Can anyone show me how to derive this summation for the value of the $n$th derivative at $x=0$ for this function:
$\frac{d}{dx^n}(\exp({\frac{x^2}{2}+x}))$ is this sum:
$\frac{d}{dx^n}(\exp({\frac{x^2}{2}+x})) =$ $\sum\limits_{k=0}^{\lfloor{\frac{n}{2}\rfloor}} \frac{1}{2^kx^{2k}k!(-2k+n)!}$
On
One way is to define the sequence of polynomials $\,P_n(x)\,$ such that $$ P_n(x)\,e^{x+x^2/2} = \left(\frac{d}{dx}\right)^n e^{x+x^2/2}. $$ The coefficients of the polynomial sequence is the OEIS sequence A111062. Please consult the OEIS entry for more details. Of course, the function $\,e^{x+x^2/2}\,$ itself is the exponential generating function for the OEIS sequence A000085. The OEIS entry gives the summation formula $$ a_n = \sum_{k=0}^{n/2} \frac{n!}{(n-2k)! 2^k k!}$$ which you already got from Wolfram|Alpha. Probably the easiest method is to find the power series for $\,e^x\,$ and $\,e^{x^2/2}\,$ and multiply them which will produce the formula.
More precisely, consider $$ \sum_{n=0}^\infty a_n\, \frac{x^n}{n!} := e^{x+x^2/2} = e^x\,e^{x^2/2}= \\ \sum_{j=0}^\infty \frac{x^j}{j!} \sum_{k=0}^\infty \frac{x^{2k}}{2^k k!} = \sum_{n=0}^\infty \frac{x^n}{n!} \sum_{k=0}^{n/2} \frac{n!}{(n-2k)! 2^k k!} $$ where $\,j = n-2k.\,$ Since we have $\,0\le j,\,0\le k,\,$ and $\,k=(n-j)/2,\,$ then we must have $\,0\le k\le n/2.$
Define $f(x) = e^{x^2/2}$ and $g(x) = e^x$. Then $$(fg)^{(n)}(0) = \sum_{k=0}^n \binom{n}{k} f^{(k)}(0) g^{(n-k)}(0)$$ by the extended product rule. Now $$f(x) = \sum_{k=0}^\infty \frac{(x^2/2)^k}{k!}, \quad g(x) = \sum_{k=0}^\infty \frac{x^k}{k!},$$ and by Taylor's theorem it follows that $$f^{(k)}(0) = \begin{cases}0, & k \text{ odd}, \newline \frac{k!}{2^{k/2} (k/2)!}, & k \text{ even}.\end{cases}$$ We also have of course $g^{(n-k)}(0) = 1$. Then $$(fg)^{(n)}(0) = \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} \frac{(2k)!}{2^k k!} = \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{n!}{k!(n-2k)!2^k}.$$