

{39,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 40% at 11:39:40 am EDT | Tuesday, June 11, 2019.

11:06:20 am EDT | Tuesday, June 11, 2019

{39,% done.}

2654208 iterations done in 2.230*10^6 seconds. Should take 64.51 days or 5.574*10^6s. Finish Sat 20 Jul 2019 03:59:36.

{40,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 41% at 2:06:20 am EDT | Wednesday, June 12, 2019.

2:23:00 am EDT | Wednesday, June 12, 2019

{40,% done.}

2719744 iterations done in 2.285*10^6 seconds. Should take 64.49 days or 5.572*10^6s. Finish Sat 20 Jul 2019 03:21:47.

{41,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 42% at 5:39:40 pm EDT | Wednesday, June 12, 2019.

5:56:20 pm EDT | Wednesday, June 12, 2019

{41,% done.}

2787328 iterations done in 2.341*10^6 seconds. Should take 64.47 days or 5.570*10^6s. Finish Sat 20 Jul 2019 02:58:29.

{42,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 43% at 9:29:40 am EDT | Thursday, June 13, 2019.

9:13:00 am EDT | Thursday, June 13, 2019

{42,% done.}

2852864 iterations done in 2.396*10^6 seconds. Should take 64.47 days or 5.570*10^6s. Finish Sat 20 Jul 2019 02:52:38.

{43,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 44% at 12:29:40 am EDT | Friday, June 14, 2019.

12:46:20 am EDT | Friday, June 14, 2019

{43,% done.}

2920448 iterations done in 2.452*10^6 seconds. Should take 64.46 days or 5.569*10^6s. Finish Sat 20 Jul 2019 02:36:26.

{44,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 45% at 4:19:40 pm EDT | Friday, June 14, 2019.

4:03:00 pm EDT | Friday, June 14, 2019

{44,% done.}

2985984 iterations done in 2.507*10^6 seconds. Should take 64.45 days or 5.568*10^6s. Finish Sat 20 Jul 2019 02:23:55.

{45,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 46% at 7:19:40 am EDT | Saturday, June 15, 2019.

7:19:40 am EDT | Saturday, June 15, 2019

{45,% done.}

3051520 iterations done in 2.562*10^6 seconds. Should take 64.44 days or 5.568*10^6s. Finish Sat 20 Jul 2019 02:20:15.

{46,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 47% at 9:13:00 pm EDT | Saturday, June 15, 2019.

11:09:40 pm EDT | Saturday, June 15, 2019

{46,% done.}

3119104 iterations done in 2.619*10^6 seconds. Should take 64.45 days or 5.569*10^6s. Finish Sat 20 Jul 2019 02:30:47.

{47,% done.}

It is completing 1% every 15 hours 50 minutes 0 seconds.

Should reach 48% at 2:59:40 pm EDT | Sunday, June 16, 2019.

2:09:40 pm EDT | Sunday, June 16, 2019

{47,% done.}

3184640 iterations done in 2.673*10^6 seconds. Should take 64.44 days or 5.568*10^6s. Finish Sat 20 Jul 2019 02:13:47.

{48,% done.}

It is completing 1% every 15 hours 50 minutes 0 seconds.

Should reach 49% at 5:09:40 am EDT | Monday, June 17, 2019.

5:43:00 am CDT | Monday, June 17, 2019

{48,% done.}

3252224 iterations done in 2.729*10^6 seconds. Should take 64.42 days or 5.566*10^6s. Finish Sat 20 Jul 2019 01:41:30.

{49,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 50% at 9:16:20 pm CDT | Monday, June 17, 2019

8:43:00 pm EDT | Monday, June 17, 2019

{49,% done.}

3317760 iterations done in 2.783*10^6 seconds. Should take 64.40 days or 5.564*10^6s. Finish Sat 20 Jul 2019 01:15:36.

{50,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 51% at 11:43:00 am EDT | Tuesday, June 18, 2019.

11:03:00 pm EDT | Tuesday, June 18, 2019 I updated B 5 million 05 2019.nb to 50% (almost 51%) complete.

11:59:40 am CDT | Tuesday, June 18, 2019

{50,% done.}

3383296 iterations done in 2.838*10^6 seconds. Should take 64.39 days or 5.564*10^6s. Finish Sat 20 Jul 2019 01:05:58.

{51,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 52% at 3:16:20 am CDT | Wednesday, June 19, 2019.

3:33:00 am EDT | Wednesday, June 19, 2019

{51,% done.}

3450880 iterations done in 2.894*10^6 seconds. Should take 64.38 days or 5.562*10^6s. Finish Sat 20 Jul 2019 00:47:09.

{52,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 53% at 7:06:20 pm EDT | Wednesday, June 19, 2019

6:33:00 pm EDT | Wednesday, June 19, 2019

{52,% done.}

3516416 iterations done in 2.948*10^6 seconds. Should take 64.37 days or 5.562*10^6s. Finish Sat 20 Jul 2019 00:32:27.

{53,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 54% at 9:33:00 am EDT | Thursday, June 20, 2019.

10:23:00 am EDT | Thursday, June 20, 2019

{53,% done.}

3584000 iterations done in 3.005*10^6 seconds. Should take 64.36 days or 5.561*10^6s. Finish Sat 20 Jul 2019 00:23:59.

{54,% done.}

It is completing 1% every 15 hours 50 minutes 0 seconds.

Should reach 55% at 2:13:00 am EDT | Friday, June 21, 2019.

1:39:40 am EDT | Friday, June 21, 2019

{54,% done.}

3649536 iterations done in 3.060*10^6 seconds. Should take 64.36 days or 5.561*10^6s. Finish Sat 20 Jul 2019 00:25:15.

{55,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 56% at 4:56:20 pm EDT | Friday, June 21, 2019.

4:39:40 pm EDT | Friday, June 21, 2019

{55,% done.}

3715072 iterations done in 3.114*10^6 seconds. Should take 64.35 days or 5.560*10^6s. Finish Sat 20 Jul 2019 00:07:55.

{56,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 57% at 7:39:40 am EDT | Saturday, June 22, 2019.

8:13:00 am EDT | Saturday, June 22, 2019

{56,% done.}

3782656 iterations done in 3.170*10^6 seconds. Should take 64.34 days or 5.559*10^6s. Finish Fri 19 Jul 2019 23:53:35.

{57,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 58% at 11:46:20 pm EDT | Saturday, June 22, 2019.

11:29:40 pm EDT | Saturday, June 22, 2019

{57,% done.}

3848192 iterations done in 3.225*10^6 seconds. Should take 64.34 days or 5.559*10^6s. Finish Fri 19 Jul 2019 23:44:51.

{58,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 59% at 2:46:20 pm EDT | Sunday, June 23, 2019.

3:19:40 pm EDT | Sunday, June 23, 2019

{58,% done.}

3915776 iterations done in 3.282*10^6 seconds. Should take 64.34 days or 5.559*10^6s. Finish Fri 19 Jul 2019 23:49:55.

{59,% done.}

It is completing 1% every 15 hours 50 minutes 0 seconds.

Should reach 60% at 7:09:40 am EDT | Monday, June 24, 2019.

6:19:40 am EDT | Monday, June 24, 2019

{59,% done.}

3981312 iterations done in 3.336*10^6 seconds. Should take 64.34 days or 5.559*10^6s. Finish Fri 19 Jul 2019 23:44:28.

{60,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 61% at 9:19:40 pm EDT | Monday, June 24, 2019.

9:53:00 pm EDT | Monday, June 24, 2019

{60,% done.}

4046848 iterations done in 3.392*10^6 seconds. Should take 64.34 days or 5.559*10^6s. Finish Fri 19 Jul 2019 23:47:44.

{61,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 62% at 1:26:20 pm EDT | Tuesday, June 25, 2019.

1:26:20 pm EDT | Tuesday, June 25, 2019

{61,% done.}

4114432 iterations done in 3.448*10^6 seconds. Should take 64.35 days or 5.559*10^6s. Finish Fri 19 Jul 2019 23:56:45.

{62,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 63% at 4:59:40 am EDT | Wednesday, June 26, 2019.

4:59:40 am CDT | Wednesday, June 26, 2019

{62,% done.}

4179968 iterations done in 3.504*10^6 seconds. Should take 64.36 days or 5.561*10^6s. Finish Sat 20 Jul 2019 00:17:28.

{63,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 64% at 8:33:00 pm CDT | Wednesday, June 26, 2019.

7:59:40 pm EDT | Wednesday, June 26, 2019

{63,% done.}

4245504 iterations done in 3.558*10^6 seconds. Should take 64.34 days or 5.559*10^6s. Finish Fri 19 Jul 2019 23:55:36.

{64,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 65% at 10:59:40 am EDT | Thursday, June 27, 2019.

11:33:00 am EDT | Thursday, June 27, 2019

{64,% done.}

4313088 iterations done in 3.614*10^6 seconds. Should take 64.33 days or 5.558*10^6s. Finish Fri 19 Jul 2019 23:37:55.

{65,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 66% at 3:06:20 am EDT | Friday, June 28, 2019.

2:33:00 am EDT | Friday, June 28, 2019

{65,% done.}

4378624 iterations done in 3.668*10^6 seconds. Should take 64.32 days or 5.557*10^6s. Finish Fri 19 Jul 2019 23:13:26.

{66,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 67% at 5:33:00 pm EDT | Friday, June 28, 2019.

6:06:20 pm EDT | Friday, June 28, 2019

{66,% done.}

4446208 iterations done in 3.724*10^6 seconds. Should take 64.31 days or 5.556*10^6s. Finish Fri 19 Jul 2019 23:02:29.

{67,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 68% at 9:39:40 am EDT | Saturday, June 29, 2019.

9:06:20 am EDT | Saturday, June 29, 2019

{67,% done.}

4511744 iterations done in 3.778*10^6 seconds. Should take 64.29 days or 5.555*10^6s. Finish Fri 19 Jul 2019 22:38:45.

{68,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 69% at 12:06:20 am EDT | Sunday, June 30, 2019.

12:23:00 am EDT | Sunday, June 30, 2019

{68,% done.}

4577280 iterations done in 3.833*10^6 seconds. Should take 64.28 days or 5.554*10^6s. Finish Fri 19 Jul 2019 22:23:18.

{69,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 70% at 3:39:40 pm EDT | Sunday, June 30, 2019.

3:56:20 pm EDT | Sunday, June 30, 2019

{69,% done.}

4644864 iterations done in 3.889*10^6 seconds. Should take 64.27 days or 5.553*10^6s. Finish Fri 19 Jul 2019 22:11:47.

{70,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 71% at 7:29:40 am EDT | Monday, July 1, 2019.

3:56:20 pm EDT | Sunday, June 30, 2019 I updated B 5 million 05 2019.nb to 70% complete.

7:13:00 am EDT | Monday, July 1, 2019

{70,% done.}

4710400 iterations done in 3.944*10^6 seconds. Should take 64.27 days or 5.553*10^6s. Finish Fri 19 Jul 2019 22:14:30.

{71,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 72% at 10:29:40 pm EDT | Monday, July 1, 2019.

10:29:40 pm EDT | Monday, July 1, 2019

{71,% done.}

4777984 iterations done in 3.999*10^6 seconds. Should take 64.26 days or 5.552*10^6s. Finish Fri 19 Jul 2019 21:50:32.

{72,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 73% at 1:46:20 pm EDT | Tuesday, July 2, 2019.

1:46:20 pm EDT | Tuesday, July 2, 2019

{72,% done.}

4843520 iterations done in 4.054*10^6 seconds. Should take 64.25 days or 5.551*10^6s. Finish Fri 19 Jul 2019 21:38:06.

{73,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 74% at 5:03:00 am EDT | Wednesday, July 3, 2019.

4:46:20 am EDT | Wednesday, July 3, 2019

{73,% done.}

4909056 iterations done in 4.108*10^6 seconds. Should take 64.24 days or 5.551*10^6s. Finish Fri 19 Jul 2019 21:29:34.

{74,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 75% at 7:46:20 pm EDT | Wednesday, July 3, 2019.

8:36:20 pm EDT | Wednesday, July 3, 2019

{74,% done.}

4976640 iterations done in 4.165*10^6 seconds. Should take 64.24 days or 5.551*10^6s. Finish Fri 19 Jul 2019 21:31:34.

{75,% done.}

It is completing 1% every 15 hours 50 minutes 0 seconds.

Should reach 76% at 12:26:20 pm EDT | Thursday, July 4, 2019.

11:36:20 am EDT | Thursday, July 4, 2019

{75,% done.}

5042176 iterations done in 4.219*10^6 seconds. Should take 64.24 days or 5.550*10^6s. Finish Fri 19 Jul 2019 21:20:17.

{76,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 77% at 2:36:20 am EDT | Friday, July 5, 2019.

3:26:20 am EDT | Friday, July 5, 2019

{76,% done.}

5109760 iterations done in 4.276*10^6 seconds. Should take 64.24 days or 5.550*10^6s. Finish Fri 19 Jul 2019 21:23:56.

{77,% done.}

It is completing 1% every 15 hours 50 minutes 0 seconds.

Should reach 78% at 7:16:20 pm EDT | Friday, July 5, 2019.

6:43:00 pm EDT | Friday, July 5, 2019

{77,% done.}

5175296 iterations done in 4.331*10^6 seconds. Should take 64.24 days or 5.550*10^6s. Finish Fri 19 Jul 2019 21:27:59.

{78,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 79% at 9:59:40 am EDT | Saturday, July 6, 2019.

9:59:40 am EDT | Saturday, July 6, 2019

{78,% done.}

5240832 iterations done in 4.386*10^6 seconds. Should take 64.25 days or 5.551*10^6s. Finish Fri 19 Jul 2019 21:40:56.

{79,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 80% at 1:16:20 am EDT | Sunday, July 7, 2019.

1:33:00 am EDT | Sunday, July 7, 2019

{79,% done.}

5308416 iterations done in 4.442*10^6 seconds. Should take 64.24 days or 5.550*10^6s. Finish Fri 19 Jul 2019 21:19:44.

{80,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 81% at 5:06:20 pm EDT | Sunday, July 7, 2019.

4:33:00 pm EDT | Sunday, July 7, 2019

{80,% done.}

5373952 iterations done in 4.496*10^6 seconds. Should take 64.23 days or 5.549*10^6s. Finish Fri 19 Jul 2019 21:10:36.

{81,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 82% at 7:33:00 am EDT | Monday, July 8, 2019.

7:33:00 am CDT | Monday, July 8, 2019

{81,% done.}

5439488 iterations done in 4.550*10^6 seconds. Should take 64.22 days or 5.549*10^6s. Finish Fri 19 Jul 2019 20:58:53.

{82,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 83% at 10:33:00 pm CDT | Monday, July 8, 2019.

11:23:00 pm EDT | Monday, July 8, 2019

{82,% done.}

5507072 iterations done in 4.607*10^6 seconds. Should take 64.22 days or 5.549*10^6s. Finish Fri 19 Jul 2019 20:58:22.

{83,% done.}

It is completing 1% every 15 hours 50 minutes 0 seconds.

Should reach 84% at 3:13:00 pm EDT | Tuesday, July 9, 2019.

2:23:00 pm EDT | Tuesday, July 9, 2019

{83,% done.}

5572608 iterations done in 4.661*10^6 seconds. Should take 64.21 days or 5.548*10^6s. Finish Fri 19 Jul 2019 20:48:27.

{84,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 85% at 5:23:00 am EDT | Wednesday, July 10, 2019.

6:13:00 am EDT | Wednesday, July 10, 2019

{84,% done.}

5640192 iterations done in 4.718*10^6 seconds. Should take 64.21 days or 5.548*10^6s. Finish Fri 19 Jul 2019 20:48:45.

{85,% done.}

It is completing 1% every 15 hours 50 minutes 0 seconds.

Should reach 86% at 10:03:00 pm EDT | Wednesday, July 10, 2019.

9:29:40 pm EDT | Wednesday, July 10, 2019

{85,% done.}

5705728 iterations done in 4.773*10^6 seconds. Should take 64.22 days or 5.548*10^6s. Finish Fri 19 Jul 2019 20:49:23.

{86,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 87% at 12:46:20 pm EDT | Thursday, July 11, 2019.

12:46:20 pm EDT | Thursday, July 11, 2019

{86,% done.}

5771264 iterations done in 4.828*10^6 seconds. Should take 64.22 days or 5.548*10^6s. Finish Fri 19 Jul 2019 20:54:21.

{87,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 88% at 4:03:00 am EDT | Friday, July 12, 2019.

4:19:40 am EDT | Friday, July 12, 2019

{87,% done.}

5838848 iterations done in 4.884*10^6 seconds. Should take 64.21 days or 5.548*10^6s. Finish Fri 19 Jul 2019 20:44:47.

{88,% done.}

It is completing 1% every 15 hours 33 minutes 20.16 seconds.

Should reach 89% at 7:53:00 pm EDT | Friday, July 12, 2019.

7:19:40 pm EDT | Friday, July 12, 2019

{88,% done.}

5904384 iterations done in 4.938*10^6 seconds. Should take 64.21 days or 5.548*10^6s. Finish Fri 19 Jul 2019 20:43:34.

{89,% done.}

It is completing 1% every 15 hours 0 minutes 0 seconds.

Should reach 90% at 10:19:40 am EDT | Saturday, July 13, 2019.

11:09:40 am EDT | Saturday, July 13, 2019

{89,% done.}

5971968 iterations done in 4.995*10^6 seconds. Should take 64.21 days or 5.548*10^6s. Finish Fri 19 Jul 2019 20:41:54.

{90,% done.}

It is completing 1% every 15 hours 50 minutes 0 seconds.

Should reach 91% at 2:59:40 am EDT | Sunday, July 14, 2019.

2:26:20 am EDT | Sunday, July 14, 2019

{90,% done.}

6037504 iterations done in 5.050*10^6 seconds. Should take 64.21 days or 5.548*10^6s. Finish Fri 19 Jul 2019 20:47:41.

{91,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 92% at 5:43:00 pm EDT | Sunday, July 14, 2019.

7:14:00 am EDT | Sunday, July 14, 2019 updated B 5 million 05 2019.nb at 91% complete.

5:43:00 pm EDT | Sunday, July 14, 2019

{91,% done.}

6103040 iterations done in 5.105*10^6 seconds. Should take 64.21 days or 5.548*10^6s. Finish Fri 19 Jul 2019 20:45:59.

{92,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 93% at 8:59:40 am EDT | Monday, July 15, 2019.

9:33:00 am EDT | Monday, July 15, 2019

{92,% done.}

6170624 iterations done in 5.162*10^6 seconds. Should take 64.22 days or 5.548*10^6s. Finish Fri 19 Jul 2019 20:51:47.

{93,% done.}

It is completing 1% every 15 hours 50 minutes 0 seconds.

Should reach 94% at 1:23:00 am EDT | Tuesday, July 16, 2019.

12:49:40 am EDT | Tuesday, July 16, 2019

{93,% done.}

6236160 iterations done in 5.217*10^6 seconds. Should take 64.22 days or 5.549*10^6s. Finish Fri 19 Jul 2019 20:59:21.

{94,% done.}

It is completing 1% every 15 hours 16 minutes 40.08 seconds.

Should reach 95% at 4:06:20 pm EDT | Tuesday, July 16, 2019.

Attachments:
5 million 01 2019.nb
5 million 05 2019.nb
B 5 million 05 2019.nb
april 2018 5 mil...nb
POSTED BY: Marvin Ray Burns
Answer

3/23/2019

Well, my mania struck again!!!!! I bought another computer that I can't afford.

Added to my cluster of

3.7 GH overclocked up to 4.7 GHz on all Intel 6 cores at 3000 MHz RAM

and 4 cores of 3.6 GHz at 2400 MHz RAM,

with optimized use of kernel priority in my programs, I should be able to compute up to 10 million digits! And, according to results from a well known cloud I tried out, my cluster should be faster than a Intel Xeon Platinum 8151 @ 3.4 GHz with 24 cores!! (in using my programs for many computations of MRB)

Here are the details of my new computer: Processor: Intel Core i9-9900K (5.0 GHz Turbo) (16-Thread) (8-Core) 3.6 GHz. CPU Boost: Stage 2: Overclock CPU - Up to 5.1GHz on All CPU Cores at 3200 Mhz RAM Extreme Cooling: H20: Stage 2: Corsair H115i PRO - 280mm Liquid CPU Cooler (Fully Sealed + No Maintenance)

Total: $2,695.00.

That is less than 1/10th the price of any single computer that can do the same as the new cluster!!!!!!

3 node MRB Digital Storm 2

Using the 3 node MRB constant super computer, I got the following red column for seconds on computing the MRB constant. Documented in the attached ":3 fastest computers together 3.nb." Pay special attention to the new record 45.5 hour million digit computation!!!!!!!

enter image description here

Attachments:
3 fastest comput...nb
POSTED BY: Marvin Ray Burns
Answer
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