Skip to main content

All Questions

Filter by
Sorted by
Tagged with
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 ...
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 $p\geq5$, ...
1 vote
2 answers
272 views

Ask for a reference or a proof of an identity involving a finite sum and the Bernoulli numbers

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)!},...
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}^...
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 ...
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 ...