Closed form for a strong recurrence relation

Question

Let $\alpha_n$ be a sequence of complex numbers and consider the sequence $b_n$ defined by the (strong) recurrence relation : $$b_{n+1} = \sum_{k=0}^n \alpha_{n-k} b_k$$ with the initial condition $b_0=1$.

Is there a closed form for $b_n$ ?

Answer 1

OK I got it. Like I said in the comments, consider the generating functions $A$ and $B$ for the $a_n$ and $b_n$. Noticing the right hand side of the recurrence relation is a Cauchy product, we get $\frac{B(z)-B(0)}{z}=A(z) B(z)$ wich gives $$B(z)=b_0 \sum_{n \geq 0} z^n A(z)^n$$

Thus $b_n = \sum_{k=1}^n$ coefficient of the term of order $n-k$ of $A(z)^k$.

This finally gives : $$b_n = \sum_{k=1}^n \sum_{p_1 + \ldots p_k=n-k} a_{p_1} \ldots a_{p_k}$$ which I verified to be correct for the first few terms. Phew!

Answer 2

Given that $b_{0} =1$ and \begin{align} b_{n+1} = \sum_{k=0}^{n} a_{n-k} b_{k} \end{align} then the first few $b_{n}$ are \begin{align} b_{1} &= a_{0} \\ b_{2} &= a_{1} + a_{0}^{2} \\ b_{3} &= a_{2} + 2 a_{0} a_{1} + a_{0}^{3} \\ b_{4} &= a_{3} + 2 a_{0} a_{2} + 2 a_{1} a_{0}^{2} + a_{1}^{2} + a_{0}^{4}. \end{align} These $b_{n}$ are not able to be put into closed form without knowing what the values of $a_{n}$ are.