Apéry's constant $\zeta(3)$ fastest convergent series

Question

UPDATE Feb.02.2024

The series below, Eq.(3) for computing and Eq.(2) for verifying, were applied by Andrew Sun on Dec.22.2023 to get over $2\cdot10^{12}$ decimal digits and break the number of currently known decimals of Apery's Constant. The whole process took about 12 days as it is reported in [1], [2] and [3].

It is worth to point out that these proven fast ζ(3) formulae Eqs.(1,2 & 3) are un-published and they have been only placed here in MO.

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By means of a generalized search of hypergeometric-type series based on Dougall's sum and applying the Wilf-Zeilberger method as implemented by Jesús Guillera to prove some classical constants relationships, three proven series are found for Apéry's constant $\zeta(3)$. These series show the lowest known computing costs to apply the binary splitting method using A. Yee's y-cruncher software, a high performance platform to compute a huge number of digits of some classical constants, having set several records in this regard. These series are currently the fastest known to calculate an arbitrary number of digits for this constant.

I am not sure if any of these series has been published, so the question is simple.

Are these series known?

The first one is $$\begin{equation*}\zeta(3) = \frac{1}{12}\sum_{n=1}^{\infty}\frac{P(n)}{n^5(2n-1)(3n-1)(3n-2)(4n-1)(4n-3)\binom{3n}{n}^2\binom{6n}{3n}^2\binom{7n}{3n}}\tag{1}\label{1} \end{equation*}$$ with $$P(n) = 376698240\,n^7 - 935880048\,n^6 + 938536090\,n^5 - 491213175\,n^4 + 144859115\,n^3 - 24133842\,n^2 + 2114280\,n - 75600$$ It has convergence rate $\rho = 1/22235661$ giving about $7.35$ decimal digits per term. The computing cost $C_s$ is measured through $$C_s = - \frac{4\,d}{\log\rho}$$ where $d$ is the denominator's polynomial degree. In this case $d = 10$. This gives $C_s=2.3645...$.

The second series has $C_s = 2.3070...$ with $d=14$ and $\rho=1/34828517376$ giving about $10.54$ decimal digits per term. It is $$\begin{equation*}\zeta(3) = \frac{1}{24}\sum_{n=1}^{\infty}\frac{P(n)}{R(n)\cdot\binom{3n}{n}\binom{6n}{3n}\binom{8n}{4n}\binom{9n}{3n}\binom{10n}{5n}}\tag{2}\label{2} \end{equation*}$$ where $$P(n)= 250765325100000\,n^{11} - 1087318449630000\,n^{10} + 2067749814046250\,n^9 - 2269551612681475\,n^8 + 1592180015776565\,n^7 - 746938801646725\,n^6 + 238210943593421\,n^5 - 51452348050672\,n^4 + 7352050259484\,n^3 - 660416507568\,n^2 + 33552610560\,n - 731566080$$ and $$R(n)=n^5\,(2n-1)\,(3n-1)\,(3n-2)\,(4n-1)\,(4n-3)\,(5n-1)\,(5n-2)\,(5n-3)\,(5n-4)$$

The 3rd series has $C_s=2.0514...$ with $d=14$ and $\rho=-1/717445350000$ giving about $11.86$ decimal digits per term. It can be expressed as $$\begin{equation*}\zeta(3) = \frac{1}{48}\sum_{n=1}^{\infty}\frac{(-1)^{(n-1)}\,P(n)}{R(n)\cdot\binom{5n}{n}\binom{5n}{2n}\binom{9n}{4n}\binom{10n}{5n}\binom{12n}{6n}}\tag{3}\label{3} \end{equation*}$$ where $$P(n)= 1565994397644288\,n^{11} - 6719460725627136\,n^{10} + 12632254526031264\,n^9 - 13684352515879536\,n^8 + 9451223531851808\,n^7 - 4348596587040104\,n^6 + 1352700034136826\,n^5 - 282805786014979\,n^4 + 38721705264979\,n^3 - 3292502315430\,n^2 + 156286859400\,n - 3143448000$$ and $$R(n)=n^5\,(2n-1)^3\,(3n-1)\,(3n-2)\,(4n-1)\,(4n-3)\,(6n-1)\,(6n-5)$$

A non-native implementation of such algorithms has been done for y-cruncher software preparing custom input files for each formula. I tested these series under this special platform against some classical formulas that has been used to break the $\zeta(3)$ digits number record. In particular the Amdeberhan-Zeilberger series AZ (1997) $C_s=2.8854...$ and Wedeniwski's series Wed (1998) $C_s=2.7554...$.

Testings were done on an Intel Core Kaby Lake i5-7300U 3.5 Ghz 32 Gb RAM standard laptop with Windows 11 Pro 64bit OS. The following Table shows timings in seconds to get 50.000.000 decimal digits of Apéry constant. $Arch$ refers to the word size of the formula's input integer polynomial coefficients. $N$ indicates if the series summand has $N$ terms grouped from the basic formula. Timings are the average of 5 runs.

\begin{align*} \begin{array}{|c|c|c|c|c|c|c|} \hline {Series} & {Time\,s} & {Arch.} & N & \frac{dec.\ digits}{term} & C_s & d \\ \hline \mathrm{Eq.(3)} & 23.361 & 64-bit & 1 & 11.86 & 2.0514 & 14\\ \hline \mathrm{Eq.(2)} & 26.387 & 64-bit & 1 & 10.54 & 2.3070 & 14\\ \hline \mathrm{Eq.(1)} & 27.465 & 32-bit & 1 & 7.35 & 2.3645 & 10 \\ \hline \mathrm{Wed-II} & 33.095 & 64-bit & 2 & 10.08 & 2.7554 & 16\\ \hline \mathrm{AZ-III} & 34.697 & 64-bit & 3 & 9.03 & 2.8854 & 15\\ \hline \mathrm{Wed-I} & 34.893 & 32-bit & 1 & 5.04 & 2.7554 & 8\\ \hline \mathrm{AZ-I} & 37.603 & 32-bit & 1 & 3.01 & 2.8854 & 5\\ \hline \end{array} \end{align*}

Considering this limited scope of testings, it is observed that, under the y-cruncher's generic algorithm custom implementation, these -I guess- new series provide a significative increment of computing speed for Apéry's $\zeta(3)$. Eqs.$(1), (2)$ and $(3)$ are about 21%, 24% and 33% faster than classical Wedeniwski's formula respectively.

Finally, the proof of these series is obtained by means of WZ certificates (too long to put them here) through the code in Guillera with input patterns $$entry = 2+7\,n,1+3\,n,1+2\,n,1+1\,n,1+1\,n$$ $$entry = 2+10\,n,1+4\,n,1+3\,n,1+2\,n,1+1\,n$$ $$entry = 2+11\,n,1+4\,n,1+3\,n,1+2\,n,1+1\,n$$ for Eqs.$(1), (2)$ and $(3)$ respectively.

Answer 1

To answer your question, a google search of part of the formula shows no other results but this one, suggesting it is original

https://www.google.com/search?q=%22376698240%22

It' possible to make arbitrary series by just coming up with bigger and bigger polynomials p(n) and r(n). Such polynomials arise by starting with a kernel, which is the original integral that yields the desired constant, and then using polynomial division on the integral to make new binomial identities. It's possible to make an arbitrary number of these, but the most impressive series are those that have a fast convergence and small polynomials . For Zeta(3), you would use three integrals for the underlying 4F3 hypergeometric function. This is done with Catalan's Constant on page 13 here, which requires two integrals. https://papers.ssrn.com/sol3/papers.cfm?abstract_id=3919892

The more interesting question is how to derive these. Just saying "WZ algorithm" is not that helpful. There needs to be a mathematically precise way to finding the fastest series constrained by a certain hypergeometric form. For example, the fastest non-reciprocal pi formula (if we limit to reciprocal of pi, Ramanujan-type formulas would obviously win) that can be expressed as the composition of two 3f2 series in which all the parameters are rational.