Binomial coefficient C(2k,n-1) alternative formula equivalent to the...
I am working with a binomial sum that arises in some combinatorial arguments (and also appears in certain generating‐function manipulations). Specifically, I have this...
View ArticleAn analogue of the sum of binomial coefficients
Let $a$ and $p$ be positive integers, and consider the polynomial $$(1+x+\cdots+x^{p-1})^a = \sum_{i=0}^{a(p-1)} a_ix^i.$$ I'm looking for an asymptotic estimate of $\sum_{i=0}^{b} a_i$. If $p=2$, this...
View ArticleA vanishing hypergeometric series
I have found a new vanishing hypergeometric series. Namely, I conjecture that$$\sum_{k=1}^\infty\frac{\binom{2k}k...
View ArticleFew conjectural series for $\zeta(5)$ and $\zeta(6)$
The usual harmonic numbers are given by$$H_n:=\sum_{0<k\le n}\frac1k\ \ \ (n=0,1,2,\ldots).$$For any integer $m>1$, those numbers$$H_n^{(m)}:=\sum_{0<k\le n}\frac1{k^m}\ \ \...
View ArticleA series related to $\zeta(3)$
Inspired by Question 508663, I discovered the following identity$$\sum_{k=1}^\infty\frac{H_{2k}-H_{k}}{k^2\binom{2k}k}=2 \zeta(3)-\frac{\pi \sqrt{3}\, \Psi^{\left(1\right)}\!...
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Generalization of Chebyshev polynomials with connection to K-bonacci sequence...
I have been exploring a combinatorial approach to express Chebyshev polynomials and generalizing them through a recurrence relation. I would like to know whether this recurrence relation can be proven...
View ArticleConjectural series for $L(s,\genfrac(){}{}{-3}\cdot)$
$\newcommand\Ksymb{\genfrac(){}{}}$For $s=1,2,3,\dotsc$, let us...
View ArticleConjectural series arising from Ramanujan's series for $1/\pi$
We can formluate new conjectural series from the classical Ramanujan-type series for $1/\pi$.For example, Ramanujan found that$$\sum_{k=0}^\infty(6k+1)\frac{\binom{2k}k^3}{256^k}=\frac{4}{\pi},\ \...
View ArticleClosed form for series involving $\binom{4n}{2n}\binom{2n}{n}$
Consider the series$$S(k)=\sum_{n=0}^{\infty}\frac{1}{(kn+1)64^n}\binom{2n}{n}\binom{4n}{2n},\qquad k\in\mathbb{Z}_{>0}.$$Two special values appear to have closed...
View ArticleIdentity with binomial coefficients and Stirling numbers of the second kind
Playing with various parameters, I came up with the following conjecture: $$ \sum\limits_{k=1}^{n} (k-1)! \binom{n+m-1}{k-1} {n \brace k} = (n+m)^{n-1}. $$Studying the base case ($m=0$), I looked...
View ArticleA series related to $\log192-\sqrt{3}\pi$
Let$$H_n:=\sum_{0<k\le n}\frac1k\ \ \ \ (n=0,1,2,\ldots).$$Inspired by Question 507603 and Question 507592, I discovered the following...
View ArticleComposite number $n$ with most $k \le n$ such that $n \mid \binom nk$
This problem arised in a local forum, proposed by a user named zxt.Let $f(n)$ be the number of nonnegative integer $k$ not greater than $n$ such that $n \mid \binom{n}{k}$. If for each positive integer...
View ArticleClosed form for a binomial product sum
Is there any closed formula for the binomial product sum\begin{align*}\sum\limits_{\substack{i_1> i_2> \cdots > i_k\\i_1, i_2, \cdots, i_k \in \{n-j+1, n-j+2, \cdots,...
View ArticleClosed form for A135494
Let$f(n)$ be an integer function such that $$ f(n) = -1, \\ f(1) = 1. $$$T(n,k)$ be A135494, i.e, an integer coefficients known as Bell transform of $f(n)$. Here $$ T(n,k) = \sum\limits_{j=1}^{n-k+1}...
View ArticleEquivalence of pair of integer coefficients
Please note that this question was completely reworked.Let$S(n,k)$ be an integer coefficients such that $$ S(n,k) = \sum\limits_{j=0}^{k} \frac{n!}{j!} {n-j \brace k-j}. $$$T(n,k)$ be an integer...
View ArticleIdentities with Stirling numbers of both kinds
With the luck of intuition, I conjecture that $$ {n+m+1 \brack m+1} = (-1)^n \sum\limits_{k=0}^{n} \left[ \left[ \sum\limits_{i=0}^{k} (-1)^{k+i} 2^{k-i} \binom{n+k}{k-i} {n+i \brace i} \right] \cdot...
View ArticleMore conjectural formulas for Riemann's zeta function (II)
Motivated by Zeilberger's series$$\sum_{k=1}^\infty\frac{21k-8}{k^3\binom{2k}k^3}=\zeta(2)$$and Questions 508743 and 508768I have discovered several identities involving higher-order derivatives of the...
View ArticleMore conjectural formulas for Riemann's zeta function (III)
This is a continuation of Questions 508768 and 508774.By Example 11 of this 2014 paper of Chu and Zhang, we have the...
View ArticleMore conjectural formulas for Riemann's zeta function (IV)
This is a continuation of Questions 508768, 508774 and 508844.In 2023 I conjectured the formula$$\sum_{k=1}^\infty\frac{145k^2-104k+18}{(2k-1)k^3\binom{2k}k\binom{3k}k^2}=\frac{\pi^2}3$$in Question...
View ArticleQuadratic polynomials and Riemann's zeta function
Motivated by my recent postings, here I propose some conjectural series for Riemann's zeta function with summands involving binomial coefficients and quadratic polynomials.By Examples 14 and 32 of this...
View ArticleMore conjectural formulas for Riemann's zeta function
Motivated by Question 508743 and the known identities$$\sum_{k=1}^\infty\frac{(-1)^k}{k^3\binom{2k}k}=-\frac25\zeta(3)\ \ \text{and}\ \ \sum_{k=1}^\infty\frac1{k^4\binom{2k}k}=\frac{17}{36}\zeta(4),$$I...
View ArticleRepresenting integers as sums of binomial coefficients and relative primeness...
Motivation: This question arose from studying an open problem in ergodic theory. Is the Pascal adic transformation weakly mixing? I tried boiling the problem down to a question about paths through...
View ArticleRational congruence of binomial coefficient matrices
Skip Garibaldi asks if there is an elementary proof of the following fact that "accidentally" fell out of some high-powered machinery he was working on.Say that two matrices $A$ and $B$ over the...
View ArticleReciprocal sum of binomials and divisibility by $3$
We all know that $\sum_{k=0}^n\binom{n}k$ is not divisible by $3$.QUESTION. Is it true that the numerator of $a_n$ (in reduced form) is never divisible by...
View ArticleSome unique coefficients with row sums equal $2$
Let$T(n,k)$ be an unique coefficients such that for all $n,m \in \mathbb{N}$ we have $$ \sum\limits_{k=1}^{2n} 2^{k-1} \binom{m+k}{k} T(n,k) = (m+2)^{2n} - 1. $$I conjecture that $$...
View ArticleSome conjectural congruences involving Domb numbers
The Domb numbers are given by$$D_n=\sum_{k=0}^n\binom{n}{k}^2\binom{2k}k\binom{2(n-k)}{n-k}\ \ \ (n=0,1,2,\ldots).$$Such numbers have combinatorial interpretation, see, e.g., http://oeis.org/A002895.I...
View ArticleAre truncated even degree binomial polynomials psd?
For integers $n$ and $k$, let $P(n,k)(x) = \sum_{i=0}^k \binom ni x^i$ be the truncated binomial polynomial. There has been work on whether $P(n,k)$ is irreducible, but this is a different question....
View ArticleOn the polynomials $\sum_{k=0}^n\binom{n+k}k^m q^k$
A sequence of polynomials$$P_0(q),\ P_1(q),\ P_2(q),\ \ldots$$with real coefficients is called $q$-log-convex if for each $n=1,2,3,\ldots$ every coefficient of the polynomial...
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Can this be done? Split Pascal's triangle (without the 1's) with a straight...
Cross-posted from MSEConsider Pascal's triangle with $n$ rows, without the $1$s, with each number corresponding to a vertex on a pyramid of equilateral triangles, as shown below with example $n=5$.Does...
View ArticleRamanujan's pi series and harmonic numbers
Inspired by Question 507666, I think that replacing the linear term $ak+b$ in Ramanujan's Pi series with a suitable linear combination of harmonic numbers can yield some interesting new series. I will...
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