**0. Background.** This question is linked to a previous one: <https://math.stackexchange.com/questions/3950321/computing-sums-of-exponential-partial-bell-polynomials>.
Based on the computation of the exponential partial Bell polynomial $B\_{n,k}(2!,\ldots,(n-k+2)!)$ there (that I hope is correct), I managed to rewrite the problem and this led me to ask this new question.


**1. The question.** Can we explicitly compute
$$S(\alpha',\beta')=\sum\_{\gamma=0}^{\min\{\beta',\,\alpha'-\beta'\}}\frac{2^{\beta'-\gamma}(-1)^\gamma}{\gamma!(\beta'-\gamma)!}\frac{(2\beta')^{\overline{\alpha'-\beta'-\gamma}}}{(\alpha'-\beta'-\gamma)!}$$
where $\beta'\geq1$ and $2\beta'\geq\alpha'\geq0$? Here $x^{\overline{k}}:=x(x+1)\ldots(x+k-1)$ denotes the rising factorial. If no closed form can be found, an estimate will be enough. Below are displayed two attempts to deal with the problem ; any advice to go further would be very appreciated.


**2.1. First attempt.** Introduce the signed Lah number (see <https://en.wikipedia.org/wiki/Lah_number>):
$$L(n,k):=(-1)^n\frac{n!(n-1)!}{k!(k-1)!(n-k)!}.$$
Then, putting $\gamma':=\alpha'-\beta'-\gamma$, we can write:
\begin{align\*}
S(\alpha',\beta')&=\sum\_{\gamma'=\min\{\alpha'-2\beta',\,0\}}^{\alpha'-\beta'}\frac{2^{2\beta'-\alpha'+\gamma'}(-1)^{\alpha'-\beta'-\gamma'}}{(\alpha'-\beta'-\gamma')!(2\beta'-\alpha'+\gamma')!}\frac{(2\beta')^{\overline{\gamma'}}}{\gamma'!}\\
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&=2^{2\beta'-\alpha'}\sum\_{\gamma'=\min\{\alpha'-2\beta',\,0\}}^{\alpha'-\beta'}\frac{2^{\gamma'}(-1)^{\gamma'}}{(\alpha'-\beta'-\gamma')!(2\beta'-\alpha'+\gamma')!}\frac{(2\beta')^{\overline{\gamma'}}}{\gamma'!}\\
%
&=\frac{2^{2\beta'-\alpha'}(-1)^{\alpha'-\beta'}}{(\alpha'-\beta')!(\alpha'-\beta'-1)!}\\&\quad\times\sum\_{\gamma'=\min\{\alpha'-2\beta',\,0\}}^{\alpha'-\beta'}\frac{2^{\gamma'}(\gamma'-1)!}{(2\beta'-\alpha'+\gamma')!}(-1)^{\gamma'}\underbrace{(-1)^{\alpha'-\beta'}\frac{(\alpha'-\beta')!(\alpha'-\beta'-1)!}{\gamma'!(\gamma'-1)!(\alpha'-\beta'-\gamma')!}}\_{=L(\alpha'-\beta',\gamma')}(2\beta')^{\overline{\gamma'}}\\
%
&=\frac{2^{2\beta'-\alpha'}(-1)^{\alpha'-\beta'}}{(\alpha'-\beta')!(\alpha'-\beta'-1)!}\sum\_{\gamma'=\min\{\alpha'-2\beta',\,0\}}^{\alpha'-\beta'}\frac{2^{\gamma'}(\gamma'-1)!}{(2\beta'-\alpha'+\gamma')!}(-1)^{\gamma'}L(\alpha'-\beta',\gamma')(2\beta')^{\overline{\gamma'}}
\end{align\*}
The term $\frac{2^{\gamma'}(\gamma'-1)!}{(2\beta'-\alpha'+\gamma')!}$ and the lower bound $\min\{\alpha'-2\beta',\,0\}$ for the sum are annoying ; here we have something pretty close to
$$\sum\_{\gamma'=0}^{\alpha'-\beta'}(-1)^{\gamma'}L(\alpha'-\beta',\gamma')(2\beta')^{\overline{\gamma'}}=(2\beta')^{\underline{\alpha'-\beta'}}.$$
Here $x^{\underline{k}}:=x(x-1)\ldots(x-k+1)$ denotes the falling factorial.


**2.2. Second attempt.** We could as well write:
\begin{align\*}
S(\alpha',\beta')&=\sum\_{\gamma=0}^{\min\{\beta',\,\alpha'-\beta'\}}\frac{2^{\beta'-\gamma}(-1)^\gamma}{\gamma!(\beta'-\gamma)!}\frac{2\beta'(2\beta'+1)\ldots(\alpha'+\beta'-\gamma)}{(\alpha'-\beta'-\gamma)!}\\
%
&=\frac{2}{(\beta'-1)!}\sum\_{\gamma=0}^{\min\{\beta',\,\alpha'-\beta'\}}2^{\beta'-\gamma}(-1)^\gamma\binom{\beta'}{\gamma}\binom{\alpha'+\beta'-\gamma}{\alpha'-\beta'-\gamma}.
\end{align\*}
Now this looks like a Vandermonde's identity: <https://en.wikipedia.org/wiki/Vandermonde%27s_identity>. This time the annoying term is $2^{\beta'-\gamma}(-1)^\gamma$.


**EDIT.** After having read some comments/answers, I think I can handle my problem **IF** I can get a bound such as $B\_{n,k}(2!,3!,\ldots,(n-k+2)!)\leq n!$ or $-$ even better $-$ $B\_{n,k}(2!,3!,\ldots,(n-k+2)!)\leq(n-k+1)!$, where I computed:




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\begin{align\*}
B\_{n,k}(2!,\ldots,(n-k+2)!)&=\sum\_{j=0}^{\min\{k,\,n-k\}}\binom{k}{j}2^{k-j}(-1)^j\frac{(2k)^{(n-k-j)}n!}{(n-k-j)!k!}.
\end{align\*}


---



Computing a sum involving factorials
Probably this is not very helpful, but it is an explicit expression after all.


I get
$$
S(a,b)=\frac{(-1)^{a+1}}{b!}{}\_2F\_1(-a+1,b+1;2b-a+1;2)\binom b{2b-a}2^{2b-a}.
$$
This follows from
$$
S(a,b)=\sum\_{\ell=2b-a}^b(-1)^\ell\binom{a+\ell-1}{2b-1}\binom b\ell2^\ell,
$$
which in turn I derived from the generating function
$$
\sum\_{a,b}S(a,b)t^au^b=e^{\frac{2-t}{(1-t)^2}tu}.
$$


Note also that $S(a,b)$ is defined for other values of $a$ and $b$. In particular, for $a\geqslant2b$ one obtains
$$
S(a,b)=\frac{(-1)^b}{b!}{}\_2F\_1(a,-b;a-2b+1;2)\binom{a-1}{2b-1}
$$


In fact, from that generating function, $b!S(a,b)$ is the coefficient of $\left(2t+3t^2+4t^3+5t^4+...\right)^b$ at $t^a$. Here is the table of few of the $b!S(a,b)$:
$$
\begin{array}{ccccccccccc}
 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 3 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 4 & 12 & 8 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 5 & 25 & 36 & 16 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 6 & 44 & 102 & 96 & 32 & 0 & 0 & 0 & 0 & 0 \\
 0 & 7 & 70 & 231 & 344 & 240 & 64 & 0 & 0 & 0 & 0 \\
 0 & 8 & 104 & 456 & 952 & 1040 & 576 & 128 & 0 & 0 & 0 \\
 0 & 9 & 147 & 819 & 2241 & 3400 & 2928 & 1344 & 256 & 0 & 0 \\
 0 & 10 & 200 & 1372 & 4712 & 9290 & 11040 & 7840 & 3072 & 512 & 0 \\
 0 & 11 & 264 & 2178 & 9108 & 22363 & 34332 & 33488 & 20224 & 6912 & 1024 \\
\end{array}
$$


Here is, in fact, a version that works for any $a$ and $b$, and does not contain any powers of $-1$ or $2$: represent
$$
\frac{2-t}{(1-t)^2}=(1-t)^{-2}+(1-t)^{-1};
$$
then from the above we obtain that $b!S(a,b)$ is the coefficient at $t^a$ of the series $t^b((1-t)^{-2}+(1-t)^{-1})^b$, i. e. of
$$
t^b\sum\_{j=0}^b\binom bj(1-t)^{-2j}(1-t)^{-(b-j)}=t^b\sum\_{j=0}^b\binom bj(1-t)^{-(b+j)}
$$
It then follows easily that
$$
S(a,b)=\frac1{b!}\sum\_{j=0}^b\binom bj\binom{a+j-1}{a-b}=\frac1{b!}{}\_2\!\!\ F\_1(a,-b;b;-1)\binom{a-1}{a-b}.
$$
Another observation that might be useful:
$$
\cosh^{a-2}(t)\sinh((a+1)t)=\sum\_{b=1}^ab!S(a,b)\sinh^{2b-1}(t)
$$


