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While working with a combinatorics problem, I conjectured that

$$ \left\{{n \atop n-k }\right\}=\sum_{p=0}^{k-1}\bigg\langle\!\!\bigg\langle{k\atop k-1-p}\bigg\rangle\!\!\bigg\rangle \binom{n+p}{2k}, $$

where $\left\{ {n \atop k} \right\}$ is the Stirling permutation numbers and $\big\langle\!\big\langle{n \atop k}\big\rangle\!\big\rangle $ denotes the Eulerian numbers of the second kind.

  • All of my motivation comes from the fact that this is known to hold for $k = 1, 2, 3$. (See this, for instance.)

  • I have little background on this topic, and I was unable to find this one from DLMF.

  • I numerically checked that this identity holds for $n = 1, \cdots, 10$ using CAS and OEIS A008517.

Although I hardly believe that this type of identity is not known, I could not find any proof or reference to this one. So any additional information will be appreciated!

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    $\begingroup$ Given that you've got an '$n-p$' term multiplied by a '$p$' term, your sum looks like a convolution of appropriate terms, and so should be straightforwardly writeable as a product of generating functions; that's absolutely the first axis I would explore along. $\endgroup$ Commented Dec 22, 2018 at 23:42
  • $\begingroup$ @StevenStadnicki Nice point! I will definitely try that. $\endgroup$ Commented Dec 22, 2018 at 23:59
  • $\begingroup$ I've just posted an induction proof for this formula as well as its twin formula for the Stirling cycle numbers, see math.stackexchange.com/a/4814252/61691 $\endgroup$
    – azimut
    Commented Nov 26, 2023 at 9:37

1 Answer 1

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"Concrete Mathematics (what else?) - Eulerian Numbers" - says:
"Second-order Eulerian numbers are important chiefly because of their connection with Stirling numbers"

Eq. (6.43) therein gives $$ \left\{ \matrix{ x \cr x - n \cr} \right\} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left\langle {\left\langle \matrix{ n \cr k \cr} \right\rangle } \right\rangle } \binom{x+n-1-k}{2n} \quad \left| \matrix{ \;0 \le n \in Z \hfill \cr \;x \in C \hfill \cr} \right. $$ which easily reduce to yours, and can be extended to define Stirling No. of 2nd kind, of the indicated form, to complex values of $x$.

Interestingly, also given is a twin one for the Stirling No. of 1st kind $$ \left[ \matrix{ x \cr x - n \cr} \right] = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left\langle {\left\langle \matrix{ n \cr k \cr} \right\rangle } \right\rangle } \binom{x+k}{2n} \quad \left| \matrix{ \;0 \le n \in Z \hfill \cr \;x \in C \hfill \cr} \right. $$

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  • $\begingroup$ Concrete Mathematics also hints at an inductive proof of these identities. I just posted a question asking for the details of the inductive proof: math.stackexchange.com/q/4813419/61691 $\endgroup$
    – azimut
    Commented Nov 24, 2023 at 19:25

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