Ramanujan'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...
View ArticleTen new conjectural series for $\frac{\log m}{\pi}$ involving harmonic numbers
Recall that the harmonic numbers are given by$$H_n:=\sum_{0<k\le n}\frac1k\ \ \ \ (n=0,1,2,\ldots).$$As a supplement to Question 436205 and my related paper Series with summands involving harmonic...
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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 ArticleDo exact collisions of the binomial “derivative fingerprint” force the...
Setup. For $0\le r<k$ write$$\mathcal H(k,r) = (-1)^{k-1-r}\,k\binom{k-1}{r}.$$This is (up to sign) $1/f_k'(r)$, the reciprocal derivative of$f_k(x)=\binom xk$ at the integer $r$, and it is the...
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...
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...
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 ArticleRecurrence equations for binomial determinants [closed]
Let $T_n$ be tridiagonal (Toeplitz) matrix of the form $$ T_n=\begin{pmatrix} 2&1&& \\ 1&\ddots&\ddots\\ &\ddots& \ddots&1\\ && 1&2\\ \end{pmatrix}_{n\times...
View ArticleSome conjectural congruences involving Domb numbers
The Domb numbers are given by$$D_n=\sum_{k=0}^n\binom{n}{k}^2\binom{2k}k\binom{2(n-k)}{n-k}\ \ \ (n=0,1,2,\ldots).$$Such numbers have combinatorial interpretation, see, e.g., http://oeis.org/A002895.I...
View ArticleSome unique coefficients with row sums equal $2$
Let$T(n,k)$ be an unique coefficients such that for all $n,m \in \mathbb{N}$ we have $$ \sum\limits_{k=1}^{2n} 2^{k-1} \binom{m+k}{k} T(n,k) = (m+2)^{2n} - 1. $$I conjecture that $$...
View ArticleThree new conjectural series involving harmonic numbers
Recall that the harmonic numbers are those$$H_n:=\sum_{0<k\le n}\frac1k\quad\ (n=0,1,2,\ldots).$$The second-order harmonic numbers are given by$$H_n^{(2)}:=\sum_{0<k\le n}\frac1{k^2}\quad\...
View ArticleReciprocal sum of binomials and divisibility by $3$
We all know that $\sum_{k=0}^n\binom{n}k$ is not divisible by $3$.QUESTION. Is it true that the numerator of $a_n$ (in reduced form) is never divisible by...
View ArticleRational congruence of binomial coefficient matrices
Skip Garibaldi asks if there is an elementary proof of the following fact that "accidentally" fell out of some high-powered machinery he was working on.Say that two matrices $A$ and $B$ over the...
View ArticleRepresenting integers as sums of binomial coefficients and relative primeness...
Motivation: This question arose from studying an open problem in ergodic theory. Is the Pascal adic transformation weakly mixing? I tried boiling the problem down to a question about paths through...
View ArticleMore conjectural formulas for Riemann's zeta function
Motivated by Question 508743 and the known identities$$\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),$$I...
View ArticleQuadratic polynomials and Riemann's zeta function
Motivated by my recent postings, here I propose some conjectural series for Riemann's zeta function with summands involving binomial coefficients and quadratic polynomials.By Examples 14 and 32 of this...
View ArticleMore conjectural formulas for Riemann's zeta function (IV)
This is a continuation of Questions 508768, 508774 and 508844.In 2023 I conjectured the formula$$\sum_{k=1}^\infty\frac{145k^2-104k+18}{(2k-1)k^3\binom{2k}k\binom{3k}k^2}=\frac{\pi^2}3$$in Question...
View ArticleMore conjectural formulas for Riemann's zeta function (III)
This is a continuation of Questions 508768 and 508774.By Example 11 of this 2014 paper of Chu and Zhang, we have the...
View ArticleMore conjectural formulas for Riemann's zeta function (II)
Motivated by Zeilberger's series$$\sum_{k=1}^\infty\frac{21k-8}{k^3\binom{2k}k^3}=\zeta(2)$$and Questions 508743 and 508768I have discovered several identities involving higher-order derivatives of the...
View ArticleIdentities with Stirling numbers of both kinds
With the luck of intuition, I conjecture that $$ {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...
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