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We have for $n>0$, $k>0$ $$\sum\limits_{j=1}^{\min(n,k)}(j!)^2{n\brace j}{k+1\brace j+1}=\sum\limits_{j=0}^{\min(n,k-1)}j!(j+1)!{n+1\brace j+1}{k\brace j+1}$$ How can we prove it?

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  • $\begingroup$ Do you have any context? $\endgroup$ – Phicar Apr 23 at 20:18
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    $\begingroup$ For the second sum "Masanobu Kaneko. Multiple zeta values, poly-Bernoulli numbers, and related zeta functions. Nagoya Math Journal, 153:189-201, 1999". First one is mine and based on previous question. $\endgroup$ – user514787 Apr 23 at 20:21
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Using the recurrence relation for the Stirling numbers of the second kind \begin{eqnarray*} {k+1\brace j+1}=(j+1){k\brace j+1}+{k\brace j}. \end{eqnarray*} The sum becomes \begin{eqnarray*} S&=&\sum\limits_{j=1}^{\min(n,k)}(j!)^2{n\brace j}{k+1\brace j+1} \\ &=& \sum\limits_{j=1}^{\min(n,k)}(j!)((j+1)!){n\brace j}{k\brace j+1} +\sum\limits_{j=1}^{\min(n,k)}(j!)^2{n\brace j}{k\brace j} \end{eqnarray*} shift the second summation variable by $1$ \begin{eqnarray*} S&=& \sum\limits_{j=0}^{\min(n,k-1)}(j!)((j+1)!){k\brace j+1} \left({n\brace j}+(j+1){n\brace j+1}\right)\\ \end{eqnarray*} Now use the recurrence formula again \begin{eqnarray*} {n+1\brace j+1}={n\brace j}+(j+1){n\brace j+1} \end{eqnarray*} & we are done.

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  • $\begingroup$ Brilliant! Thank you very much! $\endgroup$ – user514787 Apr 23 at 20:29

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