All Questions
3 questions
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:
$$\sum_{i=0}^k (-1)^{k-i}\...
18
votes
5
answers
3k
views
Bernoulli sum meets golden number
Let $B_n$ denote the Bernoulli numbers and let $\phi=\frac{1+\sqrt{5}}2$ be the golden ratio.
I encountered the following infinite sum and would like to ask:
Question. Is this true? If so, any ...
2
votes
0
answers
118
views
Analogues over finite fields of certain integers defined multiplicatively in $\mathbb Z$
For any irreducible polynomial $f$ of degree $d$ in finite field $\mathbb F$ we have that $$x^{|\mathbb F|^d}-x\equiv0\bmod f$$ and thus the polynomial $x^{|\mathbb F|^d}-x$ is the analog of factorial ...