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My question is about second kind r-Stirling numbers. Here are two important papers about it. https://www.sciencedirect.com/science/article/pii/0012365X84901614#:~:text=The%20r%2DStirling%20numbers%E2%98%86&text=The%20r%2DStirling%20numbers%20of,cycles%20and%20respectively%20distinct%20subsets. Broder/ r-Stirling numbers. https://www.sciencedirect.com/science/article/pii/S0012365X14001241#:~:text=3.&text=%2DLah%20numbers%20The%20%2DLah%20numbers,be%20in%20distinct%20ordered%20blocks. Ryul and Nacz / r-Lah numbers.

In Broder / r-Stirling numbers article, r-Stirling numbers of second kind are defined; $\left\{\begin{array}{l}n \\ m\end{array}\right\}_{r}=$ The number of partitions of the set $\{1, \ldots, n\}$ into $m$ non-empty disjoint subsets ,such that the numbers $1,2,3...,r$ are in distinct subsets.

I try to understand the definition of second kind r-Stirling numbers. For r-Stirling numbers r is a natural number. Can the range of r be extended to rational numbers or complex numbers? Thanks.

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  • $\begingroup$ When you ask a question, you should make sure it is useful for the future readers, too. You start a sentence by from what I understand ... You should write your interpretation and deduction of why $r$ is a natural number, give some context of your knowledge so that people know how to answer your question. You gave a link to two important papers , which sounds a bit like giving an assignment. Also, if you're looking for a reference, make that clear in the body, not just by a tag. $\endgroup$
    – PinkyWay
    Commented Feb 1, 2022 at 22:10
  • $\begingroup$ thanks. i will edit it:) Do you have any opinions about my question? @invisible $\endgroup$
    – user1062
    Commented Feb 1, 2022 at 22:12
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    $\begingroup$ I'm going to read it. (: I hope I didn't sound rude and that your question gets due attention. (: $\endgroup$
    – PinkyWay
    Commented Feb 1, 2022 at 22:14
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    $\begingroup$ I'm not sure about rational or complex numbers, but I think it can extended to negative integers. See this paper for a different version of a restriction that preserves the reciprocity between the restricted Stirling numbers of the first and second kinds. Hope that helps. $\endgroup$ Commented Feb 2, 2022 at 0:18

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Not sure what are your intentions, but you can use the known expression $${n\brace k}_r=\sum _{i=0}^n\binom{n}{i}{i\brace k}r^{n-i}$$ to extend it. That comes from the fact that you can choose the numbers that are going to be in the first $r$ subsets(say $n-i$ of them) in $\binom{n}{n-i}$ ways, and the rest elements have to be in $k$ blocks in ${i\brace k}$ ways. To distribute the $n-i$ elements in the $r$ first blocks, you can use any function in $r^{n-i}$ ways.

For example, for $r<0$ and $n,k$ even, one can express ${n\brace k}_r$ using Stirling numbers of higher level (see remark 3.1 here).

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    $\begingroup$ Thanks for your answer. I have two questions. The first one what about $$\left\{\begin{array}{c}n+r \\ k+r\end{array}\right\}_{r}$$ when r is negative or other conditions ? My second question the expression you wrote above are Noncentral stirling numbers S(n,k;r) to me, aren't it? So i was a little surprised. You may also look for it to Charalambides Enumerative Combinatorics, page 316. $\endgroup$
    – user1062
    Commented Feb 4, 2022 at 20:18
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    $\begingroup$ @user1062 1) That is what I mean in your notation, sorry there are two standard ways to see $r-$stirling numbers. So, what I mean by $\{n\brace k\}_r$ in your notation is actually $\{n\brace k+r\}_r$ but in your notation you want $n\geq r$ so take $n'=n-r$ and it is the same expression. For the second one, those are two different generalizations. The non-central stirling numbers are defined, in your notation, as connecting coefficients of $(x)_n$ and $(x-r)^n$ and the Stirling numbers of higher level are connecting coefficients of $x(x-1^r)(x-2^r)\cdots(x-(n-1)^r)$ and $x^n$. $\endgroup$
    – Phicar
    Commented Feb 4, 2022 at 20:38
  • $\begingroup$ That's great comment! I see it. The informations you have given are very valuable. Thank you very much. $\endgroup$
    – user1062
    Commented Feb 4, 2022 at 21:02
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    $\begingroup$ @user1062 You are welcome. Good luck! $\endgroup$
    – Phicar
    Commented Feb 4, 2022 at 21:05

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