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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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3 votes
1 answer
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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
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
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
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}...
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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 $\...
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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
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
101 votes
10 answers
16k views

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 ...
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
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

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