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Let $T(n, k)$ (A008292) be the number of permutations of length $n+1$ of distinct elements $p_1,p_2,\cdots, p_{n+1}$, where the elements belong to the set $\left\lbrace1, 2,\cdots, n+1\right\rbrace$, $p_1=1$, and for which $$(k-1)(n+1) < \sum\limits_{j=1}^{n} (p_{j+1}-p_j)\bmod (n+1) < k(n+1)$$ when $1 < k < n$ and $T(n,1)=T(n,n)=1$.

How can this be proved?

By the way, $$T(n,k)=\left\langle\! n\atop k-1 \right\rangle\!$$

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  • $\begingroup$ What exactly should be proved? The formula after "by the way"? It is equivalent to one of standard definitions of Euler numbers (via descents). $\endgroup$ Commented May 20, 2021 at 17:53
  • $\begingroup$ @FedorPetrov, thank you for commrent! Sorry, but I do not understand why is it so. $\endgroup$ Commented May 21, 2021 at 4:03
  • $\begingroup$ just substitute $(b-a) \bmod (n+1)=(b-a)+(n+1)\cdot \textbf{1}(a>b)$ for $b=p_{j+1}$, $a=p_j$ $\endgroup$ Commented May 21, 2021 at 7:26

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