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Try to beat these MRB constant records!

Posted 9 years ago
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This author is one of the average people who had a eureka moment and thought he found something special that smarter and more educated people have overlooked. However, he wants to make it clear that he is aware that there are experts in this field who have spent years studying and researching this topic (the MRB constant). While his discovery may seem novel to him, it is entirely possible that others have already explored and considered the idea. Nonetheless, he is excited to share these findings and hopes to contribute to the ongoing conversation and exploration of this topic. Join in his numeric odyssey below.

In his most recent investigation, the researcher explored the potential of Artificial Intelligence to shed light on the topic (the MRB constant). However, given his experience, he approached this with caution and followed a set of guidelines to ensure the accuracy and reliability of the AI-generated responses. While recognizing the limitations and potential errors of AI, he found that they could supply valuable original insights beyond his ability. Nevertheless, he exercised careful fact-checking and only included a small percentage of the AI-generated responses, employing techniques such as occasionally rewriting requests, assuredly referencing reliable sources, and especially using Mathematica to verify the math.

According to Bard AI,

There are a few things that are unknown about the MRB constant. First, there is no known closed-form expression for the constant. This means that it cannot be expressed in terms of a finite number of elementary functions. Second, it is not known whether the MRB constant is algebraic or transcendental. An algebraic constant is one that can be expressed as the root of a polynomial with rational coefficients. A transcendental constant is one that is not algebraic. Third, it is not known whether the MRB constant is rational or irrational. A rational constant is one that can be expressed as a fraction of two integers. An irrational constant is one that is not rational. Despite these unknowns, the MRB constant has been studied extensively by mathematicians and computer scientists. It has been shown to have a number of interesting properties, including the following:

  • The MRB constant is related to the divergent series:

\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{1/n}} = -\frac{1}{2} +\text{MRB constant}

[sic Referencing this discussion.] * The MRB constant can be used to define a new number system, called the MRB numbers.

[sic Referencing this discussion.] * The MRB constant has been used to construct new fractals.

The MRB constant is a fascinating mathematical object that is still under investigation. It is possible that further research will reveal even more interesting properties about the constant.

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

For section A, where all CMRB records have been checked, highlight text between quotes in

"§A. Possibly the most epigrammatic closed forms for checking up to 10^10 digits of CMRB!"

Then push

CTRL+f

keys on your keyboard and then

twice.



For section 0, written entirely from Ai responses, highlight text between quotes in

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

Then push

CTRL+f

keys on your keyboard and then

twice.




POSTED BY: Marvin Ray Burns.

If you see this text, the images are not showing. Refresh the page. The first 100 partial sums of {the CMRB series.}


For the best viewing, wait a minute until the word LaTeX in the LaTex script is centered below.

$$\LaTeX$$

If the phrase [Math Processing Error] is shown, or the LATEX script has vanished from the center of the above line, some of the math below might be missing or appear in the LaTex code instead of the script.

For easy navigation, use the CTRL+f keys on your keyboard. Cues in the forms of §'s and keywords in quotes are provided in the CTRL+f "Index".

If the header and the words

Reply | Flag

are shown at the same time in any of the following replies, refresh the page to see them.



That CMRB nomenclature for the MRB constant, introduced above, and used extensively below, was devised by Wolfram Alpha, as shown next.

W\A W\A2





'

An Easter egg for you to find below:

In another reality, they invented CMRB and then discovered many of its qualities.

RABBIT HOLE #1.

As a result, delving into "constructing rather than finding" math leads to RABBIT HOLE #1. Here are three unresolved inquiries:

  • If we assume that the MRB constant exists and that the author created it, did its properties exist before its invention?
  • If they did not, does that imply that the author also invented these properties?
  • If so, does the same principle hold that the originators of all mathematical constructs, such as numbers, relations, theorems, shapes, etc., also invented all unintended consequences?



Index

For section A, where all CMRB records are being checked, highlight text between quotes in

§A. Possibly the most epigrammatic closed forms for checking up to 10^10 digits of CMRB!

"Possibly the most epigrammatic closed forms used for checking up to 6,500,000 digits of CMRB!"

then push CTRL+f on keyboard.

For section 0, highlight text between quotes in

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

then push CTRL+f on keyboard.

The first post

Analyze the prototypical series for the MRB constant, Sn^(1/n)-1

(Select § with the given number or the keywords in quotes, and then press the

CTRL+f keys on your keyboard to move to that section.)

§1. Is that series convergent?

§2. Is -1 the only term that series is convergent for?

§3. Is that series absolutely convergent?

§4. Is that series "efficient?" (defined as how it compares to other series and integrals that compute CMRB in speed and computational cost.)

§5. Is there a geometric isomorphism between the many dimensions required to determine the "value" of the MRB constant to 10^34 digits (a value determined by the Planck scale),and, perhaps, the dimensions required in the prototypical problem of Graham's number?

§6. Q&A:

§7. My claim to the MRB constant (CMRB), or a case of calculus déjà vu?

§8. Where is it found?

§9. What exactly is it?

§10. How it all began,

§11. Scholarly works

§12. The why and what of the CMRB Records,



The MRB constant relates to the divergent series:

divegrent series

=DNE

The divergent sequence of its partial sums has two accumulation points with an upper limiting value or limsup known as the MRB constant (CMRB), and a liminf of CMRB-1.

So, out of the many series for CMRB, first analyze the sum prototype, i.e., the conditionally convergent series, {CMRB} =Sn^(1/n)-1

Later, we will analyze the absolutely convergent one,

enter image description here

Enter

SumConvergence[(n^(1/n) - 1 - Log[n]/n), n]

Concerning the sum prototype for CMRB

§1. Is the series convergent?

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

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 has 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.

Here is a direct proof form 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 has 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 has 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.

It converges because CMRB does

. . See the following notebook:

is the first term of the MRB constant(S) CTRL+f "is proven below by Gottfried Helms"

On a side note, there exists the following relationship between the MRB constant and log(2) and a couple of other famous constants:



§2. Next, ask and observe, question

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



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

First try the root test as shown by enter image description here

root test dicrptn

enter image description here enter image description here

However, 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.

Surprisingly, both sides of all three inequalities meet at the first Foias Constant instead of e:

In

N[FindRoot[n^(1/n) - 1 - 1/n == 0, {n, .1}, WorkingPrecision -> 38], 32]

Out[58]= {n -> 2.2931662874118610315080282912508}

In

   N[ FindRoot[n == (1 + 1/n)^n, {n, 1.2}, WorkingPrecision -> 38], 32]

Out[73]= {n -> 2.2931662874118610315080282912508}

In

      N[FindRoot[n^(1/n) == (1 + 1/n), {n, 1.2}, WorkingPrecision -> 38], 32]

Out[72]= {n -> 2.2931662874118610315080282912508}

Foias 1 Foias 2


For the series of absolute values,


Analogously to the Riemann Zeta Function , Riemann Zeta , another analogy to the harmonic series on how n below, at, and above 1 affects the convergence of the series of absolute values exists,

Similarly to the Riemann Zeta Function:

For n<1, fraction (the sum of absolute values) diverges.

For n=1, the series also diverges.

Only when n>1 does the series converge:

white backgrond  yellow, green, and bluebackgrounds

Assuming[n > 1, 
 SumConvergence[{(-1)^x (x^(1/x) - 1)/x, (x^(1/x) - 1)/
   x, (-1)^x x^(1/x)/x^1, x^(1/x)/
   x^1, (-1)^x x^(1/x)/x^(n) and x^(1/x)/x^(n)}, x]]


§4. 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.



§5.

RABBIT HOLE #2: (a conjecture, i.e., guess, for an upper limit of Graham's number) The Geometry of the MRB constant from that sum CMRB=$ \sum_{n=1}^\infty(-1)^n(n^{1/n}-1)$, Is the process that plots values from constructions arising from a peculiar non-Euclidean geometric isomorphism

between

the number of dimensions required for its partial sums (describing the length of hypercubes) to determine the "value" of the MRB constant to 10^34 digits (a value determined by the Planck scale)

and, perhaps,

the dimensions required in the prototypical problem of Graham's number

where there are the following.

From Wikipedia, Graham's number is connected to the problem in Ramsey theory:

Connect each pair of geometric vertices of an n-dimensional hypercube to obtain a complete graph on 2^n vertices. Color each of the edges of this graph either red or blue. What is the smallest value of n for which every such coloring contains at least one single-colored complete subgraph on four coplanar vertices?

enter image description here

Where for the Geometry of the MRB constant:

special hypercubes

Then find a sum to that series.

There in Diagram 3, M, at the point of the segment is where the z=MRB constant would be, and the base of that segment is the MRB constant -1.

An isomorphism is a concept in mathematics that refers to a one-to-one correspondence or mapping between two sets, preserving the binary relationships between the elements of the sets. The question at hand is whether this term is applicable in this context.

The known convergence rate of the MRB constant sum (i.e., taking n digits from $10^{n+1}$ partial sums) implies that if an isomorphism exists, it would establish a significantly smaller upper limit for Graham's number. It is currently unknown whether the conjecture that maps dimensions for the edges of hypercubes in CMRB (alternating sums of n^(1/n)) to the dimensions required for "the smallest value of n for which every such coloring contains at least one single-colored complete subgraph on four coplanar vertices in Graham's number problem (some unknown function involving 2^n)" is a plausible one or if it is merely numerological speculation.


§6.

Q&A:

Q: What can you expect from reading about CMRB and its record computations?

A:

As you see, the war treated me kindly enough, in spite of the heavy gunfire, to allow me to get away from it all and take this walk in the land of your ideas.

— Karl Schwarzschild (1915), “Letter to Einstein”, Dec 22

Q: Can you calculate more digits of CMRB?

A:

With the availability of high-speed electronic computers, it is now quite convenient to devise statistical experiments for the purpose of estimating certain mathematical constants and functions.

Copyright © 1966 ACM
(Association for Computing Machinery)

New York, NY, United States

Q: How can you compute them?

A:

The value of $\pi$ has engaged the attention of many mathematicians and calculators from the time of Archimedes to the present day, and has been computed from so many different formulae, that a complete account of its calculation would almost amount to a history of mathematics.

—  James Glaisher (1848-1928)

Q: Why should you do it?

A:

While it is never safe to affirm that the future of Physical Science has no marvels in store even more astonishing than those of the past, it seems probable that most of the grand underlying principles have been firmly established and that further advances are to be sought chiefly in the rigorous application of these principles to all the phenomena which come under our notice. It is here that the science of measurement shows its importance — where quantitative work is more to be desired than qualitative work. An eminent physicist remarked that the future truths of physical science are to be looked for in the sixth place of decimals.

 —  Albert A. Michelson (1894)

Q: Why are those digits there?

A:

[The principle, "nothing is without reason (nihil est sine ratione), or there is no effect without a cause"] must be considered one of the greatest and most fruitful of all human knowledge, for upon it is built a great part of metaphysics, physics, and moral science. (G VII 301/L 227).

—  Gottfried Wilhelm Leibniz (1646–1716)

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

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

As he went more and more public with his discoveries, making several attempts to see what portions were original, he 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 has 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 him 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 his investigations in physics and geometry. He 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 its 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

\







§10.

How it all began

From these meager beginnings:

This person's life experience demonstrates that academic grades in school may not accurately predict success in mathematics. Their report cards show poor grades in the subject. However, when given a self-paced program, the person discovered a newfound interest in math in junior high. Their grades improved with the opportunity to explore as they researched and studied more. They continued to pursue their passion for math, writing out powers of 2 and creating algebra problems for themselves to solve. They even purchased introductory calculus books. This one was the first: enter image description here Despite caring for their mother with Alzheimer's and working long hours, they attended university classes to further their knowledge. They are now listed on Wikipedia as an amateur mathematician and have studied graduate-level topics in mathematics. The reason for their slow start and eventual success is unclear and could be attributed to factors such as the educational system and psychology. However, their belief in their ability to learn and succeed in math played a crucial role in their achievements.


From these meager beginnings:

Inspired by Steven R Finch's historic essays on mathematical constants such as the one that starts out below, enter image description here

On January 11 and 23,1999, this researcher wrote,

I have started a search for a new mathematical constant! Does anyone want to help me? Consider, 1^(1/1)-2^(1/2)+3^(1/3)...I will take it apart and examine it "bit by bit." I hope to find connections to all kinds of arithmetical manipulations. I realize I am in "no man's land," but I work best there! If anyone else is foolhardy enough to come along and offer advice, I welcome you.

The point is that the author found the MRB constant (CMRB), meaning after all the giants were through roaming the forest of numbers and founding all they found, one virgin mustard seedling caught their eye. So, this writer carefully "brought it up" to a level of maturity and my understanding of math along with it! (In another reality, they invented CMRB and then discovered many of its qualities.)

In doing so found out that this constant (CMRB)

MRB math world snippet

(from https://mathworld.wolfram.com/MRBConstant.html)

was more closely related to other constants than could have been imagined.

As the apprentice of all, building upon the foundation of Chebyshev (1854–1859) on the best uniform approximation of functions, as vowed on January 23, 1999. "I took CMRB apart and examined it 'bit by bit,' finding connections to all kinds of arithmetical manipulations." Not satisfied with all too convenient constructs (half-hazardously put together formulas) that a naïve use of numeric search engines like Wolfram Alpha or the OEIS might give, this researcher set out to determine the most interesting (by being the most improbable but true) approximations for each constant in relation to it.

For example, we have a strong arithmetic relationship with the Lemniscate Constant:

According to Wikipedia

In mathematics, the lemniscate constant ϖ is a transcendental mathematical constant that is the ratio of the perimeter of Bernoulli's lemniscate to its diameter, analogous to the definition of π for the circle.


Next, consider its relationship to Liouville's (L) and the Euler Gamma constant ( $\gamma$}.


Consider its relationship to Viswanath's constant (VC)

Viswanath' math world snippit

(from https://mathworld.wolfram.com/RandomFibonacciSequence.html)

With both being functions of x1/x alone, ** we have these near-zeors of VC using C*MRB*, which have a ratio of Gelfond's constant $=e^\pi.$ =e^\pi


Then there is the Rogers - Ramanujan Continued Fraction, R(q),

of CMRB that is well-linearly approximated by terms of other terms of CMRB alone:

enter image description here



What about "?" for m= the MRB constant? We get over 7,000 digits of precision with hardly no effort at all!

enter image description here





From these meager beginnings:

On Feb 22, 2009, this researcher wrote,

It appears that the absolute value, minus 1/2, of the limit(integral of (-1)^xx^(1/x) from 1 to 2N as N->infinity) would equal the partial sum of (-1)^xx^(1/x) from 1 to where the upper summation is even and growing without bound. Is anyone interested in improving or disproving this conjecture?

enter image description here

Come to find out this discovery, a very slow-to-converge oscillatory integral, would later be further defined by Google Scholar.

Here is proof of a faster converging integral for its integrated analog (The MKB constant) by Ariel Gershon.

g(x)=x^(1/x), M1=hypothesis

Which is the same as

enter image description here because changing the upper limit to 2N + 1 increases MI by $2i/\pi.$

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

Iimofg->1

Cauchy's Integral Theorem

Lim surface h gamma r=0

Lim surface h beta r=0

limit to 2n-1

limit to 2n-

Plugging in equations [5] and [6] into equation [2] gives us:

leftright

Now take the limit as N?? and apply equations [3] and He went on to note that,

enter image description here

After this original investigator mentioned it to him, Richard Mathar published his meaningful work on it here in arxiv, where M is the MRB constant and M1 is MKB:

enter image description here

M1 has a convergent series, enter image description here which has lines of symmetry across whole-and-half number points on the x-axis, and half-periods of exactly 1, for both real and imaginary parts as in the following plots.

ReImPlot[(-1)^x (x^(1/x) - 1), {x, 1, Infinity}, PlotStyle -> Blue, 
 Filling -> Axis, FillingStyle -> {Green, Red}]

big plot small plot

Also where

Then

f[x_] = Exp[I Pi x] (x^(1/x) - 1); Assuming[
 x \[Element] Integers && x > 1, 
 FullSimplify[Re[f[x + 1/2]] - Im[f[x]]]]

gives 0


M2 and CMRB are connected:


In the complex plane, they even converge at the same rate.

Every 75 i of the upper value of the partial integration yields 100 additional digits of M2=enter image description here and of CMRB=enter image description here=enter image description here

Here is a heuristic explanation for the observed behavior.

Write the integral as an infinite series,... for each additional term.

The integrand is a quotient with numerator... This implies:

o$

Consequently

I used the fact that enter image description here

The following "partial proof of it" is from Quora.

Whileenter image description here enter image description here

enter image description here


A lot more theory behind it and ways of computing many more digits in this linked Wolfram Community post.



From these meager beginnings:

In October 2016, this researcher wrote the following here in ResearchGate:

First, we will follow the path the author took to find out that, for

ratio of a-1 to a,

the limit of the ratio of a to a - 1, as a goes to infinity, is Gelfond's Constant, (e ^ pi). We will consider the hypothesis and provide hints for proof using L’ Hospital’s Rule (since we have indeterminate forms as a goes to infinity):

The following should help in proving the hypothesis:

Cos[PiIx] == Cosh[Pix], Sin[PiIx] == I Sin-h[Pix], and Limit[x^(1/x),x->Infinity]==1.

Using L’Hospital’s Rule, we have the following:

L’ Hospital’s a's

(17) (PDF) Gelfond's Constant using MKB-constant-like integrals. Available from: https://www.researchgate.net/publication/309187705Gelfond%27sConstantusingMKBconstantlikeintegrals [accessed Aug 16 2022].

We find no limit "a" goes to infinity of the ratio of the previous forms of integrals when the "I" is left out, and we give a small proof for their divergence.

That was responsible for the integral-equation-discovery mentioned in one of the following posts, where it is written, "Using those ratios, it looks like" (There m is the MRB constant.)

enter image description here


From these meager beginnings:

In November 2013, his researcher wrote:

$C$MRB is approximately 0.1878596424620671202485179340542732. See this and this.

formally vs extend ended enter image description here

I asked, "If that's correct can you explain why?" and got the following comment. enter image description here

To see this for yourself in Mathematica enter FindRoot[NSum[(-1)^n*(n^(1/n) - x), {n, 1, Infinity}], {x, 1}] where regularization is used so that the sum that formally diverges returns a result that can be interpreted as evaluation of the analytic extension of the series.

enter image description here

See here.

Also,

enter image description here




§11.

Scholarly works about CMRB.

From these meager beginnings:

In 2015 his researcher wrote:

Mathematica makes some attempts to project hyper-dimensions onto 2-space with the Hypercube command. Likewise, some attempts at tying them to our universe are mentioned at https://bctp.berkeley.edu/extraD.html . The MRB constant from infinite-dimensional space is described at http://marvinrayburns.com/ThegeometryV12.pdf. It is my theory that like the MRB constant, the universe, under inflation, started in an infinite number of space dimensions. They almost all instantly collapsed, leaving all but the few we enjoy today.

This is not the first time for one to think the universe consists of an infinitude of dimensions. Some string theories and theorists propose it too. Michele Nardelli added a vast amount of string theory analysis and its connection to dimensions and the MRB constant.

He said,

In the following links, there are several works concerning various sectors of Theoretical Physics, Cosmology, and Applied Mathematics, in which MRB Constant is used as a "regularizer" or "normalizer". This constant allows to obtain a better approximation to the solutions obtained, developing the various equations that are analyzed. The solutions in turn, lead to four numbers that are called "recurring numbers". They are zeta (2) = Pi^2/6, 1729 (Hardy-Ramanujan number), 4096 (which multiplied by 2 gives the gauge group SO (8192)) and the Golden Ratio 1.61803398 ...

HE HAS PUBLISHED HUNDREDS OF PAPERS ON STRING THEORY AND THE MRB CONSTANT!

https://www.academia.edu/search?q=MRB%20constant https://www.researchgate.net/profile/Michele-Nardelli

Dr. Richard Crandall called the MRB constant a key fundamental constant

enter image description here

in this linked well-sourced and equally greatly cited Google Scholar promoted paper. Also here.

Dr. Richard J. Mathar wrote on the MRB constant here.

Xun Zhou, School of Water Resources and Environment, China University of Geosciences (Beijing), wrote the following in "on Some Series and Mathematic Constants Arising in Radioactive Decay" for the Journal of Mathematics Research, 2019.

A divergent infinite series may also lead to mathematical constants if its partial sum is bounded. The Marvin Ray Burns’ (MRB) constant is the upper bounded value of the partial sum of the divergent and alternating infinite series: -11/1+21/2-31/3+41/4-51/5+61/6-···=0.187859···(M. Chen, & S. Chen, 2016). Thus, construction of new infinite series has the possibility of leading to new mathematical constants.





§12.

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][204]   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


 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


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. ![CTRL+f][214]  "4th of July, 2022" for documentation.
To beat that, on the 7th of July, 2022, the same digits in  1/5 of an hour. ![CTRL+f][215]  "7th of July, 2022" for documentation (850 times as fast as the first 100,000 run!)

 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 30.5 hours of absolute time and 15.7 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.

36.7 hours million notebook

30.5 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.







POSTED BY: Marvin Ray Burns
42 Replies

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

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

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

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

POSTED BY: Daniel Lichtblau

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

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

nice system!

POSTED BY: l van Veen

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 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

While the 5,555,555 digit calculation is working, I thought I would mention my latest mini-project. I wanted to look at computing the MRB constant, (0.18785964...) through a generalized continued fraction for z^(1/z).

At Mathworld we see a suitable continued fraction here.

1 + (2 (x - 1))/(x^2 + 1 + 
    ContinuedFractionK[-(n^2 x^2 - 1) (x - 1)^2, (2 n + 1) x (x + 
        1), {n, \[Infinity]}])

.

It looks like

enter image description here

. -

With only 10 terms, I was able to calculate 4 decimals of the constant, 0.1878, the equivalent of adding up to 299,999 terms by brute force!

N[1/2 - Sum[(-1)^x*
         {(2*(x - 1))/(ContinuedFractionK[
                    (x - 1)^2*(-(n^2*x^2 - 1)), 
                    (2*n + 1)*x*(x + 1), 
                    {n, 3}] + x^2 + 1) + 1}, 
       {x, 1, 10^1}]]

. -

That gives 0.1878368104323782.

See, N[Sum[(-1)^n (n^(1/n) - 1), {n, 1, 299999}]] gives 0.18783862276177388.

It looks like the following.

enter image description here

. -

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

...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

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 (precisely nine years). That's an average of 211.5 views/hour, or one view every 17 seconds.

POSTED BY: Marvin Ray Burns

I calculated 6,500,000 digits of the MRB constant!!

The MRB constant supercomputer said,

Finished on Wed 16 Mar 2022 02 : 02 : 10. Processor and actual time were 6.2662810^6 and 1.6026403541959210^7 s.respectively Enter MRB1 to print 6532491 digits. The error from a 6, 000, 000 or more digit calculation that used a different method is 0.*10^-6029992

"Processor time" 72.526 days

"Actual time" 185.491 days

For the digits see the attached 6p5millionMRB.nb. For the documentation of the computation see 2nd 6p5 million.nb.

POSTED BY: Marvin Ray Burns

This is the second post in the MRB constant discussion, so right above this, if you can see the number of views

and the words

Reply | Flag

at the same time,

you might have to refresh the page to see the first message. The first message is crucial for understanding the MRB constant, as most of the facts are there.



The following might help anyone serious about breaking my record.

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

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

POSTED BY: Marvin Ray Burns

This is the third post in this discussion. If the messages in the first 2 posts aren't showing anything but ,

Reply|Flag

You need to refresh the page in order to view them.



The following email Crandall sent me before he died might be useful for anyone checking their results: enter image description here

POSTED BY: Marvin Ray Burns

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

enter image description here

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

POSTED BY: Marvin Ray Burns

Tue 24 Apr 2018

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

enter image description here[41]

Sat 28 Apr 2018

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

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

enter image description here

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

table

Here are the 3 programs referenced in the above table:

All MRB:

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


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


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

Crandall

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

Hybrid

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

Aug 8 2018

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

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

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

.

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

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

See attached "58 hour million."

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

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


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


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

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

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

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


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

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

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

Iterations required: 1326598

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

Actual time was 192609.7247443

Enter MRBtest2 to print 1004992 digits

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

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

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

top bottom

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

01/08/2019

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

Attachments:
POSTED BY: Marvin Ray Burns

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

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

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

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

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

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

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

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

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

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

...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

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

...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

The MRB constant supercomputer 2 is still working on 6,500,000 and 5,500,000 digit computations. The latest output on the 6,500,000 digit computation is

As of  Sun 11 Apr 2021 18:04:26 there were 5810176 iterations done in 1.2619*10^7 seconds. That is 0.46041 iterations/s. 67.37117% complete. It should take 216.766 days or 1.873*10^7s, and finish Mon 21 Jun 2021 11:03:14.

The 5,500,000 is being done by a totally different method. Its latest output is

6686208  iterations in  1.46938392445428*^7 seconds.

90.27240044247787% done.

When they are complete we will know for a fact what 5,500,000 digits of the MRB constant are! If the 6,500,000 digit computation gives the same answer as the 6,000,000 digit computation, experience has taught us, there is no reason to doubt it will be right.

POSTED BY: Marvin Ray Burns

Probably the last update on the 5,500,000 digit computation:

Print[7211520/7405568*100., "% done."]

97.37970132743364% done.

When it is done, having the same first 5,500,000 digits found in the previous 6,000,000 digit computation by a totally different method, it will prove beyond any shadow of a doubt the 5,500,000 digits are correct.

Stay tuned!

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

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

A reply a couple of places before this one has some "Programs to compute the integrated analog". Here is a recent discovery that could help in verifying the analog's digital expansions.

When f[x_] = E^(I Pi x) (1 - (1 + x)^(1/(1 + x))), the MRB constant is Sum[f[n],{n,0,Infinity}] and also, Sum[f[n],{n,1,Infinity}].

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

July 9, 2022

The new 16-core computer's computations with verification are documented in this notebook. Although I'm not making any large number of digits computations, I'm still working on speed records of MRB constant (CMRB) computations:

However, all the following timings are absolute! From the first post,

01/08/2019

Here is an update of 100k digits (in seconds):

enter image description here

Notice, I broke the 1,000-second mark!!!!!

4th of July, 2022

I did it in 861 seconds of absolute time! enter image description here with the full power of the MRB constant supercomputer 3 (MRBSC 3).

See notebook.

7th of July, 2022

I did it in 691 seconds of absolute time!

enter image description here

See notebook.

30th of July, 2022

I did it in 682 seconds of absolute time!

enter image description here See notebook.

11/27/2022

Using only 8 kernels of i9-13900K128 GB 5200MHZ RAM, only 8 kernels of i9-12900KS 64 GB 4800 MHz and only 8 kernels of i9-12900K 128 GB 5200MHZ RAM on the Wolfram Lightweight Grid, I calculated 100,493 digits of CMRB in 626.7164761s (less than 10.5 min) absolute time! And a computation time from the Timing[] command of 201.016 s (3.35 min). See the notebook here.

It ends with

131072 iterations done in 626.7 seconds. Should take 0.007341 days or 634.3s, finish Sun 27 Nov 2022 14:22:31.

Finished on Sun 27 Nov 2022 14:22:24. Processor time was 201.016 s.

Actual time was 626.7164761

Enter MRBtest2 to print 100493 digits

If you saved m3M, the difference between this and 3,014,991 known digits is 0.*10^-100494

11/16/2022

I've found ways of tweaking out new records with my computers. Here is a table of several of my speed records from my fastest computers. First a caveat, there is a discrepancy in my timings in the first of the following 2 tables: The timings at and before " 3.2 GH 6core,1666 MH RAM " were absolute timings. However, it was getting too hard to commit modern computers to just the one task of computing digits -- they are doing too many other things. So, I started recording computation time from the Timing[...]. command. However, with my new i9-13900K, this Timing[..] has led to a new problem! Timing[..] does not work well with the 9-13900K's 16 efficiency cores. Naturally, I got around that issue by doing single core operations for m2 computations. Also, computations of only a few digits of CMRB have small errors (I presume). Using the Wolfram Lightweight Grid, the 300,000-digit computation used 8 cores from the I-12900KS for efficiency.

enter image description here

This is the fairest comparison of my fastest computer's timing 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
POSTED BY: Marvin Ray Burns

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

[5]: https://www.wolframcloud.com/obj/bmmmburns/Published/late 2022 MKB(1).nb

POSTED BY: Marvin Ray Burns

If, above this, if you see the number of views, you will need to refresh the page in order to see the first post.

As I'm retiring to spend all my time tutoring, I mentioned the following before but became uncertain of the benefit for the Community Discussion Board. I still think that if someone would like to continue working on MRB constant records with Mathematica, (or a lower-level language for comparison's sake) it would be worthy of using this space!

I'm selling the latest version of the MRB constant supercomputer.

MRB constant Supercomputer.

An amazing combination of 12th and 13th Gen K's and KS's that was built to set the records for calculating records of number of digits of the MRB constant. See all the records at Wolfram Community\'s most visited post: Try to beat these MRB constant records! - Online Technical Discussion Groups—Wolfram Community

" https://community.wolfram.com/groups/-/m/t/366628 "

If you would like, I will show you how to use it as a real supercomputer (faster than many severs are) with the Wolfram Lightweight grid. The buy, search "MRB constant supercomputer" in EBAY.

POSTED BY: Marvin Ray Burns

If, above this, if you see the number of views, you will need to refresh the page in order to see the first post.

§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

"§A. Possibly the most epigrammatic closed forms for checking up to 10^10 digits of CMRB!"

Possibly the most epigrammatic closed forms used for checking up to 6,500,000 digits of CMRB!

(The proof of which remains elusive.)

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

Definition of epigrammatic as in concise marked by the use of few words to convey much information or meaning.

Clearly, the rational values for n with even more digits could be used to check much larger approximations to CMRB:

POSTED BY: Marvin Ray Burns
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