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Recently, I was reading a blog post called The P-transform by Peter Luschny, where the following formulas are given:

\begin{align*} (-1)^k\frac{n!}{k!}\mathcal P^k_n\left(1, \frac1 2, \frac2 3, \dotsc\right) & = \genfrac[]{0pt}{}n k \\ (-1)^k\frac{n!}{k!}\mathcal P^k_n(1, 1, 1, \dotsc) & = \genfrac\lvert\rvert{0pt}{}n k \\ (-1)^k\frac{n!}{k!}\mathcal P^k_n\left(1, \frac1 2, \frac1 3, \dotsc\right) & = \genfrac\{\}{0pt}{}n k \\ (-1)^k\frac{n!}{k!}\mathcal P^k_n\left(1, 0, 0, \dotsc\right) & = \delta_{n, k}. \end{align*}

The right-hand side in the first formula is the Stirling cycle number, the right-hand side in the second formula is the Lah number, right-hand side in the third formula is Stirling set number and the right-hand side in the fourth formula is Kronecker's delta.

It is obvious that $\mathcal P_{n}^{k}(1, 1, 1, ...) = \binom{n-1}{k-1}$ since the definition of the Lah numbers according to Wikipedia is

$$\genfrac\lvert\rvert{0pt}{}n k = \frac{n!}{k!} \binom{n-1}{k-1}.$$

In the blog I do not find the general formula for $\mathcal P_{n}^{k}(a_{1}, a_{2}, \dotsc)$, so my question is: what is the general formula for $\mathcal P_{n}^{k}(a_{1}, a_{2}, \dotsc)$?

Edit: (some questions I came up with after posting this question)

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    $\begingroup$ Welcome to MO! It is better to use TeX rather than images, for searchability and accessibility. I have transcribed your images—I believe accurately, but please check. $\endgroup$
    – LSpice
    Commented Jul 4 at 19:28
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    $\begingroup$ @LSpice Thank you, the equations are well transcribed. $\endgroup$
    – Math123
    Commented Jul 4 at 19:33
  • $\begingroup$ The specialization at $1$, $1/2$, $1/3, \ldots$ reminds me of the formula $h_r(1,2,3,\ldots, m) = S(r+m,m)$, where $h_r$ is the complete homogeneous symmetric function of degree $r$ and $S$ is the second kind Stirling number denoted with curly braces above. (It seems genfrac doesn't work in comments.) Dually for the (positive) first kind Stirling numbers, denoted with square brackets above, there is $e_r(1,2,3,\ldots, m) = s(m+1,m+1-r)$, where $e_r$ is the elementary symmetric function of degree $r$. $\endgroup$ Commented Jul 4 at 20:08
  • $\begingroup$ @MarkWildon, re, I believe \genfrac does work in comments (though I'll shamefacedly delete this comment if it doesn't). It takes 6 arguments, which does make it easy to miss one: $\genfrac\{\}{0pt}{}n k$ \genfrac\{\}{0pt}{}n k. $\endgroup$
    – LSpice
    Commented Jul 6 at 15:44

1 Answer 1

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$P_n$ is given as $$P_n(f) = \sum_{\lambda \,\vdash\, n} (-1)^{\lambda_1} \prod \binom{\lambda_j}{\lambda_{j+1}} f_{j}^{\lambda_j}$$ where

  • the sum is over partitions $\lambda = \lambda_1 \ge \lambda_2 \ge \cdots$ of $n$;
  • $\lambda_{j+1} = 0$ if $j+1$ is greater than the number of parts in $\lambda$;
  • I've adjusted the index from $f_{j+1}$ to $f_j$ because it appears that the increment was a hack to 1-index the variables in a 0-indexed programming language.

Then in the section Prototypic examples the text says

So let's define $P^k_n(f)$ as the $P$ transform of $f$ restricted to the partitions of $n$ with largest part $k$

Therefore $$P^k_n(f) = \sum_{\lambda' \,\vdash\, n-k} (-1)^{k} \binom{k}{\lambda'_1} f_1^k \prod \binom{\lambda'_j}{\lambda'_{j+1}} f_{j+1}^{\lambda_j}$$

A combinatorial interpretation of this can be obtained by defining a stalactite diagram (based on the English convention for Ferrers diagrams) to be a collection of cells where the top row is continuous and each cell in a lower row is directly below another cell. E.g. the stalactite diagrams with 4 cells are by columns: 4; 3,1; 2,2; 1,3; 2,1,1; 1,2,1; 1,1,2; 1,1,1,1

If we label the cells in the top row with $-f_1$ and the cells in row $k > 1$ with $f_k$ then $P_n^k(f)$ is the sum over all stalactite diagrams of $n$ cells in $k$ columns of the product of the cells in the diagram. This leads directly into the alternative formula $$P_n^k(f) = [x^n] \left(-f_1 x - f_1 f_2 x^2 - f_1 f_2 f_3 x^3 - \cdots \right)^k$$ and that answers the follow-up question as to the relationship with De Moivre polynomials: $A_{n,k}(a) = [x^n](a_1x + a_2x^2 + a_3x^3 + \cdots)^k$, so $$P_n^k(f) = A_{n,k}(-f_1, -f_1 f_2, -f_1 f_2 f_3, \ldots)$$

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    $\begingroup$ Thank you very much. Only one question: I recently discovered this article (arxiv.org/abs/2203.02868) on ArXiv where DeMoivre polynomials $A_{n, k}$ are defined (equation 1.2 and definiton 1.1). They satisfy very similar connections to Stirling numbers of both kinds (equations 2.21 and 2.22). The formula involving Stirling cycle numbers is the same, the formula involving Stirling set numbers is a bi different because of the factorials. So my question is how are $A_{n, k}(a_{1}, ...)$ and $P_{n}^{k}(a_{1}, ...)$ connected? $\endgroup$
    – Math123
    Commented Jul 5 at 8:57
  • $\begingroup$ P.S. If you think that should be a new separated question, please let me know. $\endgroup$
    – Math123
    Commented Jul 5 at 8:57
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    $\begingroup$ $A_{n,k}$ is expressed as a sum over partitions of $n$ into $k$ parts, whereas $P_n^k$ is expressed as a sum over partitions of $n$ with largest part $k$. It's possible that there's a connection through conjugation of partitions. I suggest thinking about how to express them through labellings of Ferrers diagrams, but I don't promise that that will be useful. $\endgroup$ Commented Jul 5 at 10:17
  • $\begingroup$ @PeterTaylor Thank you very much! So these stalactite diagrams are English convention of Ferrers diagrams (for example as here: edwardmpearce.github.io/tutorial-partitions/intro/visualization/…)? $\endgroup$
    – Math123
    Commented Jul 6 at 9:19
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    $\begingroup$ @Math123, no: Ferrers diagrams are a subset which is strict for $n \ge 3$. But in the French convention they would be stalagmites instead of stalactites. $\endgroup$ Commented Jul 6 at 15:31

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