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Charlier polynomials

Chihara starts with the general recurrence relation

{$P_{n+1}(x)=[x - (dn + f)]P_n(x) - [n(gn + h)]P_{n-1}(x)$} where {$g>0, g+h>0, d, f\in\mathbb{R}, d^2-4g\neq 0$}

The Charlier polynomials arise when {$d\neq 0, g=0$}.

Set {$f=\frac{h}{d}$}. In my notation, set {$cf=\frac{\alpha\beta c}{-\alpha -\beta}$}, which means {$f=\frac{\alpha\beta}{-\alpha -\beta}$}

{$P_{n+1}(x)=[x - d(n + \frac{h}{d^2})]P_n(x) - nhP_{n-1}(x)$}

Set {$a=\frac{h}{d^2}$}. In my notation, this means {$a=\frac{\alpha\beta c}{(-\alpha-\beta)^2}$}.

{$P_{n+1}(x)=[x - d(n + a)]P_n(x) - nhP_{n-1}(x)$}

Define {$P_n(x)=d^nC_n^{(a)}(x/d)$}, which is {$C_n^{(a)}(x)=\frac{1}{d^n}P_n(dx)$}.

{$\frac{1}{d^{n+1}}P_{n+1}(dx)=\frac{1}{d^{n+1}}[dx - d(n + a)]P_n(dx) - \frac{1}{d^{n+1}}nhP_{n-1}(dx)$}

{$C_{n+1}^{(a)}(x)=\frac{1}{d}[dx - d(n + a)]C_n^{(a)}(x) - \frac{1}{d^{2}}nhC_n^{(a)}(x)$}

{$C_{n+1}^{(a)}(x)=[x - (n + a)]C_n^{(a)}(x) - anC_n^{(a)}(x)$}

Kim and Zeng

Dongsu Kim, Jiang Zeng. A Combinatorial Formula for the Linearization Coefficients of General Sheffer Polynomials.

Set {$f= -\beta$}, {$-\alpha = 1$}, {$c=\frac{a}{-\beta}$}. We get

{$L(x^n)=\sum_{\sigma\in S_n}(-\beta)^{\textrm{fix}\;\sigma + \textrm{dec}\;\sigma - \textrm{cyc}\;\sigma}a^{\textrm{cyc}\;\sigma}$}

and then we set {$\beta = 0$}. Then we are summing over permutations where each cycle is a singleton or has a single decedence, thus can be identified with a part in a partition of {$[n]$}. Consequently,

{$L(x^n)=\sum_{\pi\in\Pi_n}a^{\textrm{bloc}\;\pi}$}

where {$\Pi_n$} is the set of partitions of {$[n]$} and {$\textrm{bloc}\;\pi$} is the number of parts in the partition {$\pi$}.

The recurrence relation becomes

{$P_{n+1}(x)=[x-((-\alpha)+1)n+a)]P_n(x)-[(-\alpha)n(n-1) + n\frac{a}{-\alpha}]P_{n-1}(x)$}

My notation

Start here: {$P_{n+1}(x)=[x-(((-\alpha)+(-\beta))n+cf)]P_n(x)-(-\alpha)(-\beta)[n(n-1) + nc]P_{n-1}(x)$}

Set {$f= -\beta$}, {$-\alpha = 1$}, {$c=\frac{a}{-\beta}$}. We get

{$P_{n+1}(x)=[x-((1 - \beta)n+a)]P_n(x)-[n(n-1)\beta - na]P_{n-1}(x)$}

Let {$\beta=0$}, then we have, as desired:

{$P_{n+1}(x)=[x-(n+a)]P_n(x) - naP_{n-1}(x)$}

Alternatively...

Set {$f=-\beta$} and set {$cf=a$}. We have:

{$P_{n+1}(x)=[x-(((-\alpha)+(-\beta))n+a)]P_n(x)-(-\alpha)[(-\beta)n(n-1) + an]P_{n-1}(x)$}

Let {$\beta=0$}. We have:

{$P_{n+1}(x)=[x-((-\alpha)n+a)]P_n(x)-(-\alpha)anP_{n-1}(x)$}

What happens if we simply start with setting {$\beta=0$}? Then the last term will go to zero.

{$P_{n+1}(x)=[x-(-\alpha n+cf)]P_n(x)$}

When {$\alpha = 1$}, this generates a falling factorial {$(x-cf+n-1)_n=\frac{(x-cf+n-1)!}{(x-cf-1)!}$}

So we need {$c$} to grow inversely to {$\beta$}. Thus we set {$c=\frac{a}{-\beta}$}

{$P_{n+1}(x)=[x-(-\alpha n+\frac{fa}{-\beta})]P_n(x)-(-\alpha)(-\beta)[n(n-1) + n\frac{a}{\beta}]P_{n-1}(x)$}

But now we have the problem that the term {$\frac{fa}{-\beta}$} will go to infinity if {$\beta$} goes to zero. So we need to set {$f=-\beta$}. So we need all of these conditions.

This all suggests to me that the real physics is in the combinatorics. As the quantum narrative proceeds, the recurrence relations break down - they cannot include all of the details - such as {$\alpha$}. I am curious how this manifests in the combinatorics.