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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
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
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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