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. jorge.zuniga.hansen@gmail.com"
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 \end{array} \end{align*}

Eq.$(1)$ (the only one already un-proven in this table) provides the best convergence rate for this family of series, which makes this 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), 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 is Eq.$(4)$. In fact this series has been used to compute over 1.2e+12 digits of $G$.

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



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.