Semi-Complicated recurrence relation to be solved via generating functions

Question

If $(a_n)_{n\ge 1}$ is a sequence such that $a_0=3$ and for all $n \ge 1$, \begin{equation} \sum_{k=1}^{\lceil \frac{n}{2} \rceil}a_{n-2k+1}=na_n \end{equation} , use generating functions to find an explicit formula for $a_n$.

So far I've been swimming around in the algebra a bit, and I've gotten

$$A'(x)=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=1}^\infty\left(\sum_{k=1}^{\lceil n/2 \rceil} a_{n-2k+1}\right)x^{n-1}$$

But I'm not sure how to get a nice expression in terms of $A(x)$ for the RHS of this equation.

Answer 1

The recurrence relation hints that the derivative of $A(x)$ may be involved: $$A'(x)=\sum_{n=0}^\infty na_nx^{n-1}=\sum_{n=1}^\infty\left(\sum_{k=1}^{\lceil n/2 \rceil} a_{n-2k+1}\right)x^{n-1}=A(x)+x^2A(x)+x^4A(x)+ \ldots,$$ and summing the geometric series one obtains the differential equation $$A'(x)=\frac{A(x)}{1-x^2}.$$ The solution can be obtained by separation of variables: $$A(x)=C\frac{\sqrt{x+1}}{\sqrt{x-1}}.$$ Since everything is happening for $x$ near $0$, no need to worry about convergence issues.

Answer 2

Here are the steps to obtain the differential equation: \begin{align} A'(x) &= \sum_{n=1}^\infty\left(\sum_{k=1}^{\lceil n/2 \rceil} a_{n-2k+1}\right)x^{n-1} \\ &= \sum_{k=1}^\infty x^{2k-2} \sum_{n=2k-1}^\infty a_{n-2k+1} x^{n-2k+1} \\ &= \sum_{k=1}^\infty x^{2k-2} A(x) \\ &= A(x) \sum_{k=1}^\infty (x^2)^{k-1} \\ &= \frac{A(x)}{1-x^2} \\ \end{align}