Generating function given a recurrence relation

Question

In this case I'm here again, now with a problem regarding generating functions. Now, I am asked to find the generating fuction $f(x) = \sum_{n = 0}^{\infty}a_nx^n$ such that $a_n = \frac{a_{n-3}}{n} \forall n \geq 3$ and $a_0 = 1, a_1 = 0, a_2 = 0$. I tried to solve it, but I find it impossible to reach the result without the apparition of the derivative. Here is my attempt:
Using the recurrence relation, $n\sum_{n = 3}^{\infty}a_nx^n = \sum_{n = 3}^{\infty}a_{n-3}x^n = x^3f(x) \Rightarrow x^3f(x) = x\sum_{n = 3}^{\infty}na_nx^{n-1} = x(f'(x) - b_2x^2 - b_1x - b_0)$, where $b_i$ is the $i$-th coefficient of the expansion of $f'(x)$. However, I cannot obtain that information. Do I have to to solve a differential equation. Is this related to MacLaurin series? I cannot continue, so I would appreciate someone could give me a hand.

Answer 1

All coefficients except $a_0,a_3,a_6,a_9,...$ are $0$, and $a_{3n}=\frac 1 {3^{n}n!}$. Hence, $$f(x)=\sum\limits_{n=0}^{\infty} \frac 1 {3^{n}n!}x^{3n}=e^{x^{3}/3}.$$