I try to compute -- or at least estimate -- the $n$-th derivatives of a composition of real valued smooth functions using Faà di Bruno formula. This yields formulae involving exponential partial Bell polynomials.
$\bullet$ The first task consists in computing $B_{n,k}(2!,3!,\ldots,(n-k+2)!)$ for $n\geq0$ and $0\leq k\leq n$. While a nice formula (using Lah numbers) exists for $B_{n,k}(1!,2!,\ldots,(n-k+1)!)$ (see for example Comtet book [Com] "Advanced Combinatorics: The Art of Finite and Infinite Expansions", page 135), I did not manage to find any formula for the shifted arguments. So I tried myself to find one: let $\Phi(t,u)$ be the generating function of partial Bell polynomials
\begin{align*}
\Phi(t,u)&:=\exp\left(\sum_{m=1}^{+\infty}x_m\frac{t^m}{m!}\right)=\sum_{k=0}^{+\infty}\sum_{n=0}^{+\infty}B_{n,k}(x_1,\ldots,x_{n-k+1})\frac{t^n}{n!}u^k
\end{align*}
as defined in [Com, page 133]. We compute for $x_m:=(m+1)!$:
\begin{align*}
\Phi(t,u)=\exp\left(u\sum_{m=1}^{+\infty}x_m\frac{t^m}{m!}\right)&=\exp\left(u\sum_{m=1}^{+\infty}(m+1)t^m\right)\\
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&=\exp\left(u\left[\frac{\mathrm{d}}{\mathrm{d}t}\sum_{m=0}^{+\infty}t^m-1\right]\right)\\
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&=\exp\left(u\left[\frac{1}{(1-t)^2}-1\right]\right)\\
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&=\exp\left(tu\frac{2-t}{(1-t)^2}\right)\\
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&=\sum_{k=0}^{+\infty}\frac{t^ku^k}{k!}\frac{(2-t)^k}{(1-t)^{2k}}\\
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&=\sum_{k=0}^{+\infty}\sum_{\ell=0}^{+\infty}\frac{t^ku^k}{k!}\left(\sum_{j=0}^{k}\binom{k}{j}2^{k-j}(-1)^jt^j\right)\frac{(\ell+2k-1)\ldots(\ell+1)}{(2k-1)!}t^\ell\\
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&=\sum_{k=0}^{+\infty}\sum_{\ell=0}^{+\infty}\sum_{j=0}^{k}\frac{t^{k+\ell+j}u^k}{k!}\binom{k}{j}2^{k-j}(-1)^j\frac{(\ell+2k-1)!}{\ell!(2k-1)!}\\
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&=\sum_{k=0}^{+\infty}\sum_{j=0}^{k}\sum_{n=k+j}^{+\infty}\frac{t^{n}u^k}{k!}\binom{k}{j}2^{k-j}(-1)^j\frac{(n-j+k-1)!}{(n-j-k)!(2k-1)!}.
\end{align*}
I struggle finishing the computation because of the new sum over $j\in\{0,\ldots,k\}$. In particular, I could conclude if I could have $\sum_{k=0}^{+\infty}\sum_{n=k}^{+\infty}\frac{t^{n}u^k}{k!}\ldots$ but my sum over $n$ starts with $n=k+j$ or I have a $t^{n+k}$ instead of $t^n$. Any help there would be appreciated.
EDIT. Following G Cab's nice idea there Asymptotics of a function involving factorials, I think I have obtained the answer to the first question (using the notation $x^{(n)}:=x(x+1)\ldots(x+n-1)$ for the rising factorial to fix the problem with $2k-1$ when $k=0$ in the above formula):
\begin{align*}
\Phi(t,u)&=\sum_{k=0}^{+\infty}\sum_{n=k}^{+\infty}\sum_{j=0}^{\min\{k,\,n-k\}}\frac{t^{n}u^k}{k!}\binom{k}{j}2^{k-j}(-1)^j\frac{(2k)^{(n-k-j)}}{(n-k-j)!}
\end{align*}
where the last equality above is justified by the following:
\begin{align*}
\sum_{k=0}^{+\infty}\sum_{j=0}^{k}\sum_{n=k+j}^{+\infty}f(k,j,n)&=\sum_{k=0}^{+\infty}\sum_{j=0}^{+\infty}\sum_{n=0}^{+\infty}f(k,j,n)\big[0\leq k<+\infty]\big[0\leq j\leq k\big]\big[k+j\leq n<+\infty\big]\\
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&=\sum_{k=0}^{+\infty}\sum_{j=0}^{+\infty}\sum_{n=0}^{+\infty}f(k,j,n)\big[0\leq k<+\infty]\big[0\leq j\leq k,\,k+j\leq n<+\infty\big]\\
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&=\sum_{k=0}^{+\infty}\sum_{n=0}^{+\infty}\sum_{j=0}^{+\infty}f(k,j,n)\big[0\leq k<+\infty]\big[k\leq n<+\infty\big]\big[0\leq j\leq k,\,0\leq k+j\leq n\big]\\
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&=\sum_{k=0}^{+\infty}\sum_{n=k}^{+\infty}\sum_{j=0}^{+\infty}f(k,j,n)\big[0\leq j\leq\min\{k,\,n-k\}\big]\\
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&=\sum_{k=0}^{+\infty}\sum_{n=k}^{+\infty}\sum_{j=0}^{\min\{k,\,n-k\}}f(k,j,n).
\end{align*}
I used the Iverson symbol:
\begin{align*}
\big[P(k,j,n)\big]&:=\begin{cases}
1&\text{if $P(k,j,n)$ is true},\\
0&\text{if $P(k,j,n)$ is false}.
\end{cases}
\end{align*}
Identifying the coefficient in front of $\frac{t^nu^k}{n!}$ in $\Phi(t,u)$, we find:
\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*}
I would appreciate if somebody could check this formula.
$\bullet$ Next, and this actually is the motivation of the previous part, I want to compute
\begin{align*}
\sum_{k=2}^{n}\frac{(-1)^{k-1}a^{k}(2k-3)!}{4^{k}(k-2)!}B_{n,k}\left(2!,3!,\ldots,(n-k+2)!\right)\qquad\qquad n\geq 2,
\end{align*}
for some real number $a\in(0,1)$. We can remove $a$ by bounding it by $1$ if needed, but it would be a bit better to let it. I expect that the coefficients $B_{n,k}\left(2!,3!,\ldots,(n-k+2)!\right)$ are some (kinds of) multiples of Lah numbers, but as I do not have their expression I can not go on there -- however, we could cut off the first part if a direct answer can be directly provided, while it would be interesting to have the answer to the first part as well. At the end of the course, I expect that the answer is bounded by something like $(n+1)!$. Here again, any help would be welcome.
EDIT. Using the above formula in the EDIT, the sum becomes:
\begin{align*}
\sum_{k=2}^{n}\frac{(-1)^{k-1}a^{k}(2k-3)!}{4^{k}(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!}\qquad\qquad n\geq 2,
\end{align*}
