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42 votes
4 answers
4k views

Are these fast convergent series for $\log(2)$, $\log(3)$ and $\log(5)$ already known and proven?

Now that some of the previously MSE formulae that I left here have been applied Dec.2023 to compute high precision record values ($10^{12}$ decimal digits) of trascendental constants $\Gamma(1/3)$ (Eq....
Jorge Zuniga's user avatar
  • 2,836
17 votes
1 answer
1k views

Catalan's constant fast convergent series

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 ...
Jorge Zuniga's user avatar
  • 2,836
13 votes
1 answer
1k views

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

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 ...
Jorge Zuniga's user avatar
  • 2,836
9 votes
0 answers
258 views

On a continued fraction and vector $\nu$ of length $n$

Please note that this question has been completely reworked in order not to overload it with unnecessary and useless information. Let $f(n)$ be an arbitrary function with integer values. Let $a(n)$ ...
Notamathematician's user avatar
7 votes
1 answer
268 views

Efficiently computing $\sum_k x^{k^2}$ modulo $p$

Let $p$ be prime. There is a whole host of "large" degree polynomials that can be computed efficiently modulo $p$. I was wondering if: $$q(x) = \sum_{k=0}^{p-1} x^{k^2}$$ is a polynomial ...
mtheorylord's user avatar
6 votes
0 answers
197 views

Where to cut off a double sum?

I have to compute a double infinite sum to within a given accuracy $\epsilon$. Let us say the sum is of the form $$\sum_{m\geq 1} \sum_{n\geq 1} \frac{a_{m,n}}{m^2 n^2 \max(m,n)},$$ where $|a_{m,n}|\...
H A Helfgott's user avatar
  • 20.2k
4 votes
2 answers
337 views

Is this BBP-type formula for $\ln 31$, $\ln 127$, and other Mersenne numbers also true?

In this post, a binary BBP-type formula for Fermat numbers $F_m$ was discussed as (with a small tweak), $$\ln(2^b+1) = \frac{b}{2^{a-1}}\sum_{n=0}^\infty\frac{1}{(2^a)^n}\left(\sum_{j=1}^{a-1}\frac{...
Tito Piezas III's user avatar
4 votes
1 answer
112 views

On a number of compositions of $n$ into positive triangular numbers

Let $a(n)$ be A023361 (i.e., number of compositions of $n$ into positive triangular numbers). Here $$ a(n) = \sum\limits_{i \geqslant 1, \frac{i(i+1)}{2}\leqslant n} a(n-\frac{i(i+1)}{2}), \\ a(0) = 1....
Notamathematician's user avatar
4 votes
1 answer
130 views

Intersecting algorithm for A065601

Let $a(n)$ be A065601 (i.e., number of Dyck paths of length $2n$ with exactly $1$ hill). Here $$ a(n) = \frac{1}{2(n+1)}((3n-2)a(n-1) + 2(9n-19)a(n-2) + 4(2n-3)a(n-3)), \\ a(0) = a(2) = 0, a(1) = 1. $$...
Notamathematician's user avatar
4 votes
1 answer
371 views

summation of oscillating functions

Consider series of the form $S=\sum_{n\ge1}f(n)P(n)$, where $f$ is some smooth function, and $P$ is a periodic or quasi-periodic function (e.g., $P$ can be a trigonometric function, so $S$ a Fourier ...
Henri Cohen's user avatar
  • 13.1k
3 votes
1 answer
178 views

Algorithm for the sum with binomial coefficients and Bell numbers

Let $a(n)$ be A000110 (i.e., Bell or exponential numbers: number of ways to partition a set of $n$ labeled elements). Let $b(n)$ be A355247 (i.e., expansion of exponential generating function $\exp(2(\...
Notamathematician's user avatar
3 votes
0 answers
165 views

Elegant algorithm for A140717

Let $T(n, k)$ be A140717 (i.e., triangle read by rows: $T(n,k)$ is the number of Dyck paths $d$ of semilength $n$ such that sum of peakheights of $d$ - number of peaks of $d$ equals $k$ ($n \geqslant ...
Notamathematician's user avatar
3 votes
0 answers
128 views

Fast and simple algorithm for the A329369

Let $a(n)$ be A329369 (i.e., number of permutations of $\{1,2,\cdots,m\}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \...
Notamathematician's user avatar
2 votes
0 answers
182 views

Algorithm for $\frac{1}{1-x} = \sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n}\frac{1-kx}{1+kx}$

Let $a(n)$ be A208832. Here $$ \frac{1}{1-x} = \sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n}\frac{1-kx}{1+kx}. $$ Start with vector $\nu$ of fixed length $m$ with elements $\nu_i = 1$ (that ...
Notamathematician's user avatar
2 votes
0 answers
61 views

Algorithm for main diagonal of integer coefficients associated with Schroeder numbers

Let $T_q(n, k)$ be an integer table such that $$T_q(n, k) = \begin{cases} 1 & \textrm{if } n = 0 \vee k = 0 \\ qT_q(n-1, n-1) + T_q(n, n-1) & \textrm{if } n = k > 0 \\ T_q(n, k-1) + T_q(n-1,...
Notamathematician's user avatar
2 votes
0 answers
64 views

On a $\sum\limits_{n=0}^{\infty}c_n x^n=\sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n+1}(1-f(k)x^k)$ (slightly different question)

Please note that this question differs from one of the previous questions of mine. Let $f(n)$ be an arbitrary function with integer values. Let $c_n$ be an arbitrary integer sequence. Let $a(n)$ be ...
Notamathematician's user avatar
2 votes
0 answers
100 views

Another (unique) algorithm for the A329369

Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \leqslant k ...
Notamathematician's user avatar
1 vote
1 answer
217 views

Correctness of the algorithm for the A329369, A347205 and related sequences

Let $a(n)$ be A347205. It is enough for us to know that $$ a(2^m(2k+1)) = \sum\limits_{j=0}^{m}a(2^jk), \\ a(0) = 1 $$ Let $b(n)$ be A329369. It is enough for us to know that $$ b(2^m(2k+1)) = \sum\...
Notamathematician's user avatar
1 vote
1 answer
178 views

Algorithm for A127782

Let $a(n)$ be A127782 (i.e., an integer sequence with generating function $A(x)$ such that $A(x)=1+xA(x+x^2)$). Here $$ a(n) = \sum\limits_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor} \binom{n-k-1}{...
Notamathematician's user avatar
1 vote
0 answers
57 views

Step back step forward algorithm for A108442

Let $a(n)$ be A108442. Here generating function is $\frac{1}{1-zA(z)}$ where $$ A(z) = 1 + z(A(z))^2 + z(A(z))^3. $$ Also $$ a(n) = \sum\limits_{k=1}^{n}\frac{k}{2n-k}\sum\limits_{i=0}^{n-k} \binom{2n-...
Notamathematician's user avatar
1 vote
0 answers
161 views

Efficient algorithm for A217061

Let $a(n)$ be A217061. Here $$ a(n) = \sum\limits_{m=1}^{n}\frac{1}{(m-1)!}\sum\limits_{k=0}^{n-m}(n+k-1)!\sum\limits_{j=0}^{k}\frac{1}{(k-j)!}\sum\limits_{\ell=0}^{j}\frac{2^{\ell-j}(-1)^{\ell+j}s(n-...
Notamathematician's user avatar
0 votes
0 answers
65 views

Algorithm and equivalent recursion for A258173 (related to Dyck paths)

Let $a(n)$ be A258173 i.e. sum over all Dyck paths of semilength $n$ of products over all peaks $p$ of $y_p$, where $y_p$ is the $y$-coordinate of peak $p$. A Dyck path of semilength $n$ is a $(x,y)$-...
Notamathematician's user avatar