Take the following recurrence relation into account:
$$ a_{n+2} = \frac{1}{(n+1)(n+2)} \sum_{k=0}^n (s_k - (k+1)a_{k+1})(n-k+1)a_{n-k+1} $$
I know that:
$$ s_{2m+1} = \frac{(-1)^m}{(2m+1)!} $$
and
$$ s_{2m} = 0 $$
My goal is to find out, In terms of $a_0$ and $a_1$:
$$ \sum_{n=0}^\infty a_n x^n $$
I have came out with this first few $a$ terms:
$$ a_2 = - \frac{a_1^2}{2} $$
$$ a_3 = \frac{a_1^3}{3} + \frac{a_1}{6} $$
$$ a_4 = -\frac{a_1^4}{4} $$
The $a$ terms seem to alternate signs, and it also appears there is a connection between the exponent of the $a_1$ terms and the denominator. This is all I could come up with thus far. My guess would be a natural log result of some kind due to these factors.
Edit (more on the side of how I am seeing the coefficient):
If you set $f(n) = \sum_{k=1}^{[\frac{n}{2}]-1} [s_{2k+1} a_{n-2k+1}]$ and $g(x) = \sum_{k=1}^{n-1}[k(n-k) a_k a_{n-k}]$, It seems as though $g(x)$ is just a way to convolute the terms that are already known, in which you would find a new $f(n)$ which would basically be alternating from even and odd $n$ values for corrisponding even or odd value $a_{n+2}$. This new $f(n)$ function would take the corrisponding (even or odd value that $a_{n+2}$ would take and have those terms desend, as the sequence desends, the terms would alternate signs and decrease due to the properties of $s_n$.
The purpose of doing this is to solve the following differential equation using series solutions:
$$ \sin(x) = \frac{y''}{y'} + y' $$
Setting
$$ f(x) = \sum_{n=0}^\infty s_n x^n $$
and
$$ y = \sum_{n=0}^\infty a_n x^n $$
By solving in terms of $a_1$ i can group the terms in a specific way, by doing so, with the first few terms (meaning about the first 25) I could group the $x$ terms with respect to its $a_1^n$ terms such that:
$$ y = \sum_{n=0}^\infty a_1^n f_n(x) $$ Perhaps by doing this, something interesting can come out of doing this, this intrigues me because I see that the solution to such a differential equation is:
$$ e^y = \int e^{\int \sin(x) dx} dx $$