I am trying to determine the big-$O$ size of the following double sum. I am pretty sure it is exponential in the size of $n$ (i.e. $O(c^n)$ for some constant $c$), but I would like to know how to analyze such a sum, and also if it's possible to say anything about what $c$ is.
Here it is: Let $n\geq 0$.
$$\sum_{k=0}^{n} \left( \sum_{i=1}^{k+2} (-1)^{k+i} {k+1\choose i-1} i^{n+1} \right)$$
It seems that this was too computationally intensive for WolframAlpha to calculate some examples, so I am just trying to determine the size analytically.
In case you are wondering, this expression is derived from this paper about the number of $k$-simplices in the barycentric subdivision of the standard $n$ simplex. Putting their expression into a double sum that counts $k$ from $0$ to $n$ should give the number of total simplices of all dimensions of the barycentric subdivision of the standard $n$-simplex.
Starting from $n=0$ $$\{1,5,25,149,1081,9365,94585,1091669,14174521\}$$ are the first terms of sequence $A002050$ in $OEIS$.
The exponential generating function is $$\frac{e^{2 x}-e^x}{2-e^x}$$
In the formula section, you will notice that, in year $2013$, Vaclav Kotesovec proposed (I shifted his formula by $1$) $$a_n \sim \frac {(n+1)!} { \log ^{n+2}(2)}$$ which is very good. Rounded to the next integer, it generates $$\{2,6,26,150,1082,9366,94586,1091670,14174522\}$$
As an asymptotic $$a_n=\left(\frac{n}{e \log (2)} \right)^n\,\frac{\sqrt{2 \pi n ^3} }{\log ^2(2)}\,\left(1+\frac{13}{12 n}+\frac{25}{288 n^2}+O\left(\frac{1}{n^3}\right) \right)$$ whose relative error is smaller that $0.1$% for $n \geq 7$.