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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
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
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
  • 82.7k
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
1 vote
2 answers
196 views

A question about generalized harmonic numbers modulo $p$

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
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
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
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
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
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
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
1 vote
1 answer
186 views

Claim on divisibility of a power sum

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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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