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NOTE. UPDATE 2 introduces proven series for Catalan's constant that is possibly the fastest currently known.


Working with some conjectured continued fractions that were published here, I have found the following fast monotone convergent series providing rational approximants to Catalan's constant $$\begin{equation*}G=\frac{1}{768}\sum_{n=1}^\infty\frac{4096^n\,P_0(n)}{n^3(2n-1)(3n-2)(3n-1)(4n-3)(4n-1){10n \choose 5n} {8n \choose 4n}{5n \choose 3n}}\tag{1}\label{1} \end{equation*}$$ where $P_0(n)$ is the following 6-th degree polynomial $$P_0(n)=4652032\,n^6-10340864\,n^5+8853568\,n^4-3683104\,n^3+774028\,n^2-76764\,n+2835$$ This series converges at a rate $\rho=\frac{27}{50000}=0.00054$ giving more than 3 decimal digits per term. This is just under some known BBP-type formulae for Catalan's constant (See here Eqs. 30-32) that have $\rho=\frac{1}{4096}\simeq 0.000244$ but it is well over Pilehrood's Fast formula (See here Theorem 4 pg. 234) having $\rho=\frac{1}{1024}\simeq 0.000977$ which belongs to the same class.

In fact, just to compare, Pilehrood's is $$\begin{equation*}G=\frac{1}{64}\sum_{n=1}^\infty\frac{(-1)^{n-1}256^n\,P_1(n)}{n^3(2n-1)(4n-3)^2(4n-1)^2{8n \choose 4n}^2{2n \choose n}}\tag{2}\label{2} \end{equation*}$$ where $$P_1(n) = 419840\,n^6 − 915456\,n^5 + 782848\,n^4 −332800\,n^3 + 73256\,n^2 − 7800\,n + 315$$

I have prepared the following configuration file for series Eq.$(1)$ to test it in my standard laptop with no special hardware using Alexander Yee's binary splitting y-cruncher software

{
NameShort : "Catalan"
NameLong : "Catalan's Constant. [email protected]"
AlgorithmShort : "Fast Catalan (2022). rho = 0.00054"
AlgorithmLong : "Catalan Constant Fast Hypergeometric Series (2022)"
Formula : {
    SeriesHypergeometric : {
        CoefficientP : 1
        CoefficientQ : 0
        CoefficientD : 2
        PolynomialP : [2835 -76764 774028 -3683104 8853568 -10340864 4652032]
        PolynomialQ : [99225 -2905560 33146280 -192321440 630560720 -1214720000 1360640000 -819200000 204800000]
        PolynomialR : [0 0 0 -2304 27264 -123264 266496 -276480 110592]
    }
}

It only took 28 seconds to get 50,000,000 decimal digits as it can be seen in this output details (it was verified on a 2nd run using a different —although slower— series from Jesús Guillera)

enter image description here

Now the questions

Is Eq.$(1)$ already known ?

Is there any chance to prove this conjectural series. Wilf-Zeilberger pairs perhaps? or

Is it possible, at least, to get a linear recurrence for the partial sums approximants ?

NOTE. (Just to put this topic in context) These series for Catalan's constant $G$ are linearly convergent at a high rate. They are fitted to perform a standard 3-variable Binary Splitting algorithm (with 2 variables needed for the final result), allowing to compute series even more general than hypergeometric series.

Two high performance additional series for $G$ belonging to this class are

Guillera (2019)$$\begin{equation*}G=\frac{1}{1024}\sum_{n=1}^\infty\frac{(-1)^{n-1}4096^n\,P_2(n)}{\left[n(2n-1){6n \choose 3n}{3n \choose n}{2n \choose n}^{-1}\right]^3}\tag{3}\label{3} \end{equation*}$$ where $$P_2(n) = 45136\,n^4 − 57184\,n^3 + 21240\,n^2 −3160\,n + 165$$ and Pilehrood's short (2010), —Corollary 2 pg. 233— $$\begin{equation*}G=\frac{1}{64}\sum_{n=1}^\infty\frac{256^n\,P_3(n)}{n^3(2n-1){6n \choose 3n}{6n \choose 4n}{4n \choose 2n}}\tag{4}\label{4} \end{equation*}$$ with $$P_3(n) = 580\,n^2 − 184\,n + 15$$ These expressions can be summarized in the following table

\begin{align*} \begin{array}{|c|c|c|c|c|c|c|} \hline \mathrm{Series} & \mathrm{Eq.} & \mathrm{conv. rate}\ \rho & \rho^{-1} & \frac{dec.\ digits}{term} & \mathrm{cost} & d \\ \hline \mathrm{This\ one\ (2022)} & (1) & 27/50000 & 1851.85 & 3.27 & 4.25 & 8\\ \hline \mathrm{Pilehrood\ long\ (2010)} & (2) & 1/1024 & 1024.00 & 3.01 & 4.62 & 8 \\ \hline \mathrm{Guillera\ (2019)} & (3) & 64/19683 & 307.55 & 2.49 & 4.19 & 6\\ \hline \mathrm{Pilehrood\ short\ (2010)} & (4) & 4/729 & 182.25 & 2.26 & 3.07 & 4\\ \hline \mathrm{Update\ 2\ (2023)} & (5) & 64/387420489 & 6053445.14 & 6.78 & 3.07 & 12\\ \hline \mathrm{Update\ 2\ (2023)} & (6) & 1/12500 & 12500.00 & 4.10 & 3.29 & 8\\ \hline \end{array} \end{align*}

UPDATE 1

Eq.$(1)$ has been recently proven by Jesús Guillera (Aug, 2023, pers. comm. See answer below) by finding a WZ pair (a ratio of 7th degree polynomials in $n$ and $k$) that certificates the stated series for this constant. It provides the best convergence rate for this family of series, which makes such formula the first candidate for fast computing of Catalan's constant if the number of digits needed goes to few thousands. If the number of digits required is huge (hundred thousands to millions and beyond), it is necessary to use a binary splitting algorithm. In this case the last two columns show the computing relative cost per iteration and $d$, the summand denominator's polynomial degree. Relative cost is calculated by $$cost =-\frac{ 4\,d}{\log\,\rho}$$ For these extreme cases (this is a play that some people play), the best formula so far has been Eq.$(4)$. In fact this series has been used to compute over 1.2e+12 digits of $G$. However current tests point out that Eq.$(5)$ below provides slightly lower timings.

UPDATE 2

The following Catalan's series is found and proven by using Guillera's CAS software (see here) with parameters $$entry := 12*n + 3/2, 6*n + 1, 3*n + 1, 3*n + 1, \infty$$

$$\begin{equation*}G=\frac{1}{9}\sum_{n=1}^\infty\frac{16777216^{n-1}\,Q_0(n)}{n^3(3n-2)^3(3n-1)^3(2n-1)(6n-1)(6n-5) {12n \choose 6n}{18n \choose 6n}{18n \choose 9n}}\tag{5}\label{5} \end{equation*}$$ where $Q_0(n)$ is the following $10^{th}$ degree polynomial $$Q_0(n)\,=\,133266953125232640\,n^{10} - 547282943160090624\,n^9 + 974070821452972032\,n^8 - 984148667421032448\,n^7 + 620775647093293056\,n^6 - 253085858560180224\,n^5 + 66661920081911808\,n^4 - 10984630610731008\,n^3 + 1051392967070720\,n^2 - 50559800637440\,n + 929536608000$$ This series converges at a rate $\rho=64/387420489$ giving $6.78$ decimal digits per term. It corresponds to Eq.$(4)$ but taking 3 terms at once. The intrinsic speed is the same, but it takes advantage of 64-bit integers giving a slightly better actual performance in x64 architectures.

Implemented into Alexander Yee's Y-Cruncher software that uses a specialized Binary Splitting algorithm to compute 50.000.000 decimal digits on my old Intel Core Kaby Lake - U x64 CPU, such formula performs 6.5% faster than the currently fastest known formula which is series Eq.$(4)$ above.

There is a binary splitting (BS) series for $G$ performing much faster than the one that originated this note Eq.$(1)$. It is obtained either with any of these parameters $$entry := 8*n + 1, 4*n + 1/2, 3*n + 1/2, 2*n + 1/2, \infty$$ $$entry := 7*n + 3/2, 3*n + 1, 2*n + 1, n + 1, \infty$$ which gives

$$\begin{equation*}G=\frac{16}{3}\sum_{n=1}^\infty\frac{(-4096)^{n-1}\,Q_1(n)}{n^3\,(2n-1)(3n-1)(3n-2)(6n-1)(6n-5) {5n \choose n}{10n \choose 5n}{12n \choose 6n}}\tag{6}\label{6} \end{equation*}$$ where $Q_1(n)$ is the following $6^{th}$ degree polynomial $$Q_1(n)\,=\,43203456\,n^6 - 92809152\,n^5 + 76613904\,n^4 - 30494304\,n^3 + 6004944\,n^2 - 536620\,n + 17325$$ This series converges at a rate $\rho=-1/12500$ giving about $4.10$ decimal digits per term. It has a BS cost $C_s=3.3922..$

Another interesting series, but for $\zeta(3)$ is the following found and proven with parameters $$entry := 10*n + 2, 4*n + 1, 4*n + 1, 2*n + 1, 2*n + 1$$ $$\begin{equation*}\zeta(3)=\frac{1}{64}\sum_{n=1}^\infty\frac{Q_2(n)}{n^5(2n-1)^5(4n-1)^3(4n-3)^3 {8n \choose 4n}^3{6n \choose 2n}{4n \choose 2n}}\tag{7}\label{7} \end{equation*}$$ where $Q_2(n)$ is the following $13^{th}$ degree polynomial $$Q_2(n)\,=\,229012564738048\,n^{13} - 1229189888278528\,n^{12} + 2966104641249280\,n^{11} - 4253527994044416\,n^{10} + 4037067522897408\,n^9 - 2672693698697472\,n^8 + 1267148681782880\,n^7 - 434665355558384\,n^6 + 107644209726992\,n^5 - 18970007169960\,n^4 + 2308647307806\,n^3 - 183574453491\,n^2 + 8548477560\,n - 176223600$$ This series converges at a rate $\rho=\frac{1}{12230590464}$ giving 10.09 decimal digits per term. This series corresponds to Wedeniwski's formula but taking two terms at once. See Wed-II in this MO Note (and the table therein) where the fastest and possibly new series are introduced for Apéry's Constant.


Raayoni, G., Gottlieb, S., Manor, Y. et al. Generating conjectures on fundamental constants with the Ramanujan Machine. Nature 590, 67–73 (2021). https://doi.org/10.1038/s41586-021-03229-4

Marichev, Oleg; Sondow, Jonathan; and Weisstein, Eric W. "Catalan's Constant." From MathWorld--A Wolfram Web Resource.

Khodabakhsh Hessami Pilehrood, Tatiana Hessami Pilehrood. Series acceleration formulas for beta values. Discrete Mathematics and Theoretical Computer Science, DMTCS, 2010, Vol. 12 no. 2 (2), pp.223-236. ff10.46298/dmtcs.504ff. ffhal-00990465f

Bruno Haible Thomas Papanikolaou. "Fast multiprecision evaluation of series of rational numbers" CLN Document

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1 Answer 1

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A proof of the stated formula for the Catalan constant is obtained automatically using the Maple program in arXiv 1611.04385 with entry:=1+6*n,1/2+3*n,1/2+2*n,1/2+1*n,oo;

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    $\begingroup$ Muchas gracias doctor Guillera. Formula is now proven by using Wilf-Zeilberger algorithm and finding a WZ pair. Details will be included as an update $\endgroup$ Aug 2, 2023 at 2:45

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