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37 votes
3 answers
2k views

How to prove the identity $L(2,(\frac{\cdot}3))=\frac2{15}\sum\limits_{k=1}^\infty\frac{48^k}{k(2k-1)\binom{4k}{2k}\binom{2k}k}$?

For the Dirichlet character $\chi(a)=(\frac a3)$ (which is the Legendre symbol), we have $$L(2,\chi)=\sum_{n=1}^\infty\frac{(\frac n3)}{n^2}=0.781302412896486296867187429624\ldots.$$ Note that this ...
0 votes
1 answer
129 views

Closed form for $\sum\limits_{k=0}^{n} [\operatorname{wt}(k) = m]$ where $\operatorname{wt}(n)$ is the binary weight of $n$

Let $\operatorname{wt}(n)$ be A000120 (i.e., number of $1$'s in binary expansion of $n$). Let $a(n,m)$ be the family of integer sequences such that $$ a(n,m) = \sum\limits_{k=0}^{n} [\operatorname{wt}(...
8 votes
1 answer
671 views

Infinite series and sum of two squares

Consider the following infinite sequence $a(n)$ generated by $$\sum_{n\geq0} a(n)q^n =\frac{\sum_{k\geq0}F(2k+1)q^{\binom{k+1}2}}{\sum_{k\geq0} q^{\binom{k+1}2}}$$ where the $F(2k+1)$ are the odd ...
10 votes
0 answers
350 views

How are the hypergeometric motives of WZ-Pairs connected?

If $\small{(F,G)}$ is a WZ-pair and general asymptotic conditions $\lim_{k\rightarrow\infty}\small{G(n,k)=0}$ and $\lim_{n\rightarrow\infty}\small{F(n,k)=0}$ hold, then we have the certified ...
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....
0 votes
1 answer
169 views

Partial sums of binomial coefficients and related family of polynomials

Let $a(n)$ be A302117. Here $$ a(n) = 4(n-1)a(n-1) - \frac{1}{3}\prod\limits_{k=0}^{n-1}(2k-3), \\ a(0) = 0. $$ Let $$ T(n,k) = \sum\limits_{i=0}^{k} \binom{n}{i}. $$ Let $P_n(z)$ be the family of ...
1 vote
1 answer
203 views

Structural differences between closed forms of two related infinite products?

In this question, I went quite a bit over the top, so I now tried to rephrase it in a much simpler way. Take $a \in \mathbb{R}, s \in \mathbb{C}$ and: $$\displaystyle C(s,a) := \prod_{n=1}^\infty \...
6 votes
1 answer
494 views

(Explicit) Tauberian theorems: removing $(\log x/n)$

Say that $\{a_n\}_{n\geq 1}$, $|a_n|\leq 1$, are such that $$\left|\sum_{n\leq x} a_n \log \frac{x}{n}\right|\leq \epsilon x\quad\text{for all $x\geq x_0$.}$$ What sort of bound can we deduce on $S(x)=...
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-...
4 votes
1 answer
213 views

Asymptotic behavior of weighted sums involving the fractional part function

Currently, I am studying the asymptotic behavior of sums of the form \begin{equation}\label{eq1}\tag{1} \sum_{k=1}^{n-1} f(n-k) \left\{ \frac{n}{k} \right\} \end{equation} In this context, based on ...
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 ...
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-...
9 votes
5 answers
2k views

Laplace's summation formula

I recently came across the following formula, which is apparently known as Laplace's summation formula: $$\int_a^b f(x) dx = \sum_{k=a}^{b-1} f(k) + \frac{1}{2} \left(f(b) - f(a)\right) - \frac{1}{...
2 votes
0 answers
67 views

$R$-recursion for A006351

Let $a(n)$ be A006351 (i.e., number of series-parallel networks with n labeled edges. Also called yoke-chains by Cayley and MacMahon). Here exponential generating function is $A(x)$ such that $B(x) = ...
2 votes
0 answers
59 views

$R$-recursion for A338193

Let $a(n)$ be A338193. Here generating function is $A(x)$ such that $$ A(x) = 1 + \int\frac{\left(\frac{x}{A(x)}\right)'}{\left(\frac{x}{(A(x))^2}\right)'} \, dx. $$ Let $$ R(n, q) = \begin{cases} 1 &...
16 votes
0 answers
351 views

The convergence domain of the function $\sum \{n!x\}$

This is a problem from a mathematics competition: Does there exist an irrational number $x$ such that the series $$\sum_{n=1}^{\infty}\{n!x\}<+\infty$$ where $\{ \}$ means the fractional part of a ...
1 vote
1 answer
92 views

Equivalence of sequences related to A033264

Let $a(n)$ be A033264 (i.e., number of blocks of $\{1,0\}$ in the binary expansion of $n$). Here $$ a(4n) = a(4n+1) = a(2n), \\ a(4n+2) = a(n)+1, \\ a(4n+3) = a(n), \\ a(0) = 0. $$ Let $$ \ell(n) = \...
23 votes
4 answers
2k views

Identity for an infinite product

Here is an experimental "result" exhibiting the difference of two (formal) infinite products that "almost factorizes". QUESTION. Is this true? $$\prod_{n\geq1}(1+x^{2n-1})^{24} - \...
1 vote
0 answers
89 views

Test for odd prime triples in a $2p-1$ progression

Let $a(n)$ be A057326 (i.e., first member of a prime triple in a $2p-1$ progression). Let $b(n) = B$ after $n-1$ iterations where we start with $A=n, B=1$ and for $i$ from $1$ to $n-1$ simultaneously ...
2 votes
0 answers
113 views

Numbers of positive terms in polynomials equal A069999

Let $a(n)$ be A069999 (i.e., number of possible dimensions for commutators of $n \times n$ matrices; it is independent of the field). OEIS states that no generating function is known. Let $P(n,k)$ be ...
5 votes
0 answers
285 views

How do you go about making ranges (for integer variables) independent?

Basic question: say you have a sum $$\sum_{n_1 n_2 \dotsb n_k \leq x} f(n_1,\dotsc,n_k),$$ where $f$ decomposes in some sense (say: $f(n_1,\dotsc,n_k) = g(n_1) + \dotsb + g(n_k)$, or $f(n_1,\dotsc,n_k)...
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}{...
1 vote
0 answers
82 views

Generating functions related to generating function of Catalan numbers

Let $C_n$ be A000108 (i.e., Catalan numbers). Here generating function is $C(x)$ such that $$ C(x) = \frac{1-\sqrt{1-4x}}{2x}. $$ Let $a(n)$ be an integer sequence with generating function $A(x)$ such ...
1 vote
0 answers
84 views

Coarse well-distributedness/equidistribution of Pell sequence prefixes

I am interested in the distributedness or "mixing" behavior of certain linear recurrences modulo powers of $2$. In particular, consider the Pell sequence (https://oeis.org/A000129), modulo $...
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 $(x,y)$-...
0 votes
1 answer
141 views

Property of composite numbers

Let $a(n)$ be the sequence of composite numbers (starting from $4$). Let $$b(n)=a(n-1)a(n-2) \operatorname{mod} a(n)$$ Obviously, $b(1)=b(2)=0$. I conjecture that with the only exception for the $b(3)=...
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 $a(n)$ ...
2 votes
0 answers
121 views

Solving a system of differential-like equations for reverse Euler-Maclaurin summation

Aim A particular instance of a rational zeries that has as of yet not been evaluated is: \begin{align} Z:= \sum_{n=1}^{\infty} \frac{\zeta(2n)}{(2n)!}. \label{EM1} \tag{EM1} \end{align} This sum ...
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,...
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 $$ a(n) = \frac{1}{2(n+1)}((3n-2)a(n-1) + 2(9n-19)a(n-2) + 4(2n-3)a(n-3)), \\ a(0) = a(2) = 0, a(1) = 1. $$...
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....
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(\...
2 votes
0 answers
46 views

On A088352 as an antidiagonal sums of A129179

Let $a(n)$ be A088352. Here $a(n)$ is an integer sequence with generating function $A(x)$ such that $$ A(x) = \cfrac{1}{1-x-\cfrac{x^2}{1-x^3-\cfrac{x^4}{1-x^5-\cfrac{x^6}{1-x^7-\cfrac{x^8}{\ddots}}}}}...
3 votes
2 answers
511 views

Exotic series for some mathematical constants from String Theory

Following these MO posts from Timothy Chow and Henri Cohen and their comments, based on Saha & Sinha's String Theory article, I have found three series for $\pi$, Apéry's constant $\zeta(3)$ and ...
7 votes
4 answers
793 views

Must bounded sequences be well-distributed to most *composite* moduli?

Let $\{a_n\}_{n=1}^N$, $|a_n|\leq 1$. Let $Q=\sqrt{N}$. Then $a_n$ is well-distributed modulo most prime $p\leq Q$, in the following sense: $$\sum_{p\leq Q} \frac{1}{p} \left(\frac{1}{N/p} \sum_{\...
2 votes
1 answer
310 views

Generating function for A300483 (related to Chebyshev polynomial of first kind)

Let $a(n)$ be A300483. Here $$ a(n) = 2\int\limits_{t \geqslant 0}T_n\left(\frac{t+1}{2}\right)\exp(-t)\,dt. $$ where $T_n(x)$ is $n$-th Chebyshev polynomial of first kind. Let $b(n)$ be an integer ...
10 votes
1 answer
625 views

Generating function for A261041

Let $a(n)$ be A261041 (i.e., number of partitions of subsets of $\{1,2,\dotsc,n\}$, where consecutive integers are required to be in different parts). Let $b(n)$ be an integer sequence with generating ...
2 votes
2 answers
315 views

5 different ways to define the same family of integer sequences

Let ${n \brace k}$ be a Stirling number of the second kind. Let $A_n(x)$ be an Eulerian polynomial. Here $$ A_n(x) = \sum_{i=0}^{n}i!{n \brace i}(x-1)^{n-i}. $$ Let $a_1(n,p,q)$ be the family of ...
2 votes
0 answers
157 views

What rational zeta series with non-integer arguments appear in mathematics?

Background Rational zeta series are series of the form $$\sum_{n=2}^{\infty} q_{n} \zeta(n + p, m), \label{1} \tag{1} $$ where $\zeta(x,m)$ is the Hurwitz zeta function and $q_{n}, \ p \in \mathbb{Q} \...
1 vote
1 answer
101 views

Finding $\lim_{n \to \infty} \sum_{k=2}^{n-2} \zeta(k) \zeta(n-k) x^{k-1} = x^{-1} - \psi_{0}(-x) - \gamma$ from the generating function of $\zeta(•)$

In equation (130) of this page, the identity $$\lim_{n \to \infty} \sum_{k=2}^{n-2} \zeta(k) \zeta(n-k) x^{k-1} = x^{-1} - \psi_{0}(-x) - \gamma \label{1} \tag{1} $$ is stated. Here, $\zeta(\cdot)$ is ...
1 vote
0 answers
105 views

Simpler recursion for the A358612

Let $T(n,k)$ be an integer coefficients (A358612) such that $$ T(2n+1, k) = kT(n, k) + T(n, k-1), \\ T(2n, k) = kT(n, k) + T(n, k-1) - \frac{T(2n, k-1) + T(n, k-1)}{k-1}, \\ T(n, 1) = T(0, 2) = 1 $$ ...
3 votes
1 answer
140 views

$R$-recursion for unsigned Genocchi numbers (of first kind) of even index

Let $G_n$ be A036968 (i.e., Genocchi numbers). Here $$ \frac{2t}{1+e^t}=\sum\limits_{n=0}^{\infty}G_n\frac{t^n}{n!}. $$ Also $$ t\tan\left(\frac{t}{2}\right)=\sum\limits_{n=1}^{\infty}(-1)^n G_{2n}\...
6 votes
1 answer
282 views

Integer sequences with a periodic pattern

Let $A$ and $B$ be two different integers. Let $S$ be a finite integer sequence with exactly $n_A$ $A$s and $n_B$ $B$s. By repeating $S$ infinitely many times we obtain an infinite integer sequence $P$...
7 votes
2 answers
265 views

On four Ramanujan-type "Legendrian" sequences with a 3-term recurrence?

I. Recurrences In a previous post, it was mentioned how Almkvist-Zudilin did a computer search for solutions to the recurrence relation, $$(n+1)^3s_{n+1}=(2n+1)(an^2+an+b)s_n+c\,n^3s_{n-1}$$ within a ...
6 votes
1 answer
387 views

Lucas number multiples of Fibonacci pairs

$\newcommand{\GCD}{\operatorname{GCD}}$ For $n=0,1,2,\ldots,$ let $F_n=0,1,1,2,3,5,\ldots$ and $L_n=2,1,3,4,7,11,\ldots$ be the Fibonacci and Lucas sequences. I expect the following is well known, but ...
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 ...
1 vote
0 answers
86 views

Decouple two continued-fractions

Given a three-term recurrence relation such as \begin{equation} a_nX^n+b_nX^{n+1}+c_nX^{n-1}=\zeta_n \delta_{n,0} \end{equation} it can always be written in continued-fraction form, and if you are ...
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 ...
1 vote
0 answers
32 views

On a A347205 and related row polynomials

Let $a(n)$ be A347205. Here $$ a(2^m(2k+1)) = \sum\limits_{j=0}^{m}a(2^j k), \\ a(0) = 1. $$ Let $\nu_2(n)$ be A007814 (i.e., number of trailing zeros in the binary expansion of $n$). Here $$ \nu_2(2n+...
3 votes
0 answers
161 views

On generators of the multiplicative semigroup $\{r\in\mathbb Q:\ r>1\}$

The set $M=\{r\in\mathbb Q:\ r>1\}$ is a commutative semigroup with respect to the multiplication. For any integers $a>b\ge1$, we clearly have $$\frac ab=\prod_{n=b}^{a-1}\frac{n+1}n.$$ So the ...

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