Properties of $P_{n}(x)={e}^{-x}\sum_{k=0}^{\infty}\frac{a_{k,n}{x}^{k}}{k!}$

Question

I know this will sound like a general question, but given the structure $$P_{n}(x)={e}^{-x}\sum_{k=0}^{\infty}\frac{a_{k,n}{x}^{k}}{k!}$$ where $$a_{k,n} = \frac{1}{\prod_{i=1}^{n} (k+2i) }, $$ what are some some of its properties - including recurrence relations and generating functions? Is it of bynomial type? I know that when $a_{k,n}={k}^{n}$, we have $P_{n}(x)=T_{n}(x)=$ Touchard polynomials.

Answer 1

Noticing that $$a_{k,n} = \frac1{\prod_{i=1}^n (k+2i)} = \frac1{2^n(n-1)!}\int_0^1 (1-t)^{n-1}t^{k/2}\,{\rm d}t,$$ we get $$P_n(x) = \frac{e^{-x}}{2^n(n-1)!} \int_0^1 (1-t)^{n-1} e^{x\sqrt{t}}\,{\rm d}t.$$ Maple computes intergral in terms of modified Bessel $I$ and Struve $L$ functions, giving $$P_n(x) = e^{-x}\bigg(\frac1{2^n n!} + \frac{\sqrt{2\pi}}{2}(-x)^{-n+\frac12}(L_{n+\frac12}(-x)-I_{n+\frac12}(-x))\bigg).$$

Alternatively, we can use binomial expansion and substitution $t=z^2$ to get: \begin{split} P_n(x) &= \frac{e^{-x}}{2^{n-1}(n-1)!} \sum_{k=0}^{n-1} \binom{n-1}k (-1)^k \int_0^1 z^{2k+1} e^{xz}\,{\rm d}z \\ &=\frac1{2^{n-1}(n-1)!} \sum_{k=0}^{n-1} \binom{n-1}k (-1)^k x^{-(2k+2)} (2k+1)! \bigg(e^{-x} + \sum_{i=0}^{2k+1} (-1)^{i+1} \frac{x^i}{i!}\bigg) \\ &=\frac{e^{-x}}{2^{n-1}(n-1)!} \sum_{k=0}^{n-1} \binom{n-1}k (-1)^k x^{-(2k+2)} (2k+1)! \\ &\quad + \frac1{2^{n-1}(n-1)!} \sum_{j=1}^{2n} (-1)^{j+1} (j-1)! x^{-j} \sum_{k=0}^{n-1} \binom{n-1}k (-1)^k \binom{2k+1}{j-1}. \end{split}

The inner sum can be evaluated as follows \begin{split} \sum_{k=0}^{n-1} \binom{n-1}k (-1)^k \binom{2k+1}{j-1} &= \sum_{k=0}^{n-1} \binom{n-1}k (-1)^k [z^{j-1}]\ (1+z)^{2k+1} \\ &= [z^{j-1}]\ (1+z) (1-(1+z)^2)^{n-1} \\ &= (-1)^{n-1} [z^{j-n}]\ (1+z) (2+z)^{n-1} \\ &= (-1)^{n-1} \bigg(\binom{n-1}{j-n} 2^{2n-j-1} + \binom{n-1}{j-n-1} 2^{2n-j} \bigg)\\ &=(-1)^{n-1} 2^{2n-j-1} \frac{j}n \binom{n}{j-n}, \end{split} which allows us to simplify the polynomial part of $P_n(x)$ to $$2^n\sum_{j=n}^{2n} (-1)^{n+j} \frac{j!}{(j-n)!(2j-n)!} (2x)^{-j}.$$

Answer 2

The general pattern is as follows: $$P_{n}(x)={e}^{-x}\sum_{k=0}^{\infty}\frac{{x}^{k}}{k!}\frac{1}{\prod_{i=1}^{n} (k+2i)}=\frac{1}{x^{2n}}\biggl(\frac{e^{-x}}{2^n n! }B_n(x)+C_n(x)\biggr),$$ $$B_n(x)=\sum_{p=0}^{n-1} b_{p,n} x^{2p},\;\;C_n(x)=\sum_{p=0}^n c_{p,n} x^p,$$ $$b_{p,n}=\frac{(-1)^{p+n+1}(n-1)!}{2 p!}\prod_{i=1}^{n-p}(4i-2),\;\;c_{p,n}= \frac{(-1)^{p + n}(2n - p)!}{2^{n - p}p!(n - p)!}.$$ The coefficient $c_{p,n}$ is OEIS: A001498. The closed form of $b_{p,n}$ was found by Peter Taylor.

Answer 3

We are going to expand this:

$$({e}^{x})_{n}=\int_{0}^{\nu} \mu \, \cdots \int_{0}^{\gamma} \beta \, \int_{0}^{\beta} \alpha \, \int_{0}^{\alpha} x \ {e}^{x} \,dx\,d\alpha\,d\beta\,d\gamma\cdots\,d\mu$$ where the integral symbol occurs $n$ times

So we have (you can check by yourself)

$$ \begin{split} ({e}^{x})_{1} & = {e}^{\alpha}\left(\alpha-1\right)+1 \\ ({e}^{x})_{2} & = {e}^{\beta}\left({\beta}^2-3{\beta}+3\right) + \left(\frac{{\beta}^2}{2}-3\right) \\ ({e}^{x})_{3} & = {e}^{\gamma}\left({\gamma}^3-6{\gamma}^2+15{\gamma}-15\right) + \left(\frac{{\gamma}^4}{2.4}-3\frac{{\gamma}^2}{2}+15\right)\\ \end{split} $$ and so on...

The general pattern is as follows $$ ({e}^{x})_{n}={e}^{\nu}\sum_{k=0}^{n}(-1)^{k}a(n,k){\nu}^{n-k}+\sum_{k=0}^{n-1}(-1)^{k}a(k+1,k+1)\frac{{\nu}^{2(n-k-1)}}{(2(n-k-1))!!} $$

The coefficient $a(n,k)$ is called a Bessel coefficient (https://oeis.org/A001498).

While we have also $$({e}^{x})_{n}=\int_{0}^{\nu} \mu \, \cdots \int_{0}^{\gamma} \beta \, \int_{0}^{\beta} \alpha \, \int_{0}^{\alpha} x \ \left(\sum_{k=0}^{\infty}\frac{{x}^{k}}{k!}\right) \,dx\,d\alpha\,d\beta\,d\gamma\cdots\,d\mu=\sum_{k=0}^{\infty}\frac{{\nu}^{k+2n}}{k!\prod_{i=1}^{n}(k+2i)}$$

So equating we get: $$ P_{n}(\nu)=\sum_{k=0}^{n}(-1)^{k}a(n,k)\frac{1}{{\nu}^{n+k}}+{e}^{-\nu}\sum_{k=0}^{n-1}(-1)^{k}\frac{a(k+1,k+1)}{(2(n-k-1))!!}\frac{1}{{\nu}^{2(k+1)}} $$