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'!}\\ % &=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 $2^{\beta'-\gamma}(-1)^\gamma$.

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'!}\\ % &=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: