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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}\...
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 ...
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 ...