While taking the incomplete Bell Polynomil of $x^a$ i found out that:
$$ B_{n,k}^{x^a}(x) = x^{ak-n} \sum_{m=0}^k \frac{(am)!(-1)^{k-m}}{m!(k-m)!(am-n)!} $$
Now, what i am wondering is, what is the valuue of this summation?
What i have done so far:
I figured out when $a=2$ by finding the incomplete Bell Polynomial of $x^2$ by doing the following:
$$ (x+z)^2 - x^2 = z(2x+z) = \sum_{n=1}^\infty \frac{z^n}{n!} \frac{d^n}{dx^n}[x^2] $$
$$ z^k(2x+z)^k = \sum_{n \geq k} Y^{\Delta}(n,k,x)z^n = \frac{k!}{n!}B_{n,k}^{x^2}(x) $$
Note that:
$$ Y^{\Delta}(n,k,x) = \frac{k!}{n!}B_{n,k}^{x^2}(x) $$
$$ [z^n]z^k(2x+z)^k = (2x)^{k}[z^{n-k}](1+\frac{z}{2x})^k $$
$$ [z^{n-k}](1+\frac{z}{2x})^k = [z^{n-k}]\sum_{j=0}^k {k \choose j} \left(\frac{z}{2x}\right)^j = {n-1 \choose n-k}\frac{1}{2x^{n-k}} $$ Therefore:
$$ (2x)^k[z^{n-k}](1+\frac{z}{2x})^k = (2x)^{2k-n}{n-1 \choose n-k} $$
With this info we know that $Y^{\Delta}(n,k,x) = \frac{k!}{n!}B_{n,k}^{x^2}(x)$ therefore:
$$ B_{n,k}^{x^2}(x) = \frac{n!}{k!} (2x)^{2k-n} {n-1 \choose n-k} $$
And with our prevous formula $a=2$ we find: $$ B_{n,k}^{x^2}(x) = x^{2k-n} \sum_{m=0}^k \frac{(2m)!(-1)^{k-j}}{m!(k-m)!(2m-n)!} $$
Therefore
$$ x^{2k-n} \sum_{m=0}^k \frac{(2m)!(-1)^{k-j}}{m!(k-m)!(2m-n)!} = \frac{n!}{k!} (2x)^{2k-n} {n-1 \choose n-k} $$
$$ \sum_{m=0}^k \frac{(2m)!(-1)^{k-j}}{m!(k-m)!(2m-n)!} = \frac{n!}{k!} 2^{2k-n} {n-1 \choose n-k} $$ or $$ \sum_{m=0}^k {k \choose m}{2m \choose n} (-1)^{k-j} = 2^{2k-n}{n-1 \choose n-k} $$
Also there is a problem when $a=1$, where the sum itself will be very weird...
$$B_{n,k}^{x^{a}}(x)=x^{ak-n}\sum_{\pi_{n,k}(j_{1},j_{2},...,j_{n-k+1})}n!\prod_{i=1}^{n}\frac{\left(a^{\underline{i}}\right)^{j_{i}}}{i!^{j_{i}}j_{i}!}$$
$\pi_{n,k}(j_{1},j_{2},...,j_{n-k+1})$: partitions of integer $n$ into exactly $k$ parts
$a^{\underline{i}}$: falling factorial