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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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101 votes
10 answers
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Why do Bernoulli numbers arise everywhere?

I have seen Bernoulli numbers many times, and sometimes very surprisingly. They appear in my textbook on complex analysis, in algebraic topology, and of course, number theory. Things like the criteria ...
45 votes
16 answers
8k views

What does the generating function $x/(1 - e^{-x})$ count?

Let $x$ be a formal (or small, since the function is analytic) variable, and consider the power series $$ A(x) = \frac{x}{1 - e^{-x}} = \sum_{m=0}^\infty \left( -\sum_{n=1}^\infty \frac{(-x)^n}{(n+1)!}...
Theo Johnson-Freyd's user avatar
22 votes
3 answers
1k views

Roots of Bernoulli polynomials - a pattern

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
19 votes
0 answers
649 views

Bernoulli & Betti numbers (of manifolds) and the prime 34511

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
18 votes
5 answers
3k views

Bernoulli sum meets golden number

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
18 votes
1 answer
1k views

Riemann's $\zeta$ function and the uniform distribution on $[-1,0]$

https://math.stackexchange.com/questions/64566/riemanns-zeta-function-and-the-uniform-distribution-on-1-0 Stackexchange isn't getting really excited about this, so here it is. The $n$th cumulant of ...
Michael Hardy's user avatar
16 votes
1 answer
706 views

Connection between Bernoulli numbers and Riemann-Siegel theta function?

I have come across a strange approximation for the Riemann-Siegel theta function involving the Bernoulli numbers - namely that $$\frac{1}{2} \log \left| B_{2 n}\right|\approx \vartheta (2n)\ ,\quad n ...
martin's user avatar
  • 1,903
16 votes
1 answer
584 views

What is this sequence?

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
  • 505
14 votes
1 answer
308 views

For a given even integer $k >14$ is there always a prime $p$ such that $k \leq p-3$ and $p|B_k$?

Let $k$ be a sufficiently large positive even integer. (I think $k > 14$ should do.) Can one always find a prime $p$ such that $p$ divides the numerator of the $k$-th Bernoulli number $B_k$ and $k \...
unramified's user avatar
13 votes
0 answers
766 views

A conjecture on p-divisibility of Bernoulli numbers

Is anyone aware of the history of the following conjecture on the $p$-divisibility of (the numerators) of Bernoulli numbers? CONJECTURE: For $p$ an odd prime, and $k$ even with $2 \leq k \leq p-3$, $...
unramified's user avatar
12 votes
1 answer
599 views

Fermat last theorem : proof of a criterion by Cauchy

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^{...
RUser4512's user avatar
  • 121
11 votes
3 answers
958 views

Why do $12$ and $120$ occur very often in the denominators of $\zeta(-n)$ for odd $n$?

$\zeta(-n) = - \dfrac{B_{n+1}}{n+1}$ $\zeta(-2n) = 0$ $\zeta(-1) = - \dfrac{1}{12}$ $\zeta(-3) = \dfrac{1}{120}$ $\zeta(-5) = - \dfrac{1}{252}$ $\zeta(-7) = \dfrac{1}{240}$ $\zeta(-9) = - \dfrac{...
user50746's user avatar
  • 341
10 votes
2 answers
369 views

Do Bernoulli polynomials know about face vectors?

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
8 votes
3 answers
759 views

Transformation converting power series to Bernoulli polynomial series

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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8 votes
2 answers
386 views

Coefficients of shifted Bernoulli polynomials

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
185 views

2-adic valuation of $L(0,\chi)$ for a Dirichlet character

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
  • 7,746
7 votes
0 answers
252 views

Hankel determinants for some convolutions of Catalan numbers

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
6 votes
2 answers
502 views

Any reference for the series expansion of $\Bigr[-\log(1-t)\Bigr]^x$?

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
  • 650
6 votes
1 answer
482 views

A non-$p$-adic proof of a congruence of Bernoulli numbers

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)...
Manuel Eberl's user avatar
  • 1,241
6 votes
1 answer
520 views

A novel identity connecting permanents to Bernoulli numbers

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
  • 15.6k
6 votes
1 answer
695 views

An operation is defined on polynomials. How do I generalize it to other classes of functions?

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
  • 10.1k
6 votes
0 answers
126 views

divisibility by Bernoulli numbers of discriminant of Hecke algebra over the space of modular forms of level 1

For the space of modular forms and the space of cusp forms (here I only care about the level $1$ case), we have the action by Hecke algebras. Therefore, we can calculate the discriminant of this ...
Ding's user avatar
  • 61
5 votes
2 answers
391 views

zeta(3) in Euler's Section 153

Jeffery Lagarias, in his recent article Euler's constant: Euler's work and modern developments in the AMS Bulletin, mentions that Euler obtained $\zeta(3)={{2\pi^3 b(3/2)}\over 3}$ for some "Bernoulli ...
Sang G. Hahn's user avatar
5 votes
1 answer
281 views

How naturally can functions defined by parametric integrals be interpolated from $\mathbb N$ to $\mathbb R^+$?

It is well known that several definite integrals $I_n$ containing a parameter $n\in\mathbb N$ can be expressed recursively (e.g. doing integration by parts) in terms of $I_{n-1} $ or $I_{n-2} $, and ...
Wolfgang's user avatar
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4 votes
2 answers
535 views

computing Bernoulli numbers

Is there a good way to compute the ratio ( B[n] / n! ) that occurs so often in power series coefficients? Good in the sense that you get an answer that does not overflow a double; the largest n such ...
Michael Beeson's user avatar
4 votes
1 answer
260 views

Kummer's congruence at $p=3$

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
4 votes
1 answer
243 views

Estimate of the sum Taylor's coefficients

Let $f(x) = \begin{cases}\ln\frac{x}{e^x-1}, \quad x > 0; \\ 0, \quad\qquad x=0; \\ \ln\frac{x}{e^x-1}, \quad x < 0. \end{cases}$ Power series in 0: $f(x) = \sum_{n=1}^{\infty} a_n x^n = -\frac{...
Gleb Fedorov's user avatar
4 votes
0 answers
128 views

A combinatorial triangle for the Bernoulli numbers

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
3 votes
2 answers
527 views

Coefficients in Hirzebruch polynomial and divisibility of Bernoulli numbers: reference request

I seek a reference for the fact that "coefficients of the Hirzebruch $L$-polynomial have odd denominators". The coefficients are $$\frac{2^{2k}(2^{2k-1}-1)B_k}{(2k)!}$$ where $B_k$ is the Bernoulli ...
Igor Belegradek's user avatar
3 votes
2 answers
445 views

Where to find or how to prove that the ratio of two Bernoulli polynomials is increasing?

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 ...
qifeng618's user avatar
  • 1,101
3 votes
2 answers
241 views

Sign of coefficients

Let $a_0,a_1,\dots$ be the sequence satisfying $$ \left(\sum_{n=0}^\infty a_n x^n\right)\left(\sum_{n=0}^\infty \frac{x^n}{n+1}\right)=1. $$ This means that $a_0=1$ and $a_{n+1}=-\sum_{j=0}^n\frac{a_j}...
user avatar
3 votes
1 answer
537 views

Generalized Bernoulli numbers

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
3 votes
1 answer
437 views

Identities for Bernoulli numbers

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
  • 31
3 votes
1 answer
507 views

Conjecture on bernoulli numbers and binomial coefficients

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
3 votes
1 answer
189 views

A question on Bernoulli polynomials

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
  • 1,019
3 votes
1 answer
2k views

What is $p$-adic Fourier series?

Q1: Can we define Fourier series for a function $\mathbb{Z}_p\to \mathbb{Q}_p$? Q2: There are (in a real case) Bernoulli polynomials which have the most simple Fourier expansion: $$B_n(\{x\})=-\frac{...
Alexey Ustinov's user avatar
3 votes
1 answer
149 views

Drunken X-mas polynomials for graphs

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
3 votes
2 answers
656 views

p-adic poly-Bernoulli numbers

We can define p-adic Bernoulli polynomials by using q-integral on $\mathbb{Z}_p$ and Taekyun Kim's method. But how can we define p-adic poly-Bernoulli numbers and polynomials by using integral on $\...
user avatar
3 votes
0 answers
186 views

Combinatorial interpretation of Sylvester–Lipschitz formula?

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–...
Timothy Chow's user avatar
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3 votes
0 answers
145 views

Interpreting umbral calculus in terms of some kind of extended numbers

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
  • 10.1k
3 votes
0 answers
157 views

Combinatorial interpretation for Möbius-poly-Bernoulli numbers

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 ...
user avatar
3 votes
0 answers
237 views

Values of Bernoulli polynomials at roots of unity

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
2 votes
2 answers
283 views

Ask for a proof of an identity involving the product of two Bernoulli numbers

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
  • 1,101
2 votes
1 answer
307 views

Analogue of Fermat's little theorem for Bernoulli numbers

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
2 votes
0 answers
217 views

Zeta function associated with a function $f$

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 ...
L.L's user avatar
  • 463
2 votes
0 answers
84 views

Question about infinitude of $m$-irregular primes

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
2 votes
0 answers
118 views

Analogues over finite fields of certain integers defined multiplicatively in $\mathbb Z$

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
  • 1,836
2 votes
0 answers
212 views

show that sequence $\{(-1)^n\Upsilon_n\}$ is convergent and strictly decreasing

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
1 answer
168 views

Prove $ \sum_{i=0}^{2a+1} {2a+1 \choose i} B_{2a+1-i} [ (n+1)^i+(-n)^i ] =0 $ for Bernoulli numbers $B_{n}$

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
1 vote
2 answers
211 views

Bilinear recurrence relation between even Bernoulli numbers

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