How can I find $f(a,b,c)=e^{-c^a/a}\sum\limits_{n=0}^{\infty}\left(\frac{c^a}{a}\right)^{n}\frac{(an)^{b}}{n!}$? Inspired by Dobinski formula, by lucky guess I find, that (for natural $a,b$)
$$f(a,b)=e^{-1/a}\sum\limits_{n=0}^{\infty}\frac{(an)^{b}}{a^{n}n!}=\sum\limits_{k=1}^{b}{b\brace k}a^{b-k}$$
but I have problems with
$$f(a,b,c)=e^{-c^a/a}\sum\limits_{n=0}^{\infty}\left(\frac{c^a}{a}\right)^{n}\frac{(an)^{b}}{n!}$$
For example, for $a=1$, $b=1,2,\cdots$, $c=\frac{3}{2}$, we have
$$\frac{3\cdot1}{2^1}, \frac{3\cdot5}{2^2}, \frac{3\cdot31}{2^3}, \frac{3\cdot227}{2^3}, \frac{3\cdot1897}{2^4}, \frac{3\cdot17693}{2^5}, \cdots$$
Why only denominator's degree growth?
How can I find it in general?
 A: We have that
$$
\eqalign{
  & \sum\limits_{0\, \le \,n} {x^{\,n} {{n^{\,b} } \over {n!}}} \quad \left| {\,0 \le b \in Z} \right.\quad  = \sum\limits_{0\, \le \,n} {{{x^{\,n} } \over {n!}}\sum\limits_{0\, \le \,k\,\left( { \le \,b} \right)} {\left\{ \matrix{
  b \hfill \cr 
  k \hfill \cr}  \right\}n^{\,\underline {\,k\,} } } }  =   \cr 
  &  = \sum\limits_{0\, \le \,k\,\left( { \le \,b} \right)} {\left\{ \matrix{
  b \hfill \cr 
  k \hfill \cr}  \right\}\sum\limits_{0\, \le \,n} {n^{\,\underline {\,k\,} } {{x^{\,n} } \over {n!}}} }  = \sum\limits_{0\, \le \,k\,\left( { \le \,b} \right)} {\left\{ \matrix{
  b \hfill \cr 
  k \hfill \cr}  \right\}x^{\,k} \sum\limits_{0\, \le \,n} {n^{\,\underline {\,k\,} } {{x^{\,n - k} } \over {n!}}} }  =   \cr 
  &  = \sum\limits_{0\, \le \,k\,\left( { \le \,b} \right)} {\left\{ \matrix{
  b \hfill \cr 
  k \hfill \cr}  \right\}x^{\,k} {{d^{\,k} } \over {dx^{\,k} }}\sum\limits_{0\, \le \,n} {{{x^{\,n} } \over {n!}}} }  = e^{\,x} \sum\limits_{0\, \le \,k\,\left( { \le \,b} \right)} {\left\{ \matrix{
  b \hfill \cr 
  k \hfill \cr}  \right\}x^{\,k} }  \cr} 
$$
then you can easily adapt it to your case.
