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Questions tagged [bernoulli-numbers]

The Bernoulli numbers are the rational numbers $B_n$ defined as the coefficients in the expansion $\frac{x}{e^x-1} = \sum_{n \geq 0} B_n \frac{x^n}{n!}$. They vanish when $n$ is odd and greater than $2$. They appear in the values at integers of the Riemann $\zeta$ function. These classical numbers play an important role in number theory and in several other places in mathematics.

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I conjecture that For every integer $k>78$, there exists an odd prime $p$ such that the sum of last two base-$p$ digits of $k$ is $\geq p$. We may additionally assume that $k+1$ is a prime, a ...
Max Alekseyev's user avatar
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This is a categorification question: The sequence of exponential generating functions (indexed by $n$) given by $(e^{x} -1)^n$ allow you to count the number of surjective maps from a $k$ element set ...
Sidharth Ghoshal's user avatar
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I recently discovered the following identity involving Bernoulli numbers: $$ B_{n+m+1} = -\sum_{k=0}^{n} \sum_{v=0}^{m} \frac{(n+m+1)! \, B_k \, B_v}{(n+m-k-v+2)! \, k! \, v!} $$ This holds for all ...
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Let $B_n$ be Bernoulli number. I conjecture that $$ B_{2n} = \frac{1}{(2n+1) \binom{4n+1}{2n}} \left( (n+1)(2n+1) - \sum\limits_{k=0}^{n-1} \binom{2n+1}{2k} \binom{2(n+k)+1}{2k} B_{2k} \right). $$ ...
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While working through some calculations, I discovered the following intriguing relation involving Bernoulli numbers, Stirling numbers of the first kind, and Bessel numbers of the first kind: $$ \sum_{...
Abdelhay Benmoussa's user avatar
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Recently, I've come back again to one of the prominent articles in area of polynomials, power sums etc. That is Johann Faulhaber and sums of powers: https://arxiv.org/abs/math/9207222. Indeed, a great ...
Petro Kolosov's user avatar
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Denote by $S(n,k)$ the Stirling number of the second kind, i.e. $S(n,k)\cdot k!$ is the number of surjections among $k^n$. Let $$Y_k(n):=S(n,n-k)\frac{(n-k)!}{n!},$$ and denote by $d_k$ the ...
Bad English's user avatar
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Given a finite connected graph $\Gamma$ with vertices $\lbrace 1,\ldots,N\rbrace$, we can consider the polynomial $$\sum_{\pi\in\mathcal S_N}x^{\sum_{j=1}^Nd_\Gamma(j,\pi(j))}$$ where $\mathcal S_N$ ...
Roland Bacher's user avatar
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In 13 Lectures on Fermat's Last Theorem, Ribenboim states the following theorem (on page 7) attributed to Cauchy: If the first case of Fermat's theorem fails for the exponent $p$, then the sum: $$ 1^{...
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In Euler–Maclaurin formula Bernoulli numbers express a finite sum through the integral. In my generalization a finite sum is expressed through another finite sum with a different step. All that is ...
Марат Рамазанов's user avatar
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It is well known that the classical Bernoulli polynomials $B_j(t)$ are generated by \begin{equation*} \frac{s\operatorname{e}^{ts}}{\operatorname{e}^s-1}=\sum_{j=0}^{\infty}B_j(t)\frac{s^j}{j!}, \quad ...
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I arrived at this formula by inductive reasoning, but I don’t know how to prove it. For any natural numbers $m$ and $k=0,1,2,\ldots, m-1$, $B_i$ - Bernoulli numbers we have: $$\sum_{i=0}^k (-1)^{k-i}\...
juna's user avatar
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Let $B_{2k}$ be the Bernoulli numbers of even index and $\varphi(n)$ be Euler's totient function. We recall one instance of Kummer's congruences: for each integer $m\geq1$ and a prime number $p\geq5$, ...
T. Amdeberhan's user avatar
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If we denote the Bernoulli numbers by $B_n$, then $$ {a^{k+1}(a^{2k} - 1) B_{2k} \over 2k}\in \mathbb{Z}$$ for any $a\in \mathbb{Z}$ and $k\in\mathbb{N}$. This fact is often called the Sylvester–...
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As well known, Bernoulli number $B_{2n}$ is a rational number which can always be shown as $B_{2n}=V_{2n}/A_{2n}$ , where $\gcd(V_{2n}, A_{2n})=1$. For example, $B_{16}=\frac{3617}{-510}$ and $3617$ ...
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Let the function $f(t) = \cos(at)$, where ($0 < a < 1$). Let us define $$\zeta(z, f) = \frac{1}{\Gamma(z)} \int_0^{+\infty} \frac{t^{z-1}\cos(at)}{e^t-1}\, dt. $$ Is there a general formula that ...
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Motivation: We informally call an infinite lower triangular matrix $\operatorname{T}(n, k)$ of integers a combinatorial triangle of a sequence of integers or rational numbers if it can be obtained ...
Peter Luschny's user avatar
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Suppose $\{ Y_i\}_{i = 1}^n$ is i.i.d. Bernoulli distribution with mean $p$. Denote the sample of $\{ Y_i\}_{i = 1}^n$ as $\overline{Y} = \frac{1}{n} \sum_{i = 1}^n Y_i$. I want to know where there ...
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Let $B_k$ be the Bernoulli numbers and let \begin{equation} T_k=\frac{2^{2k}}{(2k)!}|B_{2k}|, \quad k\ge1. \end{equation} Prove the inequality \begin{equation*} \frac{\frac{1}{k+2}\sum_{j=0}^{k+1}\...
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Let $c(x)=\frac{1-\sqrt{1-4x}}{2x}$ be the generating function of the Catalan numbers and let $$x^k c(x)^{2k}=(c(x)-1)^k =\sum_{n\geq0}c(k,n)x^n.$$ Consider the determinants $$D(k,n,m)= \det\left(c(k,...
Johann Cigler's user avatar
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In A Multimodular Algorithm for Computing Bernoulli Numbers, Harvey uses the following congruence for Bernoulli numbers: $$B_k \equiv \frac{k}{1-c^k} \sum_{x=1}^{p-1} x^{k-1} h_c(x)\quad(\text{mod}\ p)...
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Let $p \equiv 1 \pmod{3}$ be a prime and denote $H_{n,m} = \sum_{k = 1}^n 1/k^m$ as the $n,m$-th generalized harmonic number. I'm interested in computing $H_{(p-1)/3,\, 2}$ and $H_{(p-1)/6,\,2}$ ...
matt stokes's user avatar
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It is well known that the Bernoulli numbers $B_{k}$ for $k\in\{0,1,2,\dotsc\}$ can be generated by \begin{equation*} \frac{z}{\textrm{e}^z-1}=\sum_{k=0}^\infty B_k\frac{z^k}{k!}=1-\frac{z}2+\sum_{k=1}^...
qifeng618's user avatar
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Let $B_{n}$ for $n\ge0$ denote the Bernoulli number generated by \begin{equation*} \frac{z}{\textrm{e}^z-1}=\sum_{n=0}^\infty B_n\frac{z^n}{n!}=1-\frac{z}2+\sum_{n=1}^\infty B_{2n}\frac{z^{2n}}{(2n)!},...
qifeng618's user avatar
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I know that currently umbral calculus is developed as some kind of theory of operators and functionals but were there any attempts to put it on a more solid philosophical grounds as study of functions ...
Anixx's user avatar
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Let $p>2$ be a prime, $\zeta$ a primitive $p$th root of unity, and $m >0$ a square-free integer such that $m \not\equiv 3 \mod 4$ and $\gcd(m,p)=1$. Let $\chi$ be the imaginary quadratic ...
matt stokes's user avatar
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Is the following analogue of Fermat's Little Theorem for Bernoulli numbers true? Let $D_{2n}$ be the denominator of $\frac{B_{2n}}{4n}$ where $B_n$ is the $n$-th Bernoulli number. If $\gcd(a, D_{2n}) ...
Nilotpal Kanti Sinha's user avatar
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Faulhaber's formula expresses a sum over some finite number of naturals to the $m^{th}$ power in terms of the Bernoulli numbers $B_{j}$ (using the $B_{1} = 1/2$ convention) or polynomials $\hat{B}_{j}$...
Matthew Cassell's user avatar
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1 answer
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I am currently researching divergent integrals. Definition. An extended number is an expression of the form $\int_a^b f(x)\,dx$, where $a,b\in \overline{\mathbb{R}}$ and function $f(x)$ is defined ...
Anixx's user avatar
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For a matrix $[a_{j,k}]_{1\le j,k\le n}$ over a field, its permanent is defined by $$\mathrm{per}[a_{j,k}]_{1\le j,k\le n}:=\sum_{\pi\in S_n}\prod_{j=1}^n a_{j,\pi(j)}.$$ In a recent preprint of mine, ...
Zhi-Wei Sun's user avatar
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So, can we transform an even function into an odd function and vice versa? Let's consider this method: Transformation even->odd: Suppose $f_{even}(x)$ is a function which satisfies the following ...
Anixx's user avatar
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Prove for the Bernoulli numbers $B_n$, that for all $a \in \mathbb{N}$, that $ \sum_{i=0}^{2a+1} {2a+1 \choose i} B_{2a+1-i} [ (n+1)^i+(-n)^i ] =0 $. As much as this is a neat identity, it's a crucial ...
Benjamin L. Warren's user avatar
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Crossposted from https://math.stackexchange.com/questions/4116414/conjecture-on-bernoulli-numbers-and-binomial-coefficients In playing around with some formulas, I have come up with the following ...
Fox Mulder's user avatar
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Do you know of a text where I can find a nicely motivated proof of the formula for $1^{k}+2^{k}+\cdots+n^{k}$? At the very beginning of page 68 of Professor H. S. Wilf's generatingfunctionology, one ...
Jamai-Con's user avatar
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The purpose of this question is to resolve a mystery surrounding the prime 34511 that has got me bogged down for a while now. If you only care about the number theory and not the motivation coming ...
Jens Reinhold's user avatar
12 votes
2 answers
477 views

This question is grounded firmly in numerology. It originates in an observation about some Bernoulli polynomials and the regular icosahedron. Let $F_{k+1}(n)=\sum_{i=1}^n i^k$ be the sum of the ...
David Richter's user avatar
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The Möbius-Bernoulli numbers ,are related to Dedekind Sums $$\sum_{d|n}\frac{t\mu(d)}{e^{td}-1}=\sum_{k=0}^\infty M_k(n)\frac{t^k}{k!}$$ where $|t|<\frac{2\pi}{n}$, and $M_k(1)=B_k$. We define the ...
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1 vote
1 answer
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Let $x,y,z$ are integer and $x,y>0$ Define $S(x,y)=1^y+2^y+3^y+...+x^y$ Can it be shown that If given $z\ne0$ then $S(x,y)\equiv z\pmod{x}$ have finitely many solution of $x$ with respect to $y$. ...
Pruthviraj's user avatar
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For any irreducible polynomial $f$ of degree $d$ in finite field $\mathbb F$ we have that $$x^{|\mathbb F|^d}-x\equiv0\bmod f$$ and thus the polynomial $x^{|\mathbb F|^d}-x$ is the analog of factorial ...
VS.'s user avatar
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Edit: Few years ago, I have posted my claim on $\Upsilon$ function regarding prime number but recently I have observed, last observation turns false that's way, (by putting $\Upsilon$ value in ...
Pruthviraj's user avatar
1 vote
2 answers
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Throughout this question $n$ is a positive integer greater than 1. Consider the following well-known identity by Euler, $$\sum_{k=1}^{n-1} \binom{2n}{2k}B_{2k}B_{2n-2k}=-(2n+1)B_{2n}.$$ Rather ...
bryanjaeho's user avatar
3 votes
1 answer
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Denote by $B_r$ the $r$-th Bernoulli polynomial. Are there any positive integers $r, x$ such that. $B_r(x)$ divides $B_r(x+1)$ or vice versa ?
Q_p's user avatar
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8 votes
2 answers
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I stumbled across the following curious empirical properties of the Bernoulli polynomials $B_n(x)$. Can anyone provide a reference or proof? Let $k\in\mathbb{Z}$, $k\geq 2$. Then (empirically): The ...
Richard Stanley's user avatar
7 votes
1 answer
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Let $\chi : (\mathbb Z/f\mathbb Z)^\times \to K = \mathbb Q(\mu_{\phi(f)})$ be a primitive Dirichlet character. Assume moreover that it is not quadratic, that is, $\chi^2$ is not the trivial character....
Asvin's user avatar
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16 votes
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This is again a question that I asked at Stack Exchange, but got no answer so far, so I am trying here. Let: $$ a_n=\sum_{k\ge0}(k+1) {n+2\brack k+2}(n+2)^kB_k$$ $B_k$ is the Bernoulli number. ${n\...
René Gy's user avatar
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Let $B_n$ denote the Bernoulli numbers and let $\phi=\frac{1+\sqrt{5}}2$ be the golden ratio. I encountered the following infinite sum and would like to ask: Question. Is this true? If so, any ...
T. Amdeberhan's user avatar
22 votes
3 answers
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Contemplating a question on math.SE, I have stumbled on this: Here, the point labeled $n$ is that root of the $n$th Bernoulli polynomial which has smallest positive imaginary part. Does anyone know ...
მამუკა ჯიბლაძე's user avatar
8 votes
3 answers
858 views

I wonder, can anyone describe an expression or formula of a transform that converts $$\sum_{k=0}^\infty \frac{a_k x^k}{k!}$$ into $$\sum_{k=0}^\infty \frac{a_k B_k(x)}{k!}$$ where $B_k(x)$ are ...
Anixx's user avatar
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6 votes
2 answers
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Any reference that we can find the following $$\Bigr[-\log(1-t)\Bigr]^x = t^x + x t^x \sum_{k=0}^\infty \psi_k(x+k)\,t^{k+1}; \quad \mbox{for all} \, x\in \mathbb R, \, |t|<1$$ where $\psi_k(.)$ ...
Z. Alfata's user avatar
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I am wondering if there are any nice results on the values of Bernoulli polynomials at roots of unity, besides those at 1 or -1.
Staideal's user avatar