For an analytic function with the property $f^{(n)}(0)=a_n$, we have $f(x)=\sum\limits_{n=0}^\infty \frac{a_n x^n}{n!}$. This can be extended to $f^{(n)}(x)=\sum\limits_{n=0}^\infty \frac{a_{n+1} x^n}{n!}$. This is useful for solving a recurrence relation using theory of differential equations: for example,
$a_{n+2}=a_{n+1}+a_n+2^n$
can be solved by multiplying both sides by $\frac{x^n}{n!}$ and summing from $n=0$ to $n=\infty$ yields $f''(x)=f'(x)+f(x)+e^{2x}$, with $f(x)=\sum\limits_{n=0}^\infty \frac{a_n x^n}{n!}$. The solution of this is $f(x)=C_1e^{\phi x}+C_2e^{(1-\phi)x}+e^{2x}$, where $\phi=\frac{1+\sqrt{5}}{2}$. This means the solution of the recurrence relation is
$a_n=C_1\phi^n+C_2(1-\phi)^n+2^n$
This is quite handy, but I only know how it works when the highest "term" of the recurrence is linear; for example, $a_n$ maps to $f(x)$, and $na_n$ maps $x f'(x)$, but I have no idea what $a_n^2$ maps to. My question is, if $f(x)=\sum\limits_{n=0}^\infty \frac{a_n x^n}{n!}$, is there some kind of relationship between $\sum\limits_{n=0}^\infty \frac{a_n x^n}{n!}$ and $\sum\limits_{n=0}^\infty \frac{a_n^2 x^n}{n!}?$If so, does this generalise to higher powers of $a_n?$