1-Associated Stirling Number of the Second Kind identity verification

Question

I recently posted this in regards to Associated Stirling Numbers of the Second Kind (SNSK) and I was trying to fix my equations to find and identity, and am now looking for verification that this identity is correct.

Let $$\sum_{n=0}^\infty A_n^{(m)}\frac{x^n}{n!}=\frac{(x^2/2!)^{m/2}}{(e^x-1-x)^m}$$ I also have that $$\sum_{n=0}^\infty b(1;n,j)\frac{x^n}{n!}=\frac{(e^x-1-x)^j}{j!}$$ Using these two identities, I came up with the following; $$\sum_{n=0}^\infty A_n^{(-m)}\frac{x^n}{n!}=\frac{(e^x-1-x)^m}{(x^2/2!)^{m/2}}$$ $$=\frac{(\sqrt{2})^mm!}{x^{m}}\frac{(e^x-1-x)^m}{m!}$$ $$=\frac{(\sqrt{2})^mm!}{x^m}\sum_{n=0}^\infty b(1;m+n,m)\frac{x^{m+n}}{(m+n)!}$$ $$\sum_{n=0}^\infty b(1;m+n,m)\frac{(\sqrt{2})^mm!n!}{(m+n)!}\frac{x^n}{n!}$$ Therefore, I have that $$A_n^{(-m)}=(\sqrt{2})^m b(1;m+n,m)\frac{m!n!}{(m+n)!}$$ Rearranging, I have $$b(1;m+n,n)=\binom{m+n}{m}\frac{A_n^{(-m)}}{(\sqrt{2})^m}$$ First, can anyone verify that this is correct? Second, and I'd like to have this resolved...I really want to be able to define my original polynomial as $$\sum_{n=0}^\infty A_n^{(m)}\frac{x^n}{n!}=\left(\frac{x^2/2!}{e^x-1-x}\right)^m$$ However, my problem comes at the place where I get $$\sum_{n=0}^\infty A_n^{(-m)}\frac{x^n}{n!}=\left(\frac{e^x-1-x}{x^2/2!}\right)^m$$ $$=\frac{2^m(2m)!}{x^{2m}}\frac{(e^x-1-x)^m}{(2m)!}$$ Because I have that $(2m)!$ in the demonimator and the $x^{2m}$, there is no substituting the formula for the exponential expression.

I am thinking that if there is a connection, then my identity should be of the form: $$A_n^{(-m)}=k^m b(1;2m+n,2m)\frac{(2m)!n!}{(2m+n)!}$$ where $k$ is an unknown constant or variable term. Any insight?

Answer 1

Note: The answer to your first question is affirmative: Yes, your calculations are correct.

The answer to your second question is: Sorry, no deep insight from my side, but hopefully some helpful hints for further investigations.

A short look at your first part:

First part:

Since $A_n^{(m)}$ is used with a slightly different meaning in OPs first part and in OPs second part I will use here $\widetilde{A}_n^{(m)}$ and keep the original notation $A_n^{(m)}$ for the second part, which seems to be of more interest to the OP.

With respect to OPs referenced question he introduced a sort of generalisation of Norlund Polynomials, denoted $A_n^{(m)}$. A slight modification of these polynomials are $\widetilde{A}_n^{(m)}$ given via the following generating function:

Let \begin{align*} \sum_{n=0}^{\infty}\widetilde{A}_n^{(m)}\frac{x^n}{n!}=\frac{\left(\frac{x^2}{2}\right)^{\frac{m}{2}}}{\left(e^x-1-x\right)^m}\tag{1} \end{align*}

Next, we look at certain Associated Stirling Numbers of the Second Kind $b(1;n,m)$ given via $$\sum_{n=0}^{\infty}b(1;n,m)\frac{x^n}{n!}=\frac{\left(e^x-1-x\right)^m}{m!}$$ We observe, since $$\left(e^x-1-x\right)^m=\left(\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\right)^m=2^{-m}x^{2m}\left(1+\frac{1}{12}x+\ldots\right)^m$$ the right hand side starts with $x^{2m}$ and so:

The following is valid for the numbers $b(1;n,m)$ $$b(1;n,m)=0\qquad\qquad n<2m$$ and we can write \begin{align*} \sum_{n=2m}^{\infty}b(1;n,m)\frac{x^n}{n!}=\frac{\left(e^x-1-x\right)^m}{m!}\tag{2} \end{align*}

We can now proceed with:

First part: Calculation

We use $-m$ in (1) and we observe: \begin{align*} \sum_{n=0}^{\infty}\widetilde{A}_n^{(-m)}\frac{x^n}{n!}&=\frac{\left(e^x-1-x\right)^m}{\left(\frac{x^2}{2}\right)^{\frac{m}{2}}}\\ &=\frac{\left(\sqrt{2}\right)^mm!}{x^m}\frac{\left(e^x-1-x\right)^m}{m!}\\ &=\frac{\left(\sqrt{2}\right)^mm!}{x^m}\sum_{n=2m}^{\infty}b(1;n,m)\frac{x^n}{n!}\qquad\qquad\qquad\qquad\text{by (2)}\\ &=\frac{\left(\sqrt{2}\right)^mm!}{x^m} \sum_{n=m}^{\infty}b(1;n+m,m)\frac{x^{n+m}}{(n+m)!}\qquad\qquad n\rightarrow n+m \\ &=\left(\sqrt{2}\right)^mm!\sum_{n=m}^{\infty}b(1;n+m,m)\frac{x^{n}}{(n+m)!}\\ &=\left(\sqrt{2}\right)^m\sum_{n=m}^{\infty}b(1;n+m,m)\binom{n+m}{n}^{-1}\frac{x^{n}}{n!} \end{align*}

Comparison of coefficients gives:

The following is valid: \begin{align*} \widetilde{A}_n^{(-m)}&=\left(\sqrt{2}\right)^mb(1;n+m,m)\binom{n+m}{n}^{-1}& n\geq m\tag{3}\\ &=0& n < m \end{align*}

and we conclude with:

First Part: Result:

We see, that (3) coincides with the result of OPs first part and we therefore conclude that OPs calculations are correct (besides missing a range specification).

We further see that $\widetilde{A}_n^{(-m)}=0$ if $n<m$.

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Now, let's have a look at the second part. Here we use $A_n^{(m)}$ as it was given by the OP in the second part, namely

Let \begin{align*} \sum_{n=0}^{\infty}A_n^{(m)}\frac{x^n}{n!}=\frac{\left(\frac{x^2}{2}\right)^{m}}{\left(e^x-1-x\right)^m}\tag{4} \end{align*}

We can already start with:

Second part: Calculation

We use $-m$ in (4) and we observe: \begin{align*} \sum_{n=0}^{\infty}A_n^{(-m)}\frac{x^n}{n!}&=\frac{\left(e^x-1-x\right)^m}{\left(\frac{x^2}{2}\right)^{m}}\\ &=2^mm!x^{-2m}\frac{\left(e^x-1-x\right)^m}{m!}\\ &=2^mm!x^{-2m}\sum_{n=2m}^{\infty}b(1;n,m)\frac{x^n}{n!}\qquad\qquad\qquad\text{by (2)}\\ &=2^mm!\sum_{n=2m}^{\infty}b(1;n,m)\frac{x^{n-2m}}{n!}\\ &=2^mm!\sum_{n=0}^{\infty}b(1;2m+n,m)\frac{x^{n}}{(2m+n)!}\quad\qquad n\rightarrow 2m+n\tag{5} \\ &=2^mm!\sum_{n=0}^{\infty}b(1;2m+n,m)\frac{n!}{(2m+n)!}\frac{x^{n}}{n!} \\ \end{align*}

Similarly to part 1 we compare the coefficients of corresponding powers of $x^n$ to get:

\begin{align*} A_n^{(-m)}&=2^m\frac{m!n!}{(2m+n)!}b(1;2m+n,m)\qquad\qquad n\geq 0\\ &=\frac{2^mm!}{(2m)!}\binom{2m+n}{n}^{-1}b(1;2m+n,m)\tag{6}\\ \end{align*}

Since $$(2m)!=(2m)!!(2m-1)!!=2^mm!(2m-1)!!$$ we observe:

Second part: Result

The following identity is valid: \begin{align*} b(1;2m+n,m)=(2m-1)!!\binom{2m+n}{n}A_n^{(-m)}\qquad\qquad n \geq 0\tag{7} \end{align*}

Note: It seems, that the difference of the result (7) to OPs suggestion $$k^mb(1;2m+n,2m)\binom{2m+n}{n}^{-1}$$ besides the factor $k^m$ is due to the fact, that the index shift of the sum in the RHS was $n \rightarrow n+2m$ instead of $n \rightarrow n+m$ as it was used in the first part according to OPs calculation.

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Hint: I suppose that the paper Explicit formulas for the Nörlund polynomials $B_n^{(x)}$ and $b_n^{(x)}$ is helpful for further investigations. The polynomial $b(1;n,m)$ used here corresponds with $b(n,m)$ in the paper. Interesting formulas with $b(n,m)$ are stated in (1.14) and in the proof section.