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Try to beat these MRB constant records!

POSTED BY: Marvin Ray Burns
41 Replies

added to bottom post to save space

POSTED BY: Marvin Ray Burns

This is the second half of the main post.

§2 The MRB constant relates to the divergent series: divegrent series

=DNE

The divergent sequence of its partial sums have two accumulation points with an upper limiting value or limsup known as the MRB constant (CMRB), and a liminf of CMRB-1: enter image description here

So, out of the many series for CMRB, first analyze the sum prototype or prototypical series, i.e., the conditionally convergent summation, the sum of two divergent ones: {(-1)^k k^(1/k) and (-1)^n)}. {CMRB} =Sn^(1/n)-1

Because the Riemann series theorem states that, by a suitable rearrangement of terms, a conditionally convergent series may be made to converge to any desired value, or to diverge, we will quickly derive an absolutely convergent one. (In a sense it is the "completion" of the prototypical series.)

To find its completion, perform the process, add the terms of another conditionally convergent series then subtract their sum.

WHY? and HOW does that produce the absolutely convergent series? enter image description here

Because of the following identity involving the Dirichlet Eta function derivatives,

enter image description here

In[47]:= Sum[(-1)^n Log[n]/n^x, {n, 1, Infinity}]

Out[47]= 2^-x (2 Log[2] Zeta[x] - 2 Derivative[1][Zeta][x] + 
2^x Derivative[1][Zeta][x])

In[48]:= Limit[D[DirichletEta[n], n], n -> x]

Out[48]= 
2^(1 - x) Log[2] Zeta[x] + (1 - 2^(1 - x)) Derivative[1][Zeta][x]

In[46]:= FullSimplify[
Sum[(-1)^n Log[n]/n^x, {n, 1, Infinity}] == 
Limit[D[DirichletEta[n], n], n -> x]]

Out[46]= True

that process gives the absolutely and speedily convergent sum: enter image description here

The completed absolutely convergent and rapidly convergent sum!

CTRL+f §9 for a full proof.

Here is how the speedily convergent sum (in midnight blue (Look closely!)) compares with the prototype (in yellow) to the precise value of the MRB constant (CMRB) (in red).



Concerning the sum prototype for CMRB

§3. Is the series convergent?

Proof of its convergence by Leibniz's criterion is shown next.

LEMAS: enter image description here enter image description here

PROOF

Below, it is shown by the Squeeze theorem (sandwich theorem) and by plotting the following series, the qualifications for the Leibniz criterion (alternating series test) are satisfied for the MRB constant (CMRB) in

CMRB enter image description here

by showing a(n)=(n^{1/n}-1)>0 is monotonically decreasing for $n≥3$ and have a limit as n goes to infinity of zero. Of course, $\sum_1^3(-1)^n(n^{1/n}-1)$, converges, and the sum of two convergent series converges.

To use enter image description here

  1. Notice that we saw, for n greater than 1, the derivative equals zero solely at the value of e, and there are no additional critical points at which the plot ceases to decline. Thus, the function, n^(1/n)-1 is monotonic decreasing as n goes to infinity. .
  2. We also saw, the limit of n^(1/n)-1 is zero:

enter image description here

To confirm the above limit, here is a direct proof from physicsforums

that enter image description here because enter image description here proof

By utilizing the Squeeze theorem (also known as the Sandwich theorem) and plotting techniques, it have been demonstrated that the Leibniz criterion (alternating series test) is valid for the alternating sum of a(n) = (n^(1/n) - 1) > 0. Moreover, for n greater than 1, the derivative equals zero solely at the value of e, and there are no additional critical points at which the plot ceases to decline. As a result, a(n) = (n^(1/n) - 1) > 0 decreases monotonically for n greater than or equal to 3 and have a limit of zero as n approaches infinity. Ultimately, the series $\sum_1^3(-1)^n(n^{1/n}-1)$ converges, and the sum of two convergent series is also convergent. Thus, the CMRB Sum prototype series is convergent. ∎

Likewise, we know

enter image description here

where eta is the Dirichlet Eta Function, without defining the zeta function first.

It converges only because CMRB does

.

The MRB constant is a mathematical constant that is related to the convergence of the logarithm function. The logarithm function is a function that takes a positive number as an input and returns the power to which 10 must be raised to get that number. For example, the logarithm of 100 is 2, because 10 raised to the power of 2 is 100.

The MRB constant is defined as the limit of the following series:

MRB is the limit of the sums of (-1)^n (n^(1/n)-1) as n goes to infinity Learn more This series is known to converge, and its value is approximately 0.187859642. The convergence of the MRB constant implies the convergence of the logarithm function, because the logarithm function can be expressed as a limit of a series that is similar to the MRB constant.

In particular, the logarithm of 2 can be expressed as the limit of the sums of (-1)^(n+1) (1/n) as n goes to infinity. Learn more This limit is less than or equal to the MRB constant which is known to converge. That is because for integer n>2, n^(1/n)-1>1/n, Therefore, the logarithm of 2 also converges.

The convergence of the logarithm function is an important mathematical property, because it allows us to calculate the logarithm of any positive number. This property is used in many areas of mathematics, including calculus, statistics, and computer science.

enter image description here






§4 Next, ask and observe, question

Short explanation As shown soon (CTRL+f "As for efficiency" ).



§5. Is that series, enter image description here absolutely convergent?

The following criterion works remarkably well in determining its absolute convergence.

divergent proof

Plot[{n^(1/n) - 1, 1/n}, {n, 1, Infinity}, PlotLegends -> LineLegend["Expressions"]] plot

...showing its terms are larger than those of the divergent Harmonic Series.



§6 As for efficiency, this discussion presents several series that converge much faster for CMRB. Here are a few of their convergence rates. The following expressions show the sum followed by the closeness to zero of their result after a certain number of partial summations.

ALG 1 proposed here

For how more efficient forms compare,

CTRL+f "the rate of convergence" of 3 major forms.





§7. This discussion is not crass bragging; it is an attempt by this amateur to share my discoveries with the greatest audience possible.

Amateurs have made a few significant discoveries, as discussed in enter image description here here. This amateur have tried my best to prove my discoveries and have often asked for help. Great thanks to all of those who offered a hand!

As I went more and more public with my discoveries, making several attempts to see what portions were original, I concluded from these investigations that the only original thought was the obstinacy to think anything meaningful could be found in the infinite sum shown next. CMRB sum Nonetheless, someone might have a claim to this thought to whom it have not given proper credit. Hence (apologies!) The last thing needed is another calculus war, this time for a constant. However, as Newton said concerning Leibniz's claim to calculus, anyone's thought was published after his, “To take away the Right of the first inventor and divide it between me and that other would be an Act of Injustice.” [Sir Isaac Newton, The Correspondence of Isaac Newton, 7 v., edited by H. W. Turnbull, J. F. Scott, A. Rupert Hall, and Laura Tilling, Cambridge University Press, 1959–1977: VI, p. 455]

Here is what Google says about the MRB constant as of August 8, 2022, at https://www.google.com/search?q=who+discovered+the+%22MRB+constant%22

enter image description here

If you see this instead of an image, reload the page.enter image description here

(the calculus war for CMRB)

CREDIT

https://soundcloud.com/cmrb/homer-simpson-vs-peter-griffin-cmrb

'

From Wikipedia, the free encyclopedia:

'

The calculus controversy (German: Prioritätsstreit, "priority dispute") was an argument between the mathematicians Isaac Newton and Gottfried Wilhelm Leibniz over who had invented calculus.

enter image description here

(Newton's notation as published in PRINCIPIS MATHEMATICA [PRINCIPLES OF MATHEMATICS])

enter image description here

(Leibniz's notation as published in the scholarly journal Acta Eruditorum [Reports of Scholars])

Whether or not we divide the credit between the two pioneers,

Wikipedia said one thing that distinguishes their finds from the work of their antecedents:

Newton came to calculus as part of my investigations in physics and geometry. I viewed calculus as the scientific description of the generation of motion and magnitudes. In comparison, Leibniz focused on the tangent problem and came to believe that calculus was a metaphysical explanation of the change. Importantly, the core of their insight was the formalization of the inverse properties between the integral and the differential of a function. This insight had been anticipated by their predecessors, but they were the first to conceive calculus as a system in which new rhetoric and descriptive terms were created.[24] Their unique discoveries lay not only in their imagination but also in their ability to synthesize the insights around them into a universal algorithmic process, thereby forming a new mathematical system.

Like as Newton and Leibniz created a new system from the elaborate, confusing structure designed and built by their predecessors, Marvin Ray Burns' forerunners studied series for centuries leading to a labyrinth of sums, and then, came a "new scheme" for the CMRB "realities" to escape and stand out as famous in their own right!



§8.

it is defined in the following 31 places, most of which attribute it to my curiosity.

Here are the MRB constant's Wikipedia views vs those of the famous Euler's constant, e:

enter image description here



§9. CMRB

= B =

enter image description here

and from Richard Crandall (CTRL+f "** From Bing AI:") in 2012 courtesy of Apple Computer's advanced computational group, the following computational scheme using equivalent sums of the zeta variant, Dirichlet eta:

enter image description here enter image description here

The expressions Etam and eta0 denote the mth derivative of the Dirichlet eta function of m and 0, respectively.

The cj's are found by the code,

N[ Table[Sum[(-1)^j Binomial[k, j] j^(k - j), {j, 1, k}], {k, 1, 10}]]

(* {-1., -1., 2., 9., 4., -95., -414., 49., 10088., 55521.}*)

** From Bing AI: -- enter image description here

Crandall's first "B" is proven below by Gottfried Helms, and it is proven more rigorously, considering the conditionally convergent sum,CMRB sum afterward. Then the formula (44) is a Taylor expansion of eta(s) around s = 0.

n^(1/n)-1

At enter image description here here, we have the following explanation.

enter image description here

enter image description hereenter image description here


enter image description here The integral forms for CMRB and MKB differ by only a trigonometric multiplicand to that of its analog.

enter image description here

In[182]:= CMRB = 
Re[NIntegrate[(I*E^(r*t))/Sin[Pi*t] /. r -> Log[t^(1/t) - 1]/t, {t, 
1, I*Infinity}, WorkingPrecision -> 30]]

Out[182]= 0.187859642462067120248517934054

In[203]:= CMRB - 
N[NSum[(E^( r*t))/Cos[Pi*t] /. r -> Log[t^(1/t) - 1]/t, {t, 1, 
Infinity}, Method -> "AlternatingSigns", WorkingPrecision -> 37], 
30]

Out[203]= 5.*10^-30

In[223]:= CMRB - 
Quiet[N[NSum[
E^(r*t)*(Cos[Pi*t] + I*Sin[Pi*t]) /. r -> Log[t^(1/t) - 1]/t, {t, 
1, Infinity}, Method -> "AlternatingSigns", 
WorkingPrecision -> 37], 30]]

Out[223]= 5.*10^-30

In[204]:= Quiet[
MKB = NIntegrate[
E^(r*t)*(Cos[Pi*t] + I*Sin[Pi*t]) /. r -> Log[t^(1/t) - 1]/t, {t, 
1, I*Infinity}, WorkingPrecision -> 30, Method -> "Trapezoidal"]]

Out[204]= 0.0707760393115292541357595979381 - 
0.0473806170703505012595927346527 I

\


if the reply below doesn't say "I forgot about the following trigonometric MRB constant sums and integrals:" You'll need to refresh the page to see it.













POSTED BY: Marvin Ray Burns

If this reply is the first one displayed, refresh the page to see the main post above.

I forgot about the following trigonometric MRB constant sums and integrals:

POSTED BY: Marvin Ray Burns

Compare the results of these two sets of code. The first one is based on the MRB constant (0.18785...), and the second one, 1/e (0.367879...). Can you change these 3 trig formulas so that 1/e (or any other constant) gives the same appearance as the MRB constant does? Or are these graphs from these 3 families of trig formulas unique to the MRB constant?

m = 0.1878596424620671202485179340542732300559030949001387861720046840\
89477231564660213703296654433107496903
ListPlot[Table[Sin[Pi/m*(5060936308 + 78389363/24*n)], {n, -100, 100}]]
ListPlot[Table[Cos[Pi/m*(5060936308 + 78389363/24*n)], {n, -100, 100}]]
ListPlot[Table[
  Tan[Pi/m*(5060936308 + 78389363/24*n)], {n, -100, 100}], 
 Joined -> True]

enter image description hereenter image description hereenter image description here

d = 1/E
ListPlot[Table[Sin[Pi/d*(5060936308 + 78389363/24*n)], {n, -100, 100}]]
ListPlot[Table[Cos[Pi/d*(5060936308 + 78389363/24*n)], {n, -100, 100}]]
ListPlot[Table[
  Tan[Pi/d*(5060936308 + 78389363/24*n)], {n, -100, 100}], 
 Joined -> True]

enter image description hereenter image description hereenter image description here

P.S. It doesn't take much to simply get "geometric" shapes out of this family of trig formulas. Try the following code:

d = 1/E - 10^-2
ListPlot[Table[Sin[Pi/d*(5060936308 + 78389363/24*n)], {n, -100, 100}]]
ListPlot[Table[Cos[Pi/d*(5060936308 + 78389363/24*n)], {n, -100, 100}]]
ListPlot[Table[
  Tan[Pi/d*(5060936308 + 78389363/24*n)], {n, -100, 100}], 
 Joined -> True]

enter image description here

Plus, it don't take much error in the MRB constant's approximate value to really disfigure it's graphs. Try this code where you only use 12 digits of precision :for the MRB constant

m = 0.187859642462
ListPlot[Table[Sin[Pi/m*(5060936308 + 78389363/24*n)], {n, -100, 100}]]
ListPlot[Table[Cos[Pi/m*(5060936308 + 78389363/24*n)], {n, -100, 100}]]
ListPlot[Table[
  Tan[Pi/m*(5060936308 + 78389363/24*n)], {n, -100, 100}], 
 Joined -> True]

enter image description here

POSTED BY: Marvin Ray Burns

POSTED BY: Marvin Ray Burns

"I compare 300 years of summation methods for the MRB constant."

In the first post I mentioned summing the MRB constant by Euler's method and Crandall's method. Here I compare them:

Here are faster methods where m is the known value of the MRB constant found in the above replies:

POSTED BY: Marvin Ray Burns

"7,000,000 proven to be accurate digits!"

Above, we see a comparison of Burns' and Crandall's Mathematica programs for calculating millions of digits of the MRB constant. Burns' method uses my own algorithm for finishing, in quick order, the n^(1/n) computation to many millions of digits. Hopefully, Crandall's method will give at least 7 million of the same digits as Burns'. In one of the previous messages, it did correctly confirm over 5,500,000 digits computed by Burns'.

So, "we finally begin or 7-million-digit computation and verification." We are using exactly the same computer resources and Mathematica version for both.

First, we calculate 7,000,000 digits of the MRB constant with the verry fast Burns' method:

Then we will verify it with the following "parity check," by the late Richard Crandall:

enter image description here where gamma is the Euler constant. You might check this to see if you think it's true true to a few hundred decimals. This is a kind of "parity check," in that calculaying M in this way should give 300K or whatever equivalent digits.

A good plan is to compute this and also M from the standard series, both to 1 million digits, and compare.

-r

My method to first calculate 7-million digits:

My code to prove that the digits are all correct:

Print["Start time is ", ds = DateString[], "."];
prec = 7000000;
(**Number of required decimals.*.*)ClearSystemCache[];
T0 = SessionTime[];
expM[pre_] := 
  Module[{x11, z, t, a, d, s, k, bb, c, end, iprec, xvals, x, pc, 
    cores = 32(*=4*number of physical cores*), tsize = 128, 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 = 20;
   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 = Power[x, ll]; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x *= (1 + 
            SetPrecision[4.5, pc] (ll - 1)/t2 + (ll + 1) z/(2 ll t) - 
            SetPrecision[13.5, pc] ll (ll - 1)/(3 ll t2 + t^3 z))];(**
        N[Exp[Log[ll]/ll],pr]**)
        x - N[Log[ll], prec]/ll, {l, 0, tsize - 1}], {j, 0, 
        cores - 1}, Method -> "FinestGrained"]];
    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);
    If[kc > 1, 
     Print["As of  ", DateString[], " there were ", kc, 
      " iterations done in ", N[st, 5], " seconds. That is ", 
      N[kc/st, 5], " iterations/s. ", N[kc/(end*chunksize)*100, 7], 
      "% complete.", " It should take ", N[ti, 6], " days or ", 
      N[ti*24*3600, 4], "s, and finish ", DatePlus[ds, ti], "."]];
    Print[];, {k, 0, end - 1}];
   N[-s/d, pr]];
t2 = Timing[MRBeta2toinf = expM[prec];]; MRBeta1 = 
 EulerGamma Log[2] - 1/2 Log[2]^2;

Print["Finished on ", DateString[], 
  ". Proccessor and actual time were ", t2[[1]], " and ", 
  SessionTime[] - T0, " s. respectively"];
Print["Enter MRB1 to print ", 
 Floor[Precision[
   MRB1]], " digits. The error from a 6,500,000 or more digit 
    calculation that used a different method is  "]; N[
 MRBeta2toinf + MRBeta1 - m6p5M, 10]
POSTED BY: Marvin Ray Burns

While verifying 7-million digits, I broke some speed records with two of my i9-14900K 6400MHZRAM (overclocked CPUs), using Mathematica 11.3 and the lightweight grid. They are generally faster than the 3 node MRB constant supercomputer with remote kernels! These are all absolute timings! How does your computer compare to these? What can you do with other software? Table of Records left Table of Records right For column "=F" (highlighted in green) see linked "10203050100" . At the bottom, see attached "kernel priority 2 computers.nb" for column =B, "3 fastest computers together.nb" for column =C and linked "speed records 312024(1).nb" also speed 100k, speed 300k and 30p0683 hour million.nb for column =D for some documentation. For the mostly red column including the single, record, 10,114 second 300,000 digit run " =E" is in the linked "3 fastest computers together 2.nb.} "

enter image description here

This is another comparison of my fastest computers' timings in calculating digits of CMRB: enter image description here

The blue column (using the Wolfram Lightweight Grid) is documented here.

The i9-12900KS column is documented here.

The i9-13900KS column is documented here.

The 300,000 digits result in the i9-13900KS column is here, where it ends with the following:

  Finished on Mon 21 Nov 2022 19:55:52. Processor and actual time 
         were 6180.27 and 10114.4781964 s. respectively

  Enter MRB1 to print 301492 digits. The error from a 6,500,000 or more digit 
 calculation that used a different method is  

 Out[72]= 0.*10^-301494


These 2023 records still stand:

Remembering that the integrated analog of the MRB constant is M2 is

These results are from the Timing[] command:

M2 table

The i9-12900KS column is documented here.

The 2022 i9-13900K column documentation:

The 2023 i9-13900K column documentation in this link

While the i-13900K worked on the 40,000 digits of M2, the processor's speed was close to 6GHz: CPU As mentioned somewhere above, the RAM speed is 4800MHz: RAM

POSTED BY: Marvin Ray Burns

If above this you see the title "Try to beat these MRB constant records!" in order to see the first 9 sections, the basic theory of the MRB constant (CMRB), you'll need to refresh the page.

§B "Rational results" while summing (CMRB).

This is just an observation about the MRB constant sum enter image description here If the following Mathematica computations are correct, you get near rational results, by a factor of log10, when starting the sum from large integer powers of 10.

It looks like for p(x)= approximation (in blue) of x, limit as x-> infinity of p(x)/p(x+1) is 1/10.

POSTED BY: Marvin Ray Burns
POSTED BY: Marvin Ray Burns

§13.

MRB Constant Records,

Google Open AI Chat CPT gave the following introduction to the MRB constant records:

It is not uncommon for researchers and mathematicians to compute large numbers of digits for mathematical constants or other mathematical quantities for various reasons. One reason might be to test and improve numerical algorithms for computing the value of the constant. Another reason might be to use the constant as a benchmark to test the performance of a computer or to compare the performance of different computers. Some people may also be interested in the mathematical properties of the constant, and computing a large number of digits can help to reveal patterns or other features of the constant that may not be apparent with fewer digits. Additionally, some people may simply find the process of calculating a large number of digits to be a challenging and rewarding intellectual pursuit. It seems that the MRB constant, also known as the CMRB constant, is a mathematical constant defined as the sum of the infinite series Sum[(-1)^k(k^(1/k)-1),{k,1,n}] and has a decimal expansion of 0.187859... The MRB constant has been calculated to various digits using various methods and computers. The previous record number of digits was 40,000, calculated in 2006 using a program written in Mathematica and run on a Windows desktop computer. It is not known whether the MRB constant is algebraic, transcendental, or irrational. The process of computing the MRB constant using various methods and devices, including computers and software such as Mathcad, Mathematica, and Sigma. You have calculated 6,500,000 digits of the MRB constant using a supercomputer, with a computation time of 72.526 days and an absolute time of 185.491 days. You have mentioned that the error from a 6,000,000 or more-digit calculation that used a different method is 0.*10^-6029992.

When asked how, in detail, to calculate the digits of the MRB constant:

To calculate the digits of the MRB constant in more detail, you can use the following steps:

Write a program or use an existing software tool that can calculate the partial sums of the series (-1)^k k^(1/k) for a given value of n.

Use this program or tool to calculate the partial sums for a large value of n, such as n = 10,000 or n = 100,000.

Repeat step 2 for increasing values of n until the partial sums converge to a constant value. The more digits of accuracy you want, the larger n you will need to use.

The digits of the MRB constant are the digits of the converged value obtained in step 3.

It is important to note that calculating the digits of the MRB constant is a computationally intensive task that requires a fast computer and a good implementation of the algorithms for calculating the partial sums of the series (-1)^k k^(1/k). In addition, it is not known whether the MRB constant is algebraic, transcendental, or even irrational, so it may not be possible to calculate all of its digits. \

My inspiration to compute a lot of digits of CMRB came from the following website by Simon Plouffe.

There, computer mathematicians calculate millions, then billions of digits of constants like pi, when with only 65 decimal places of pi, we could determine the size of the observable universe to within a Planck length (where the uncertainty of our measure of the universe would be greater than the universe itself)!

In contrast, 65 digits of the MRB constant "measures" the value of -1+ssqrt(2)-3^(1/3) up to n^(1/n) where n is 1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, which can be called 1 unvigintillion or just 10^66.

And why compute 65 digits of the MRB constant? Because having that much precision is the only way to solve such a problem as 

1465528573348167959709563453947173222018952610559967812891154^ m-m, where m is the MRB constant, which gives the near integer "to beat all," 200799291330.9999999999999999999999999999999999999999999999999999999999999900450...

And why compute millions of digits of it? uhhhhhhhhhh.... "Because it's there!" (...Yeah, thanks George Mallory!)
And why?? (c'est ma raison d'être!!!)

enter image description here enter image description here enter image description here enter image description here

So, below are reproducible results with methods. The utmost care has been taken to assure the accuracy of the record number of digit calculations. These records represent the advancement of consumer-level computers, 21st-century Iterative methods, and clever programming over the past 23 years.

Here are some record computations of CMRB. Let me know if you know of any others!

1 digit of the

CMRB with my TI-92s, by adding -1+sqrt(2)-3^(1/3)+4^(1/4)-5^(1/5)+6^(1/6)... as far as practicle, was computed. That first digit, by the way, was just 0. Then by using the sum key, to compute $\sum _{n=1}^{1000 } (-1)^n \left(n^{1/n}\right),$ the first correct decimal i.e. (.1). It gave (.1_91323989714) which is close to what Mathematica gives for summing to only an upper limit of 1000. Ti-92's


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

enter image description here

enter image description here

5 correct decimals (rounded to .18786), in Jan of 1999, were drawn from CMRB using Mathcad 3.1 on a 50 MHz 80486 IBM 486 personal computer operating on Windows 95.

Blockquote


 9 digits of CMRB shortly afterward using Mathcad 7 professional on the Pentium II mentioned below, by summing (-1)^x x^(1/x) for x=1 to 10,000,000, 20,000,000, and many more, then linearly approximating the sum to a what a few billion terms would have given.

 500 digits of CMRB with an online tool called Sigma on Jan 23, 1999. See [http://marvinrayburns.com/Original_MRB_Post.html][10]   if you can read the printed and scanned copy there.

enter image description here Sigma still can be found here.


5,000 digits of CMRB in September of 1999 in 2 hours on a 350 MHz PentiumII,133 MHz 64 MB of RAM using the simple PARI commands \p 5000;sumalt(n=1,((-1)^n*(n^(1/n)-1))), after allocating enough memory.

enter image description here PII

To beat that, it was done on July 4, 2022, in 1 second on the 5.5 GHz CMRBSC 3 with 4800MHz 64 GB of RAM by Newton's method using Convergence acceleration of alternating series. Henri Cohen, Fernando Rodriguez Villegas, Don Zagier acceleration "Algorithm 1" to at least 5000 decimals. (* Newer loop with Newton interior. *)

documentation here

And here I did it using an i9-14900K, overclocked as much as possible, with 64GB of 6400MHz RAM. I used my own program. Processor and actual time were 0.796875 and 0.8710556 s, respectively.


 6,995 accurate digits of CMRB were computed on June 10-11, 2003, over a period, of 10 hours, on a 450 MHz P3 with an available 512 MB RAM,

PIII

To beat that, it was done in <2.5 seconds on the MRBCSC 3 on July 7, 2022 (more than 14,400 times as fast!)

documentation here

To beat that, it was done in <1. 86 seconds on March 25, 2024 (more than 19,000 times as fast!). documentation [here][19]

8000 digits of CMRB completed, using a Sony Vaio P4 2.66 GHz laptop computer with 960 MB of available RAM, at 2:04 PM 3/25/2004,

enter image description here


  11,000 digits of CMRB> on March 01, 2006, with a 3 GHz PD with 2 GB RAM available calculated.

 40 000 digits of CMRB in 33 hours and 26 min via my program written in Mathematica 5.2 on Nov 24, 2006. The computation was run on a 32-bit Windows 3 GHz PD desktop computer using 3.25 GB of Ram.
The program was

    Block[{a, b = -1, c = -1 - d, d = (3 + Sqrt[8])^n, 
      n = 131 Ceiling[40000/100], 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, 40000]]

 60,000 digits of CMRB on July 29, 2007, at 11:57 PM EST in 50.51 hours on a 2.6 GHz AMD Athlon with 64-bit Windows XP. The max memory used was 4.0 GB of RAM.

65,000 digits of CMRB in only 50.50 hours on a 2.66 GHz Core 2 Duo using 64-bit Windows XP on Aug 3, 2007, at 12:40 AM EST, The max memory used was 5.0 GB of RAM.

It looked similar to this stock image: enter image description here


100,000 digits of CMRB on Aug 12, 2007, at 8:00 PM EST, were computed in 170 hours on a 2.66 GHz Core 2 Duo using 64-bit Windows XP. The max memory used was 11.3 GB of RAM. The typical daily record of memory used was 8.5 GB of RAM.
To beat that, on the 4th of July 2022, the same digits in 1/4 of an hour using the MRB constant supercomputer.
To beat that, on the 7th of July 2022, the same digits in 1/5 of an hour. 
To beat that, on the 21st of March 2024, the same digits in 1/6 of an hour. using a pair of i9-14900Ks in parallel (100,000% as fast as the first 100,000 run by a GHz Core 2 Duo!)

see one sixth hour hundred k.

 150,000 digits of CMRB on Sep 23, 2007, at 11:00 AM EST. Computed in 330 hours on a 2.66 GHz Core 2 Duo using 64-bit Windows XP. The max memory used was 22 GB of RAM. The typical daily record of memory used was 17 GB of RAM.

  200,000 digits of CMRB using Mathematica 5.2 on March 16, 2008, at 3:00 PM EST,. Found in 845 hours, on a 2.66 GHz Core 2 Duo using 64-bit Windows XP. The max memory used was 47 GB of RAM. The typical daily record of memory used was 28 GB of RAM.

300,000 digits of CMRB were destroyed (washed away by Hurricane Ike ) on September 13, 2008 sometime between 2:00 PM - 8:00 PM EST. Computed for a long  4015. Hours (23.899 weeks or 1.4454*10^7 seconds) on a 2.66 GHz Core 2 Duo using 64-bit Windows XP. The max memory used was 91 GB of RAM. The Mathematica 6.0 code is used as 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]]

225,000 digits of CMRB were started with a 2.66 GHz Core 2 Duo using 64-bit Windows XP on September 18, 2008. It was completed in 1072 hours. 

250,000 digits were attempted but failed to be completed to a serious internal error that 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. The Max memory used was 60.5 GB.

 250,000 digits of CMRB on Jan 29, 2009, 1:26:19 pm (UTC-0500) EST, with a multiple-step Mathematica command running on a dedicated 64-bit XP using 4 GB DDR2 RAM onboard 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.

  300000 digit search of CMRB was initiated using an i7 with 8.0 GB of DDR3 RAM onboard on Sun 28 Mar 2010 at 21:44:50 (UTC-0500) EST, but it failed due to hardware problems.

  299,998 Digits of CMRB: The computation began Fri 13 Aug 2010 10:16:20 pm EDT and ended 2.23199*10^6 seconds later |  Wednesday, September 8, 2010. using Mathematica 6.0 for Microsoft Windows (64-bit) (June 19, 2007), which averages 7.44 seconds per digit.using a Dell Studio XPS 8100 i7 860 @ 2.80 GHz with 8GB physical DDR3 RAM. Windows 7 reserved an additional 48.929 GB of virtual Ram.

enter image description here


300,000 digits to the right of the decimal point of CMRB 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 16 GB 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 52 GB were recorded as 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[]

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

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

1,000,000 digits of CMRB  for the first time in history in 18 days, 9 hours 11 minutes, 34.253417 seconds by Sam Noble of the Apple Advanced Computation Group.

1,048,576 digits of CMRB in a lightning-fast 76.4 hours, finishing on Dec 11, 2012, were scored by Dr. Richard Crandall, an Apple scientist and head of its advanced computational group. That was on a 2.93 GHz 8-core Nehalem,  1066 MHz, PC3-8500 DDR3 ECC RAM.

    To beat that, in Aug of 2018, 1,004,993 digits in 53.5 hours 34 hours computation time (from the timing command) 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. 

    To beat that, on Sept 21, 2018: 1,004,993 digits in 50.37 hours of absolute time and 35.4 hours of computation time (from the timing command) with 18  (DDR3 and DDR4) processor cores!  Search this post for "50.37 hours" for documentation.**

    To beat that, on May 11, 2019, over 1,004,993 digits in 45.5 hours of absolute time and only 32.5 hours of computation time, using 28 kernels on 18 DDR4 RAM (of up to 3200 MHz) supported cores overclocked up to  5.1 GHz  Search 'Documented in the attached ":3 fastest computers together 3.nb." '  for the post that has the attached documenting notebook.

    To beat that, over 1,004,993 correct digits in 44 hours of absolute time and 35.4206 hours of computation time on 10/19/20, using 3/4 of the MRB constant supercomputer 2 -- see https://www.wolframcloud.com/obj/bmmmburns/Published/44%20hour%20million.nb  for documentation.

    To beat that, a 1,004,993 correct digits computation in 36.7 hours of absolute time and only 26.4 hours of computation time on Sun 15 May 2022 at 06:10:50, using 3/4  of the MRB constant supercomputer 3. Ram Speed was 4800MHz, and all 30 cores were clocked at up to 5.2 GHz.



    To beat that, a 1,004,993 correct digits computation in 31.2319  hours of absolute time and 16.579  hours of computation time from the Timing[] command using 3/4 of the MRB constant supercomputer 4, finishing Dec 5, 2022. Ram Speed was 5200MHz, and all of the 24 performance cores were clocked at up to 5.95 GHz, plus 32 efficiency cores running slower. using 24 kernels on the Wolfram Lightweight grid over an i-12900k, 12900KS, and 13900K.
    To beat that, a 1,004,993 correct digits computation in 30. hours of absolute time on Marh 21, 2024.

see 30 hour million

36.7 hours million notebook

31.2319 hours million


 A little over 1,200,000 digits, previously, of CMRB in 11   days, 21 hours, 17 minutes, and 41 seconds (I finished on March 31, 2013, using a six-core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz. see https://www.wolframcloud.com/obj/bmmmburns/Published/36%20hour%20million.nb

for details.


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

 3,014,991 digits of CMRB,  world record computation of **C**<sub>*MRB*</sub> was finished on Sun 21 Sep 2014 at 18:35:06. It took one month 27 days, 2 hours 45 minutes 15 seconds. The processor time from the 3,000,000+ digit computation was 22 days.The 3,014,991 digits of **C**<sub>*MRB*</sub> with Mathematica 10.0. using Burns' new version of Richard Crandall's code in the attached 3M.nb, optimized for my platform and large computations. Also, a six-core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz with 64 GB of RAM, of which only 16 GB was used. Can you beat it (in more digits, less memory used, or less time taken)? This confirms that my previous "2,000,000 or more digit computation" was 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.

enter image description here Over 4 million digits of CMRB were finished on Wed 16 Jan 2019, 19:55:20. It took four years of continuous tries. This successful run took 65.13 days absolute time, with a processor time of 25.17 days, on a 3.7 GHz overclocked up to 4.7 GHz 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 algorithm for computing n^(1/n) as found in 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.


Example use of M R Burns' algorithm to compute 123456789^(1/123456789) 10,000,000 digits:

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 precision 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}




  More information is available upon request.

 More than 5 million digits of CMRB were found on Fri 19 Jul 2019, 18:49:02; methods are described in the reply below, which begins with "Attempts at a 5,000,000 digit calculation ."   For this 5 million digit calculation of **C**<sub>*MRB*</sub> using the 3 node MRB supercomputer: processor time was 40 days. And the actual time was 64 days.   That is in less absolute time than the 4-million-digit computation, which used just one node.

Six million digits of CMRB after eight tries in 19 months. (Search "8/24/2019 It's time for more digits!" below.) finishing on Tue, 30 Mar 2021, at 22:02:49 in 160 days.
    The MRB constant supercomputer 2 said the following:
    Finished on Tue 30 Mar 2021, 22:02:49. computation and absolute time were
    5.28815859375*10^6 and 1.38935720536301*10^7 s. respectively
    Enter MRB1 to print 6029991 digits. The error from a 5,000,000 or more-digit calculation that used a different method is      
    0.*10^-5024993.

That means that the 5,000,000-digit computation Was accurate to 5024993 decimals!!!

enter image description here


5,609,880, verified by two distinct algorithms for x^(1/x), digits of CMRB on Thu 4 Mar 2021 at 08:03:45. The 5,500,000+ digit computation using a totally different method showed that many decimals are in common with the 6,000,000+ digit computation in 160.805 days.

6,500,000 digits of CMRB on my second try,

Successful code was:

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

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

In[4]:= Print["Start time is ", ds = DateString[], "."];
prec = 6500000;
(**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/396288];
   Do[xvals = Flatten[Parallelize[Table[Table[ll = start + j*tsize + l;
         x = N[E^(Log[ll]/(ll)), iprec];
         pc = iprec;
         While[pc < pr/65536, pc = Min[3 pc, pr/65536];
          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/99072]**)
         x = SetPrecision[x, pr/16384];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/16384] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/16384] ll (ll - 1) 1/(3 ll t2 + t^3 z));(*N[Exp[Log[
         ll]/ll],pr/4096]*)x = SetPrecision[x, pr/4096];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/4096] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/4096] ll (ll - 1) 1/(3 ll t2 + t^3 z));(*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);
    If[kc > 1, 
     Print["As of  ", DateString[], " there were ", kc, 
      " iterations done in ", N[st, 5], " seconds. That is ", 
      N[kc/st, 5], " iterations/s. ", N[kc/(end*chunksize)*100, 7], 
      "% complete.", " It should take ", N[ti, 6], " days or ", 
      N[ti*24*3600, 4], "s, and finish ", DatePlus[ds, ti], "."]];
    Print[];, {k, 0, end - 1}];
   N[-s/d, pr]];
t2 = Timing[MRB1 = expM[prec];]; Print["Finished on ", 
 DateString[], ". Proccessor and actual time were ", t2[[1]], " and ",
  SessionTime[] - T0, " s. respectively"];
Print["Enter MRB1 to print ", 
 Floor[Precision[
   MRB1]], " digits. The error from a 5,000,000 or more digit \
calculation that used a different method is  "]; N[M6M - MRB1, 20]

enter image description here

The MRB constant supercomputer replied,

Finished on Wed 16 Mar 2022 02: 02: 10. computation and absolute time
were 6.26628*10^6 and 1.60264035419592*10^7s respectively Enter MRB1
to print 6532491 digits. The error from a 6, 000, 000 or more digit
the calculation that used a different method is 
0.*10^-6029992.

"Computation time" 72.526 days.

 "Absolute time" 185.491 days.

It would have taken my first computer, a TRS-80 at least 4307 years with today's best mathematical algorithms. 15 GHz/1.77 MHZ 185.491 days1 year/(365 days) enter image description here

It was checked in 6.9 hours: CTRL+f "Where MRB 1 is my record 6,500,000 digit computation of MRB constant digits, I found out that"


Next is "§0. Wolfram+AI is the quintessential team evaluating the MRB constant." Else refresh.






POSTED BY: Marvin Ray Burns

If, above this, you don't see "Next is "§0. Wolfram+AI is the quintessential team evaluating the MRB constant.", you'll need to refresh the page to see the MRB constant records.

§0. Wolfram+AI is the quintessential team evaluating the MRB constant.

partner

(CC)

Rhapsody in MRB by GPT

AI generated slide show with poetry.

Read aloud by Marvin Ray Burns

here enter image description here

```




The MRB constant CMRB equals enter image description here.

Bard AI pointed out a connection from Riemann Hypothesis to CMRB.

enter image description here

**In conclusion, the MRB constant and the Riemann hypothesis are two related but distinct concepts. The MRB constant is a mathematical constant that is defined in terms of the prime number distribution. The Riemann hypothesis is a conjecture about the distribution of the prime numbers. If the Riemann hypothesis is true, then it would have implications for the MRB constant. However, it is also possible that the Riemann hypothesis is false. If this is the case, then it would not be possible to say anything about the MRB constant based on the Riemann hypothesis. **


enter image description here

Example

Then concerning the whole series GPT 4 said, enter image description here



enter image description here FACTS Explain

CMRB

MTHWLD IMG MTWLD TXT


starting Ok. Here: BingAI should have said, 2^(1/2) - 1 = 0.4142135.

Instead, it calculated (1/2)^(1/2) - 1 = -1 + 1/Sqrt[2]=-0.292893 .


A haiku


Compare GPT with Bing in recognizing the MRB constant. CHAT Bing


Compare GPT with Bing in writing Mathematica code.

: GPT GPT faster1 GPT faster2 It showed "N[MRB[100], 100]" GPT altsgns1 GPT altsgns2 GPT altsgns3

Now, it's Bing AI's turn. enter image description here enter image description here enter image description here

With just a little tweaking and a lot of automating, the program says, It takes around 10^(n+/-0.2) partial sums s for n digits.


After receiving Wikipedia's and MathWorld's articles on the MRB constant (the upper limit point of the sequence of the above partial sums), Google OpenAI Chat CPT described the constant as follows.

...

Burns: Write a Wikipedia article for the MRB constant.

I'm sorry, as an AI language model, I cannot create or modify articles on Wikipedia. However, I can provide a draft of an article on the MRB constant that you can use as a starting point:

Chat AI 2 References:

  • Burns, M. R. "An Alternating Series Involving n^(th) Roots." Unpublished note, 1999.
  • Burns, M. R. "Try to Beat These MRB Constant Records!"
  • http://community.wolfram.com/groups/-/m/t/366628.
  • Crandall, R. E. "Unified Algorithms for Polylogarithm, L-Series, and Zeta Variants." 2012a.
  • http://www.marvinrayburns.com/UniversalTOC25.pdf.
  • Crandall, R. E. "The MRB Constant." §7.5 in Algorithmic Reflections: Selected Works. PSI Press, pp. 28-29, 2012b.
  • Finch, S. R. Mathematical Constants. Cambridge, England: Cambridge University Press, p. 450, 2003.
  • Plouffe, S. "MRB Constant." http://pi.lacim.uqam.ca/piDATA/mrburns.txt.
  • Sloane, N. J. A. Sequences A037077 in "The On-Line Encyclopedia of Integer Sequences."

GPT4 MRB1 GPT4 MRB2 GPT4 MRB3

POSTED BY: Marvin Ray Burns

Along the lines of how normal the MRB constant is, I noticed the following about its original sum,

$\displaystyle{\sum_{n=1}^{\infty}(-1)^n(n^{1/n}-1)}.$

I wondered, (what would happen to the sum if you added smaller and smaller steps?)

 ReImPlot[ NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/n},  WorkingPrecision -> 30, Method ->"AlternatingSigns"], {n, 1,  Infinity}]

enter image description here

`

  In[35]:= Re[
   NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^20}, 
    WorkingPrecision -> 70, Method -> "AlternatingSigns", 
    NSumTerms -> 2000]]

  Out[35]= -1.\
  999999749999999973333339999999998228052447182132376098872359604490555*\
  10^-14

  In[36]:= Re[
   NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^30}, 
    WorkingPrecision -> 70, Method -> "AlternatingSigns", 
    NSumTerms -> 2000]]

  Out[36]= -1.\
  999999749999999999999999997333333999999999999999999982280524444387501*\
  10^-24

  In[37]:= Re[
   NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^40}, 
    WorkingPrecision -> 70, Method -> "AlternatingSigns", 
    NSumTerms -> 2000]]

  Out[37]= -1.\
  999999749999999999999999999999999999733333399999999999999990562538808*\
  10^-34

  In[38]:= Re[
   NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^50}, 
    WorkingPrecision -> 70, Method -> "AlternatingSigns", 
    NSumTerms -> 2000]]

  Out[38]= -1.\
  999999749999999999999999999999999999999999999973243568040505517332039*\
  10^-44

In this case, finding exactly where the error starts is difficult for me. But let's assume the following is true.

In[5]:= Re[ NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^30}, WorkingPrecision -> 40, Method -> "AlternatingSigns"]]

Out[5]= -1.12250000000000000000000000002637761741*10^-28

In[6]:= Re[NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 2/10^40},  WorkingPrecision -> 50, Method -> "AlternatingSigns"]]

Out[6]= -2.244999999999999999999999999999999999982964553611*10^-38

  In[39]:= p =  Re[NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^200}, WorkingPrecision -> 1000, Method -> "AlternatingSigns"]]

Out[39]= \ -1.1224999999999999999999999999999999999999999999999999999999999999999\ 9999999999999999999999999999999999999999999999999999999999999999999999\ 9999999999999999999999999999999999999999999999999999999999999887999999\ 9999999999999999999999999999999999999999999999999999999999999999999999\ 9999999999999999999999999999999999999999999999999999999999999999999999\ 9999999999999999999999999999999999999999999999999951865743238994910613\ ...*10^-198

In[41]:= Rationalize[N[Q100 = 10^98 p + 1, 200], 0]]

$$\frac{8908685968819599109131403118040089086859688195991091314031180400890868596881959910913140311804008907}{8908685968819599109131403118040089086859688195991091314031180400890868596881959910913140311804008908}$$

I think it's worth pointing out that the rational approximation consists, entirely, of repeating $$89086859688195991091314031180400's$$ except for the last digit of the numerator.

Some steps of multiples of powers of 10 also give solutions with many repeating 9s.

In[59]:= N[ Re[NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^40}, WorkingPrecision -> 1000, Method -> "AlternatingSigns"]], 100]

Out[59]=-1.122499999999999999999999999999999999998879999999999999999999999999999999999995186574323899491061367*10^-38

In[54]:= N[Re[NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/20^40}, WorkingPrecision -> 1000, Method ->"AlternatingSigns"]], 55]

Out[54]= \ -1.020907802740111947059631347656249999999999999999999074*10^-50

In[69]:= N[Re[NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/50^40}, WorkingPrecision -> 1000, Method -> "AlternatingSigns"]], 150]

Out[69]= \ -1.2342018021785599999999999999999999999999999999999999999999999999998\ 6460030820316153243290828799999999999999999999999999999999999999993601\ 859835697267*10^-66

In[84]:= N[ Re[NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/200^40}, WorkingPrecision -> 1000, Method -> "AlternatingSigns"]], 200]

Out[84]= \ -1.0209078027401119470596313476562499999999999999999999999999999999999\ 9999999999999999999999990735577139406090041440222648816416040062904357\ 91015624999999999999999999999999999999999999637878099817017722*10^-90

I posted it at https://math.stackexchange.com/questions/4575530/sum-infty-textstylex-11-k-atopk-in-1-ldots-infty-1x-x1 to see if anyone can explain it.

I found the following that shows a little bit of what happens for smaller and smaller steps of powers 2 and 5, which are related to powers 10.

By a factor of pi, the imaginary part has at least as many repetitions of decimals.

In

N[Table[Im[
    NSum[(-1)^x (x^(1/x) - 1), {x, 1, Infinity, 1/10^n}, 
     WorkingPrecision -> 180, Method -> "AlternatingSigns"]], {n, 20, 
    50}], 100]/Pi

Out[216]= \
{-1.014999999999999999889749999999999999985380680380635051103404220561\
131468438815891294117640691851961*10^-37, \
-1.0149999999999999999889749999999999999998538068038063505110089362727\
15953551936726556669955860834527*10^-39, \
-1.0149999999999999999988974999999999999999985380680380635051100642567\
94264174386914539123932192756267*10^-41, \
-1.0149999999999999999998897499999999999999999853806803806350511006174\
62009746382736692535387904947180*10^-43, \
-1.0149999999999999999999889749999999999999999998538068038063505110061\
49514164568466234472485086767530*10^-45, \
-1.0149999999999999999999988974999999999999999999985380680380635051100\
61470035712789301212271980781307*10^-47, \
-1.0149999999999999999999998897499999999999999999999853806803806350511\
00614675251194997650990266937597*10^-49, \
-1.0149999999999999999999999889749999999999999999999998538068038063505\
11006146727406017081148770216506*10^-51, \
-1.0149999999999999999999999988974999999999999999999999985380680380635\
05110061467248954237916126569712*10^-53, \
-1.0149999999999999999999999998897499999999999999999999999853806803806\
35051100614672464436446265904565*10^-55, \
-1.0149999999999999999999999999889749999999999999999999999998538068038\
06350511006146724619258529763685*10^-57, \
-1.0149999999999999999999999999988974999999999999999999999999985380680\
38063505110061467246167479364741*10^-59, \
-1.0149999999999999999999999999998897499999999999999999999999999853806\
80380635051100614672461649687715*10^-61, \
-1.0149999999999999999999999999999889749999999999999999999999999998538\
06803806350511006146724616471771*10^-63, \
-1.0149999999999999999999999999999988974999999999999999999999999999985\
38068038063505110061467246164693*10^-65, \
-1.0149999999999999999999999999999998897499999999999999999999999999999\
85380680380635051100614672461647*10^-67, \
-1.0149999999999999999999999999999999889749999999999999999999999999999\
99853806803806350511006146724616*10^-69, \
-1.0149999999999999999999999999999999988974999999999999999999999999999\
99998538068038063505110061467246*10^-71, \
-1.0149999999999999999999999999999999998897499999999999999999999999999\
99999985380680380635051100614672*10^-73, \
-1.0149999999999999999999999999999999999889749999999999999999999999999\
99999999853806803806350511006147*10^-75, \
-1.0149999999999999999999999999999999999988974999999999999999999999999\
99999999998538068038063505110061*10^-77, \
-1.0149999999999999999999999999999999999998897499999999999999999999999\
99999999999985380680380635051101*10^-79, \
-1.0149999999999999999999999999999999999999889749999999999999999999999\
99999999999999853806803806350511*10^-81, \
-1.0149999999999999999999999999999999999999988974999999999999999999999\
99999999999999998538068038063505*10^-83, \
-1.0149999999999999999999999999999999999999998897499999999999999999999\
99999999999999999985380680380635*10^-85, \
-1.0149999999999999999999999999999999999999999889749999999999999999999\
99999999999999999999853806803806*10^-87, \
-1.0149999999999999999999999999999999999999999988974999999999999999999\
99999999999999999999998538068038*10^-89, \
-1.0149999999999999999999999999999999999999999998897499999999999999999\
99999999999999999999999985380680*10^-91, \
-1.0149999999999999999999999999999999999999999999889749999999999999999\
99999999999999999999999999853807*10^-93, \
-1.0149999999999999999999999999999999999999999999988974999999999999999\
99999999999999999999999999998538*10^-95, \
-1.0149999999999999999999999999999999999999999999998897499999999999999\
99999999999999999999999999999985*10^-97}

Here is an answer to the previous question.

While mathematicians try to crack this nut, here's a physicist's point of view. I will focus on how to calculate this integral, keeping things as simple as possible, probably approximately.

his answer

This is an example of that how we can use diverging series to compute values. It is only important to guess where to truncate the series.

POSTED BY: Marvin Ray Burns

Programs to compute the integrated analog

The efficient programs

Wed 29 Jul 2015 11:40:10

From an initial accuracy of only 7 digits,

0.07077603931152880353952802183028200137`19.163032309866352 - 
 0.68400038943793212918274445999266112671`20.1482024033675 I - \
(NIntegrate[(-1)^t (t^(1/t) - 1), {t, 1, Infinity}, 
    WorkingPrecision -> 20] - 2 I/Pi)

enter image description here

we have the first efficient program to compute the integrated analog (MKB) of the MRB constant, which is good for 35,000 digits.

Block[{$MaxExtraPrecision = 200}, prec = 4000; f[x_] = x^(1/x);
 ClearAll[a, b, h];
 Print[DateString[]];
 Print[T0 = SessionTime[]];

 If[prec > 35000, d = Ceiling[0.002 prec], 
  d = Ceiling[0.264086 + 0.00143657 prec]];

 h[n_] := 
  Sum[StirlingS1[n, k]*
    Sum[(-j)^(k - j)*Binomial[k, j], {j, 0, k}], {k, 1, n}];

 h[0] = 1;
 g = 2 I/Pi - Sum[-I^(n + 1) h[n]/Pi^(n + 1), {n, 1, d}];

 sinplus1 := 
  NIntegrate[
   Simplify[Sin[Pi*x]*D[f[x], {x, d + 1}]], {x, 1, Infinity}, 
   WorkingPrecision -> prec*(105/100), 
   PrecisionGoal -> prec*(105/100)];

 cosplus1 := 
  NIntegrate[
   Simplify[Cos[Pi*x]*D[f[x], {x, d + 1}]], {x, 1, Infinity}, 
   WorkingPrecision -> prec*(105/100), 
   PrecisionGoal -> prec*(105/100)];

 middle := Print[SessionTime[] - T0, " seconds"];

 end := Module[{}, Print[SessionTime[] - T0, " seconds"];
   Print[c = Abs[a + b]]; Print[DateString[]]];


 If[Mod[d, 4] == 0, 
  Print[N[a = -Re[g] - (1/Pi)^(d + 1)*sinplus1, prec]];
  middle;
  Print[N[b = -I (Im[g] - (1/Pi)^(d + 1)*cosplus1), prec]];
  end];


 If[Mod[d, 4] == 1, 
  Print[N[a = -Re[g] - (1/Pi)^(d + 1)*cosplus1, prec]];
  middle;
  Print[N[b = -I (Im[g] + (1/Pi)^(d + 1)*sinplus1), prec]]; end];

 If[Mod[d, 4] == 2, 
  Print[N[a = -Re[g] + (1/Pi)^(d + 1)*sinplus1, prec]];
  middle;
  Print[N[b = -I (Im[g] + (1/Pi)^(d + 1)*cosplus1), prec]];
  end];

 If[Mod[d, 4] == 3, 
  Print[N[a = -Re[g] + (1/Pi)^(d + 1)*cosplus1, prec]];
  middle;
  Print[N[b = -I (Im[g] - (1/Pi)^(d + 1)*sinplus1), prec]];
  end];]

May 2018

I got substantial improvement in calculating the digits of MKB by using V11.3 in May 2018, my new computer (processor Intel(R) Core(TM) i7-7700 CPU @ 3.60GHz, 3601 MHz, 4 Core(s), 8 Logical Processor(s) with 16 GB 2400 MH DDR4 RAM):

Digits  Seconds
2000    67.5503022
3000    217.096312
4000    514.48334
5000    1005.936397
10000   8327.18526
 20000  71000

They are found in the attached 2018 quad MKB.nb.

They are twice as fast,(or more) as my old records with the same program using Mathematica 10.2 in July 2015 on my old big computer (a six-core Intel(R) Core(TM) i7-3930K CPU @ 3.20 GHz 3.20 GHz with 64 GB of 1066 MHz DDR3 RAM):

digits          seconds

2000    256.3853590 
3000    794.4361122
4000       1633.5822870
5000        2858.9390025
10000      17678.7446323 
20000      121431.1895170
40000       I got error msg

May 2021

After finding the following rapidly converging integral for MKB, enter image description here

(See Primary Proof 3 in the first post.)

I finally computed 200,000 digits of MKB (0.070776 - 0.684 I...) Started ‎Saturday, ‎May ‎15, ‎2021, ‏‎10: 54: 17 AM, and finished at 9:23:50 am EDT | Friday, August 20, 2021, for a total of 8.37539*10^6 seconds or 96 days 22 hours 29 minutes 50 seconds.

The full computation, verification to 100,000 digits, and hyperlinks to various digits are found below at 200k MKB A.nb. The code was

g[x_] = x^(1/x); u := (t/(1 - t)); Timing[
 MKB1 = (-I Quiet[
      NIntegrate[(g[(1 + u I)])/(Exp[Pi u] (1 - t)^2), {t, 0, 1}, 
       WorkingPrecision -> 200000, Method -> "DoubleExponential", 
       MaxRecursion -> 17]] - I/Pi)]

enter image description here

After finding the above more rapidly converging integral for MKB, In only 80.5 days, 189,330 real digits and 166,700 imaginary were confirmed to be correct by the following different formula. as Seen at https://www.wolframcloud.com/obj/bmmmburns/Published/2nd%20200k%20MRB.nb

All digits at

https://www.wolframcloud.com/obj/bmmmburns/Published/200K%20confirmed%20MKB.nb (Recommended to open in desktop Mathematica.)

N[(Timing[
   FM2200K - (NIntegrate[(Exp[Log[t]/t - Pi t/I]), {t, 1, Infinity I},
        WorkingPrecision -> 200000, Method -> "Trapezoidal", 
       MaxRecursion -> 17] - I/Pi)]), 20]

enter image description here

I've learned more about what MaxRecusion is required for 250,000 digits to be verified from the two different formulas, and they are being computed as I write. It will probably take over 100 days.

Laurent series for the analog

I've not perfected the method, but here is how to compute the integrated analog of the MRB constant from series.

$f = (-1)^z (z^(1/z) - 1); MKB = 
 NIntegrate[$f, {z, 1, Infinity I}, WorkingPrecision -> 500]; 
Table[s[x_] = Series[$f, {z, n, x}] // Normal; 
  Timing[Table[
    MKB - Quiet[ 
      NIntegrate[s[x] /. z -> n, {n, 1, Infinity I}, 
       WorkingPrecision -> p, Method -> "Trapezoidal", 
       MaxRecursion -> Ceiling[Log2[p/2]]]], {p, 100, 100 x, 
     100}]], {x, 1, 10}] // TableForm

enter image description here

Table[Short[s[n]], {n, 1, 5}] // TableForm enter image description here

Attachments:
POSTED BY: Marvin Ray Burns

Time for a quick memorial:

This discussion began on 1/20/2014.

"This MRB records posting reached a milestone of over 120,000 views on 3/31/2020, around 4:00 am."

"As of 04:00 am 1/2/2021, this discussion had 300,000 views!"

"And as of 08:30 pm 2/3/2021, this discussion had 330,000 views!"

"7:00 pm 10/8/2021 it had 520,000 views!"

1:40 am 3/2/2022 600,000 views

8:25 pm 5/4/2022 650,000 views

In the last seven months, this discussion has had as many visitors as it did in its first seven years!

1/20/2023 695,000 views in nine years. That's an average of 8.8 views/hour, or one view every 6.8 minutes.

1/15/2024 810,223 views in ten years. That's an average of 9.24 views/hour, or one view every 6.49 minutes.

POSTED BY: Marvin Ray Burns

...including all the methods used to compute CMRB and their efficiency.

While waiting for results on the 2nd try of calculating 6,500,000 digits of the MRB constant (CMRB), I thought I would compare the convergence rate of 3 different primary forms of it. They are listed from slowest to fastest.

POSTED BY: Marvin Ray Burns

WOW!!!!

I discovered a non-trivial infinitude of proper integrals that all equal the MRB constant (CMRB): enter image description here

Maybe a few more restrictions, like a≠b.

See cloud notebook.

enter image description here

g[x_] = x^(1/x); CMRB = NSum[(-1)^k (g[k] - 1), {k, 1, Infinity}, 
     WorkingPrecision -> 100, Method -> "AlternatingSigns"];

In[239]:= g[x_] = x^(1/x); Table[w = (I (t - b))/(t - a);
 CMRB - NIntegrate[
   Re[g[(1 + w)] Csc[\[Pi] w]] (t - a)^-2*(b - a), {t, a, b}, 
      WorkingPrecision -> 100], {a, 0, 5}, {b, a + 1, 6}]

Out[239]= {{-9.3472*10^-94, -9.3472*10^-94, -9.3472*10^-94, \
-9.3472*10^-94, -9.3472*10^-94, -9.3472*10^-94}, {-9.3472*10^-94, \
-9.3472*10^-94, -9.3472*10^-94, -9.3472*10^-94, -9.3472*10^-94}, \
{-9.3472*10^-94, -9.3472*10^-94, -9.3472*10^-94, -9.3472*10^-94}, \
{-9.3472*10^-94, -9.3472*10^-94, -9.3472*10^-94}, {-9.3472*10^-94, \
-9.3472*10^-94}, {-9.3472*10^-94}}

In[240]:= g[x_] = x^(1/x); Table[w = (I (t - b))/(t - a);
 CMRB - NIntegrate[
   Re[g[(1 + w)] Csc[\[Pi] w]] (t - a)^-2*(b - a), {t, a, b}, 
      WorkingPrecision -> 100], {a, 4/10, 5}, {b, a + 1, 6}]

Out[240]= {{-9.3472*10^-94, -9.3472*10^-94, -9.3472*10^-94, \
-9.3472*10^-94, -9.3472*10^-94}, {-9.3472*10^-94, -9.3472*10^-94, \
-9.3472*10^-94, -9.3472*10^-94}, {-9.3472*10^-94, -9.3472*10^-94, \
-9.3472*10^-94}, {-9.3472*10^-94, -9.3472*10^-94}, {-9.3472*10^-94}}

In[234]:= a = E; b = Pi;

In[254]:= a = E; b = Pi; g[x_] = x^(1/x); (w = (I (t - b))/(t - a);
 Print[CMRB - 
   NIntegrate[
    Re[g[(1 + w)] Csc[\[Pi] w]] (t - a)^-2*(b - a), {t, a, b}, 
    WorkingPrecision -> 100]]); Clear[a, b]

During evaluation of In[254]:= -9.3472*10^-94

In[260]:= a = 1; b = I; g[x_] = x^(1/x); (w = (I (t - b))/(t - a);
 Print[CMRB - 
   NIntegrate[
    Re[g[(1 + w)] Csc[\[Pi] w]] (t - a)^-2*(b - a), {t, a, b}, 
    WorkingPrecision -> 100]]); Clear[a, b]

During evaluation of In[260]:= -9.3472*10^-94+0.*10^-189 I
POSTED BY: Marvin Ray Burns

Beyond any shadow of a doubt, I verified 5,609,880 digits of the MRB constant on Thu 4 Mar 2021 08:03:45. The 5,500,000+ digit computation using a totally different method showed about that many decimals in common with the 6,000,000+ digit computation. The method for the 6,000,000 run is found in a few messages above in the attached notebook titled "MRBSC2 6 million...nb."

Print["Start time is ", ds = DateString[], "."];
prec = 6000000;
(**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/396288];
   Do[xvals = Flatten[Parallelize[Table[Table[ll = start + j*tsize + l;
         x = N[E^(Log[ll]/(ll)), iprec];

       (**N[Exp[Log[ll]/ll],pr/396288]**)


         pc = iprec;
         While[pc < pr/65536, pc = Min[3 pc, pr/65536];
          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/65536]**)

         x = SetPrecision[x, pr/16384];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/16384] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/16384] ll (ll - 1) 1/(3 ll t2 + t^3 z));

             (*N[Exp[Log[ll]/ll],pr/16384]*)


          x = SetPrecision[x, pr/4096];
         xll = x^ll; z = (ll - xll)/xll;
         t = 2 ll - 1; t2 = t^2;
         x = 
          x*(1 + SetPrecision[4.5, pr/4096] (ll - 1)/
               t2 + (ll + 1) z/(2 ll t) - 
             SetPrecision[13.5, 
               pr/4096] ll (ll - 1) 1/(3 ll t2 + t^3 z));

         (*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);
    If[kc > 1, 
     Print["As of  ", DateString[], " there were ", kc, 
      " iterations done in ", N[st, 5], " seconds. That is ", 
      N[kc/st, 5], " iterations/s. ", N[kc/(end*chunksize)*100, 7], 
      "% complete.", " It should take ", N[ti, 6], " days or ", 
      N[ti*24*3600, 4], "s, and finish ", DatePlus[ds, ti], "."]];
    Print[];, {k, 0, end - 1}];
   N[-s/d, pr]];
t2 = Timing[MRB1 = expM[prec];]; Print["Finished on ", 
 DateString[], ". Proccessor and actual time were ", t2[[1]], " and ",
  SessionTime[] - T0, " s. respectively"];
Print["Enter MRB1 to print ", 
 Floor[Precision[
   MRB1]], " digits. The error from a 5,000,000 or more digit \
calculation that used a different method is  "]; N[MRB - MRB1, 20]

The 5,500,000+digit run is found below in the attached "5p5million.nb," including the verified 5,609,880 digits.

(*Fastest (at RC's end) as of 30 Nov 2012.*)prec = 5500000;(*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];

      (*N[Exp[Log[ll]/ll], pr/27]*)

        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];];
N[MRBtest2 - MRB, 20]
Attachments:
POSTED BY: Marvin Ray Burns

...including the dispersion of the 0-9th decimals in CMRB decimal expansions.

Distribution of digits

Here is the distribution of digits within the first 6,000,000 decimal places (.187859,,,), "4" shows up more than other digits, followed by "0," "8" and "7."

enter image description here

Here is the distribution of digits within the first 5,000,000 decimal places (.187859,,,), "4" shows up a lot more than other digits, followed by "0," "8" and "6." enter image description here

Here is a similar distribution over the first 4,000,000: enter image description here

3,000,000 digits share a similar distribution:

enter image description here

Over the first 2 and 1 million digits "4" was not so well represented. So, the heavy representation of "4" is shown to be a growing phenomenon from 2 million to 5 million digits. However, "1,2,and 5" still made a very poor showing: enter image description here

I attached more than 6,000,000 digits of the MRB constant.

Attachments:
POSTED BY: Marvin Ray Burns

I DECLARE VICTORY!

I computed 6,000,000 digits of the MRB constant, finishing on Tue 30 Mar 2021 22:02:49. The MRB constant supercomputer 2 said the following:

  Finished on Tue 30 Mar 2021 22:02:49. Processor and actual time were 5.28815859375*10^6 and 1.38935720536301*10^7 s. respectively

  Enter MRB1 to print 6029991 digits. The error from a 5,000,000 or more digit calculation that used a different method is  

  0.*10^-5024993

That means that the 5,000,000 digit computation was actually accurate to 5024993 decimals!!!

For the complete blow-by-blow see MRBSC2 6 million 1st fourth.nb.

Attachments:
POSTED BY: Marvin Ray Burns

In the same vein of a non-trivial approximation:

The MRB constant =m.

m = NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, 
   WorkingPrecision -> 30, Method -> "AlternatingSigns"];

We know from previous posts, M=integral

To find how fast this integral converges to m, I entered

lsmall = Table[
  m - NIntegrate[Csch[\[Pi] t] Im[(1 + I t)^(1/(1 + I t))], {t, 0, x},
     WorkingPrecision -> 40], {x, 1, 20}]; N[Ratios[1/lsmall], 15]

followed by the slower

lbig = Table[
  m - NIntegrate[Csch[\[Pi] t] Im[(1 + I t)^(1/(1 + I t))], {t, 0, x},
     WorkingPrecision -> 240], {x, 101, 120}]; N[Ratios[1/lbig], 15]

and ended up with values that converged to

23.2958716448535.

Using those ratios, it looks like

m-int/m-int

Since that is a form of 0/0, we can use L'Hospital's Rule and see that it is true:

Limit[(Csch[\[Pi] t] Im[(1 + I t)^(1/(
     1 + I t))])/(Csch[\[Pi] (t + 1)] Im[(1 + I (t + 1))^(1/(
     1 + I ( t + 1)))]), t -> Infinity]

gives

etopi

Now for the part that is similar to the approximations mentioned in the immediate, previous post.

You may know that

N[20/(E^Pi - Pi), 9]

gives

1.00004500.

While

N[2 - 100/(E^Pi/m - E^Pi), 8]

gives

1.0004005.

Furthermore,

N[2 - 100/(E^Pi/m - E^Pi), 12]/N[20/(E^Pi - Pi), 12]

gives

1.00035544632.

While

N[3 - 100/(E^Pi/m - E^Pi), 12] - N[20/(E^Pi - Pi), 12]

gives

1.00035546231.

Holy double, double vision Batman!

There is a way to make the approximations of 20/(E^Pi - Pi) vs 2 - 100/(E^Pi/x - E^Pi), where x is a relation to m, equal to 11 digits, instead of just similar to 9,

using the parking constant. E^Pi is involved in a term again! (The term is a Cosmological constant, of sorts.)

(*parking constant,c*)
c = NIntegrate[
   E^(-2*(EulerGamma + Gamma[0, t] + Log[t])), {t, 0, Infinity}, 
   WorkingPrecision -> 50, MaxRecursion -> 20];

Let ma be the MRB constant with that term:

ma = m + (2/3 - E^Pi)/(1/3 - c)/10^6;

Calculating more digits of 20/(E^Pi - Pi):

N[20/(E^Pi - Pi), 12]

gives

1.00004500307.

Calculating the new sum with ma instead of m:

N[2 - 100/(E^Pi/ma - E^Pi), 12]

gives

1.00004500308.

What about the second of the similar-looking approximated sums,

N[2 - 100/(E^Pi/x - E^Pi), 15]/N[20/(E^Pi - Pi), 15] vs.N[3 - 100/(E^Pi/x - E^Pi), 16] - N[20/(E^Pi - Pi), 16]?

So we have the new x, ma. Then

N[2 - 100/(E^Pi/ma - E^Pi), 15]/N[20/(E^Pi - Pi), 15]

gives

1.00000000001744.

While

N[3 - 100/(E^Pi/ma - E^Pi), 16] - N[20/(E^Pi - Pi), 16]

gives the exactly the same value of

1.000000000017440.

Holy corrected vision Batman!

POSTED BY: Marvin Ray Burns

...including arbitrarily close approximation formulas for CMRB

m=the MRB constant. We looked at how n^m-m is similar to E^Pi-Pi (a near integer). One might think this is off the subject of breaking computational records of the MRB constant, but it also could help show whether a closed-form exists for computing and checking the digits of m from n^m-m=a near integer and n is an integer.

So, I decided to make an extremely deep search of the n^m-m=a near integer, and n is an integer field. Here are the pearls I gleaned:

In[35]:= m = 
  NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, WorkingPrecision -> 100,
    Method -> "AlternatingSigns"];

In[63]:= 225897077238546^m - m

Out[63]= 496.99999999999999975304752932252481772179797865

In[62]:= 1668628852566227424415^m - m

Out[62]= 9700.9999999999999999994613109586919797992822178

In[61]:= 605975224495422946908^m - m

Out[61]= 8019.9999999999999999989515156294756517433387956

In[60]:= 3096774194444417292742^m - m

Out[60]= 10896.0000000000000000000000096284579090392932063

In[56]:= 69554400815329506140847^m - m

Out[56]= 19549.9999999999999999999999991932013520540825206

In[68]:= 470143509230719799597513239^m - m

Out[68]= 102479.000000000000000000000000002312496475978584

In[70]:= 902912955019451288364714851^m - m

Out[70]= 115844.999999999999999999999999998248770510754951

In[73]:= 2275854518412286318764672497^m - m

Out[73]= 137817.000000000000000000000000000064276966095482

In[146]:= 2610692005347922107262552615512^m - m

Out[146]= 517703.00000000000000000000000000000013473353420

In[120]:= 9917209087670224712258555601844^m - m

Out[120]= 665228.00000000000000000000000000000011062183643

In[149]:= 19891475641447607923182836942486^m - m

Out[149]= 758152.00000000000000000000000000000001559954712

In[152]:= 34600848595471336691446124576274^m - m

Out[152]= 841243.00000000000000000000000000000000146089062

In[157]:= 543136599664447978486581955093879^m - m

Out[157]= 1411134.0000000000000000000000000000000035813431

In[159]:= 748013345032523806560071259883046^m - m

Out[159]= 1498583.0000000000000000000000000000000031130944

In[162]:= 509030286753987571453322644036990^m - m

Out[162]= 1394045.9999999999999999999999999999999946679646


In[48]:= 952521560422188137227682543146686124^m - m

Out[48]=5740880.999999999999999999999999999999999890905129816474332198321490136628009367504752851478633240


In[26]:= 50355477632979244604729935214202210251^m - m

Out[26]=12097427.00000000000000000000000000000000000000293025439870097812782596113788024271834721860892874


In[27]:= 204559420776329588951078132857792732385^m - m

Out[27]=15741888.99999999999999999999999999999999999999988648448116819373537316944519114421631607853700001


In[46]:= 4074896822379126533656833098328699139141^m - m

Out[46]= 27614828.00000000000000000000000000000000000000001080626974885195966380280626150522220789167201350


In[8]:= 100148763332806310775465033613250050958363^m - m

Out[8]= 50392582.999999999999999999999999999999999999999998598093272973955371081598246


In[10]=  116388848574396158612596991763257135797979^m - m

Out[10]=51835516.000000000000000000000000000000000000000000564045501599584517036465406


In[12]:= 111821958790102917465216066365339190906247589^m - m

Out[12]= 188339125.99999999999999999999999999999999999999999999703503169989535000879619


In[33] := 8836529576862307317465438848849297054082798140^m - m

Out[33] = 42800817.00000000000000000000000000000000000000000000000321239755400298680819416095288742420653229


In[71] := 532482704820936890386684877802792716774739424328^m - m

Out[71] =924371800.999999999999999999999999999999999999999999999998143109316148796009581676875618489611792


In[21]:= 783358731736994512061663556662710815688853043638^m - m

Out[21]= 993899177.0000000000000000000000000000000000000000000000022361744841282020


In[24]:= 8175027604657819107163145989938052310049955219905^m - m

Out[24]= 1544126008.9999999999999999999999999999999999999999999999999786482891477\
944981


19779617801396329619089113017251584634275124610667^m - m
gives
1822929481.00000000000000000000000000000000000000000000000000187580971544991111083798248746369560.


130755944577487162248300532232643556078843337086375^m - m

gives 

2599324665.999999999999999999999999999999999999999999999999999689854836245815499119071864529772632.
i.e.2, 599, 324, 665. 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 689

(51 consecutive 9 s)

322841040854905412176386060015189492405068903997802^m - m

gives

3080353548.000000000000000000000000000000000000000000000000000019866002281287395703598786588650156

i.e. 3, 080, 353, 548. 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000,019

(52 consecutive 0 s)


310711937250443758724050271875240528207815041296728160^m - m

gives

11195802709.99999999999999999999999999999999999999999999999999999960263763...
i.e. 11,195,802,709. 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 602, 637,63

(55 consecutive 9s)

1465528573348167959709563453947173222018952610559967812891154^ m - m  
gives 
200799291330.9999999999999999999999999999999999999999999999999999999999999900450730197594520134278  
i. e., 200, 799, 291, 330.999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 999, 99 

(62 consecutive 9 s).

Here is something that looks like it might lead to another form of arbitrarily close approximations.

enter image description here as in https://www.wolframcloud.com/obj/7238e6f0-6fa5-4015-aaf5-1cca5c2670ca

POSTED BY: Marvin Ray Burns

On 2/24/2020 at 4:35 pm, I started a 10,000 digit calculation of the MRB constant using the integral

enter image description here

Here is the code:

First, compute 10,000 digits using Mathematica's "AlternatingSigns" option.

ms = NSum[(-1)^n (n^(1/n) - 1), {n, 1, Infinity}, 
   Method -> "AlternatingSigns", WorkingPrecision -> 10000];

Then compute the integral.

Timing[mi = 
NIntegrate[
Csch[\[Pi] t] E^((t ArcTan[t])/(1 + t^2)) (1 + 
t^2)^(1/(2 + 2 t^2)) Sin[(2 ArcTan[t] - t Log[1 + t^2])/(2 + 
2 t^2)], {t, 0, \[Infinity]}, WorkingPrecision -> 5000, 
Method -> "Trapezoidal", PrecisionGoal -> 10000, 
MaxRecursion -> 50]]

It is still working now on 2/26/2020 at 6:05 pm.

I messed up, but I'll let the computation complete anyway.

(My integral's result will only have around 5000 digits of precision -- so I should expect it to only be that accurate when I compare it to the sum.) But, this computation will give the approximate time required for a 10,000 digit calculation with that MaxRecursion (which might be way more than enough!)

It is still running at 7:52 am on 2/27/2020. The computer has been running at right around 12 GB of RAM committed and 9 GB of RAM in use, since early in the computation.

I started a second calculation on a similar computer. This one will be faster and give us a full 10,000 digits. But I reduced the MaxRecursion somewhat significantly. We'll see if all 10 k digits are right...

code

Timing[mi = 
  NIntegrate[
   Csch[\[Pi] t] E^((t ArcTan[t])/(1 + t^2)) (1 + 
       t^2)^(1/(2 + 2 t^2)) Sin[(2 ArcTan[t] - t Log[1 + t^2])/(2 + 
        2 t^2)], {t, 0, \[Infinity]}, WorkingPrecision -> 10000, 
   Method -> "Trapezoidal", PrecisionGoal -> 10000, 
   MaxRecursion -> 35]]

That lower threshold for MaxRecursion worked just fine!!!!!!!!!!!!!!! It took only 7497.63 seconds (roughly 2 hours) to calculate 10,000 accurate digits of the MRB constant from the integral.

2/27/2020 at 9:15 PM:

I just now started15,000 and a 20,000 digit computations of the integral form of the MRB constant. The 15,000 digit calculation of the MRB constant through the integral,enter image description here finished in 15,581s (4.328 hours) and was correct to all 15,000 digits!!!!!!!

I also calculated 20,000 correct digits in 51,632s (14.34 hr) using the integral code

Timing[mi = 
  NIntegrate[
   Csch[\[Pi] t] E^((t ArcTan[t])/(1 + t^2)) (1 + 
       t^2)^(1/(2 + 2 t^2)) Sin[(2 ArcTan[t] - t Log[1 + t^2])/(2 + 
        2 t^2)], {t, 0, \[Infinity]}, WorkingPrecision -> 20000, 
   Method -> "Trapezoidal", PrecisionGoal -> 20000, 
   MaxRecursion -> 30]]

Furthermore, I calculated 25,000 correct digits in 77,212.9s (21.45 hr) using the integral code

Timing[mi = 
  NIntegrate[
   Csch[\[Pi] t] E^((t ArcTan[t])/(1 + t^2)) (1 + 
       t^2)^(1/(2 + 2 t^2)) Sin[(2 ArcTan[t] - t Log[1 + t^2])/(2 + 
        2 t^2)], {t, 0, \[Infinity]}, WorkingPrecision -> 25000, 
   Method -> "Trapezoidal", PrecisionGoal -> 25000, 
   MaxRecursion -> 30]]

I think that does wonders to confirm the true approximated value of the constant. As calculated by both

enter image description here and enter image description here to at least 25,000 decimals, the true value of the MRB constant is

ms=mi≈ [Attached "MRB to 25k confirmed digits.txt"].

Computation and check of 25k digit integral calculation found in "comp of 25k confirmed digits.nb".

As 0f March 2, 2020, I'm working on timed calculations of 30k,50k and 100k digits of the integral. enter image description here

I finished a 30,000 accurate digit computation of the MRB constant via an integral in 78 hours. See "comp of 25k and 30k confirmed digits b.nb" for the digits and program.

Also, I finished a 50,000 accurate digit computation of the MRB constant via an integral in 6.48039 days. See "up to 50k digits of a MRB integral.nb" for the digits and program.

POSTED BY: Marvin Ray Burns

In 38 1/2 days, I computed 100,000 digits of the MRB constant from the enter image description here

Here is the code:

Timing[mi = 
  NIntegrate[
   Csch[\[Pi] t] E^((t ArcTan[t])/(1 + t^2)) (1 + 
       t^2)^(1/(2 + 2 t^2)) Sin[(2 ArcTan[t] - t Log[1 + t^2])/(2 + 
        2 t^2)], {t, 0, \[Infinity]}, WorkingPrecision -> 100000, 
   Method -> "Trapezoidal", PrecisionGoal -> 100000, 
   MaxRecursion -> 30]]

I attached the notebook with the results.

Attachments:
POSTED BY: Marvin Ray Burns

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:
POSTED BY: Marvin Ray Burns

nice system!

POSTED BY: l van Veen

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

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

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

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:
POSTED BY: Marvin Ray Burns

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

Upon reading them, Google OpenAI Chat CPT wrote the following reply:

enter image description here

I computed and checked 500 digits of the MRB constant, 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 want 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

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

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
POSTED BY: Marvin Ray Burns

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

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

Nice work. Worth a bit of excitement, I' d say.

POSTED BY: Daniel Lichtblau

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

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

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

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
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