Assume the following relation sequence:
$a_n = \sum_{i=2}^{n-2} a_{i} a_{n-i}$
for $n \geq 3$, where $a_0 = a_1 = a_2 = 1$. How can we find a closed form expression for $a_n$?
I tried to use generating function as $f(x) = \sum_{n=0}^{\infty} a_n x^n$. However, I do not know how to proceed?
I appreciate any help in this regard!
Too long for a comment.
Assuming that there is no typo in $$a_n = \sum_{i=2}^{n-2} a_{i}\,a_{n-i}$$ I see a problem already for $n=3$ since $$a_3 = \sum_{i=2}^{\color{red}{1}} a_{i}\,a_{3-i}\qquad a_4 = \sum_{i=2}^{\color{red}{2}} a_{i}\,a_{4-i}=a_2^2$$ So, I shall suppose that $a_3=0$ and, if this is true, then $a_{2n+1}=0$ and then the sequence would be $$\{1,1,1,0,1,0,2,0,5,0,14,0,42,0,132,0,429,0,1430,0,4862\}$$ and $a_{2n}$ seems to be the Catalan number $C_n=\frac{(2 n)!}{n! (n+1)!}$.
If this is true, we do not care about the values of $a_0$ and $a_1$ and the sequence would be $$\left\{{a_0},{a_1},{a_2},0,{a^2_2},0,2 {a^3_2},0,5 {a^4_2},0,14 {a^5_2},0,42 {a^6_2},0,132 {a^7_2},0,429 {a^8_2},0,1430 {a^9_2},0,4862 {a^{10}_2}\right\}$$