Premise
I have been playing around for fun with the following recursive relation:
$$ f(1) = \frac{a_1}{2} $$ $$ f(n+1) = \frac{a_{n+1} - \sum^n_{k=1} f(k) * f(n-k+1)}{2} $$
Where $a_n$ is a sequence of integers, this generates a new sequence with some properties I have been interested in. The following are the first few terms:
$$ \frac{a_1}{2},\, \frac{a_2}{2} - \frac{a_1^2}{8},\, \frac{a_3}{2} - \frac{a_1*a_2}{4} + \frac{a_1^3}{16},\,… $$
I had been attempting to remove the recursive relation, but I am not sure where I would even start or if it is even possible.
Question
I was hoping for some guidance/answer on how (if possible) I might be able to describe the function $f$ without being recursive.
Let me allow to re-index your sequences for convenience, such that the considered recurrence relation becomes $$ f(n) = \frac{1}{2}\left(a_n - \sum_{k=0}^{n-1} f(k)f(n-k)\right), \; f(0) = \frac{a_0}{2}. $$ The sum corresponds to a (truncated) discrete convolution product. In the continuous case, it would be handled with the help of the Laplace transform, in order to transform it into a standard product, that is why we will use its discrete analog, namely the (unilateral) Z-transform, defined as follows : $$ F(z) := \sum_{n=0}^\infty f(n)z^{-n} $$ A convolution theorem can be straighforwardly deduced from this definition, so that $(f*g)(n) \mapsto F(z) \cdot G(Z)$, hence $F(z) = \frac{1}{2}(A(z) - F(z)^2)$ in the present case, where $A(z)$ is the Z-transform of the sequence $(a_n)_n$, and finally $F(z) = -1 \pm \sqrt{1+A(z)}$, which still has to be inverse-transformed $-$ but the result will depend on $A(z)$.
N.B. : it is also possible to use the more common generating function $\tilde{F}(z) = \sum_{n\ge0} f(n)z^n$, but it turns out to be harder to invert in most cases, while retaining the same kind of properties as the unilateral Z-transform, since $\tilde{F}(z) = F(1/z)$.