Summation and factorial. If $$2^{n}a_{n}=\sum_{k=0}^{n}{a_{k}a_{n-k}},\phantom{a}\forall n\in\mathbb{N},$$ with $a_{0}=1,$ then $$a_{n}=\dfrac{a_{1}^{n}}{n!},\phantom{a}\forall n\in\mathbb{N}.$$
Attempt: Notice that
(1) For $ n=2t + 1, $ we have $$2^{2t+1}a_{2t+1}=2(a_{0}a_{2t+1}+a_{1}a_{2t}+a_{2}a_{2t-1}+\cdots+a_{t-1}a_{t+2}+a_{t}a_{t+1}).$$
Then
\begin{align}
(2^{2t}-1)a_{2t+1}&=a_{1}a_{2t}+a_{2}a_{2t-1}+\cdots+a_{t-1}a_{t+2}+a_{t}a_{t+1}\\
&=a_{1}^{2t+1}\left(\dfrac{1}{(2t)!}+\dfrac{1}{2!(2t-1)!}+\cdots+\dfrac{1}{(t-1)!(t+2)!}+\dfrac{1}{t!(t+1)!}\right)\\
&=\dfrac{a_{1}^{2t+1}}{(2t)!}\left(1+\dfrac{2t}{2!}+\cdots+\dfrac{(t+3)(t+4)\cdots(2t-1)(2t)}{(t-1)!}+\dfrac{(t+2)(t+3)(t+4)\cdots(2t-1)(2t)}{t!}\right).
\end{align}
(2) For $n=2t,$ we have $$2^{2t}a_{2t}=2a_{0}a_{2t}+2a_{1}a_{2t-1}+2a_{2}a_{2t-2}+\cdots+2a_{t-1}a_{t+1}+a_{t}^{2}.$$
Then
\begin{align}
2(2^{2t-1}-1)a_{2t}&=2a_{1}a_{2t-1}+2a_{2}a_{2t-2}+\cdots+2a_{t-1}a_{t+1}+a_{t}^{2}\\
&=a_{1}^{2t}\left(\dfrac{2}{(2t-1)!}+\dfrac{2}{2!(2t-2)!}+\cdots+\dfrac{2}{(t-1)!(t+1)!}+\dfrac{1}{t!t!}\right)\\
&=\dfrac{a_{1}^{2t}}{(2t-1)!}\left(2+\dfrac{2(2t-1)}{2!}+\cdots+\dfrac{2(t+2)(t+3)\cdots(2t-1)}{(t-1)!}+\dfrac{(t+1)(t+2)(t+3)\cdots(2t-1)}{t!}\right).
\end{align}
Is there a formula to reduce the above expressions?
Any hints would be appreciated.

The sum is a convolution of $a_n$ with itself, so the generating function of the RHS is $A(z)^2$, where $A(z)=\sum_n a_n z^n$ is the generating function of $a_n$. The generating function of $2^n a_n$ is $A(2z)$, so we have $A(2z)=A(z)^2$.  Hence $A(z)=\exp(c z)$ for some constant $c$, which implies that $a_n=c^n/n!$. Now evaluate at $n=1$ to obtain $a_1=c^1/1!=c$, so $a_n=a_1^n/n!$.
