**10**

votes

**3**answers

813 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) = - ...

**55**

votes

**9**answers

6k 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 ...

**25**

votes

**11**answers

3k 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 ...

**1**

vote

**0**answers

64 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 ...

**4**

votes

**1**answer

124 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 = ...

**0**

votes

**1**answer

185 views

### Probability of the maximum of a throw of an infinite number of $n$-sided dice being $k$ [closed]

Let $X$ be the random variable obtained as the maximum of a throw of $m$ dice (each of which is $n$-sided). In other words, $X = \max\{l_1,\cdots, l_m\}$ where $l_i$ can take any value between $1$ and ...

**10**

votes

**0**answers

170 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 ...

**5**

votes

**2**answers

366 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 ...

**12**

votes

**0**answers

584 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 $2 \leq k \leq p-3$, ...

**0**

votes

**0**answers

258 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:
...

**3**

votes

**2**answers

224 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 ...

**3**

votes

**2**answers

215 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 ...

**1**

vote

**0**answers

321 views

### P-Adic poly Bernoulli numbers

we can define p-adic Bernoulli polynomials by using q-integral on $Z_p$ and T.Kim's method, But how can we define p-adic poly-Bernoulli numbers and polynomials by using integral on $Z_p$ ?

**15**

votes

**1**answer

775 views

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

http://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 ...