Let $\mathfrak{S}\_n$ be the permutation group on an $n$-element set. For each fixed $k\in\mathbb{N}$, consider the two sets
$$A\_n(k)=\{\sigma\in\mathfrak{S}\_n\vert\,\, \text{$\exists i,\,\, 1\leq i\leq n\,$ such that $\,\sigma(i)-i=k$}\}$$
and 
$$B\_n(k)=\{\sigma\in\mathfrak{S}\_n\vert\,\, \text{$\exists i,\,\, 1\leq i\leq n\,$ such that $\,\sigma(i+1)-\sigma(i)=k$}\}.$$ 



> 
> **QUESTION.** I believe the following is true, for each $n$ and $k$:
>  $$\# A\_n(k)=\# B\_n(k).$$
>  Is there a **combinatorial** proof of this? If it is known, then can you provide references?
> 
> 
> 



Two statistics on the permutation group
A simple variant of the "transformation fondamentale" of Rényi and
of Foata-Schützenberger does the trick. Write a permutation $\sigma$ in
disjoint cycle form, with the smallest element of each cycle first,
and the cycles arranged in decreasing order of the smallest element,
e.g., $(7,8)(5,6,9)(3)(1,4,2)$. Erase the parentheses to get another
permutation $\hat{\sigma}$, written as a word, e.g., $785693142$. This gives
a bijection $\mathfrak{S}\_n\to\mathfrak{S}\_n$ with the property that
$\sigma(i)-i=k>0$ if and only if
$\hat{\sigma}(j+1)-\hat{\sigma}(j)=k$, where $\hat{\sigma}(j)=i$. 


