Start once, second half is easier

Define $t_n = a_n - a_{n-1} , n \geq 2$ with $, t_2 = 2$ . Then,

$a_{n+2} = (n+3)a_{n+1} - (n+2)a_n$

$\implies a_{n+2} - a_{n+1} = (n+2)(a_{n+1} -a_n)$

$\implies t_{n+2} = (n+2)t_{n+1}$

Replacing $n + 2$ with $n$

$\implies t_{n} = (n)t_{n-1}$

Which is the definition of the Factorial Function.Since $t_2 = 2!$ , we have $t_n = n!$

Now, $\sum_{i = 2}^n t_1 = \sum_{i = 2}^n a_1 - a_{1-1} = a_n - a_1 = a_n - 1$

$\sum_{i = 2}^n i!= a_n - 1 \implies \sum_{i = 1}^n i!= a_n$

Therefore we have to find sum of $n$ such that $\sum_{i = 1}^n i! = n^2$

For $n \geq 5$ ,

$\sum_{i = 1}^n i! \equiv 1! + 2! + 3! + 4! +5! + ... \equiv 1 + 2 + 6 + 24 + 0 + ... \equiv 3 \pmod{10}$

But $x^2 \equiv 1,4,9,6,5,0 \pmod{10}$ only. Therefore, $\sum_{i = 1}^n i!$ can not be a square.

Thus we only have to check from $1$ to $5$ .We get two possible solutions, $1$ and $3$ . Sum of these are $\boxed{4}$

Straight to the series evaluate-able without solving it:

1
2
3
4
5
6
7

    1   1   TRUE    1
    3   4   FALSE   
1   9   9   TRUE    3
2   33  16  FALSE   
3   153 25  FALSE   
4   873 36  FALSE   
...

The series grows very fast comparatively. Therefore only 1 + 3 = 4.

Answer: $\boxed{4}$

On placing the values of of n in the equation we get a1 = 1, a2 = 3, a3= 9, a4= 33 which implies the series is always increasing hence we are left with only a3 and a1. Therefore 3 + 1 = 4

Ans. 4