It's well known that generating functions are great to solve recurence relations in form $$a_n = A*a_{n-1} + B*a_{n-2} + \dots$$
But i was wondering what happens if recurence relation contains division in subscripts? It is: $$a_n = A*a_{\lfloor \frac{n}{2} \rfloor} + B*a_{\lfloor \frac{n}{3} \rfloor} + \dots$$ where $A$ and $B$ are some contants. On this stage, let's ignore boundary cases (n=0, 1 etc.) and assume $A$ and $B$ are equal to $1$.
Noting $A(x)$ as generating function for sequence $<a_n>_{0}^\infty$ we have $$A(x) = (1+x)A(x^2) + (1+x+x^2)A(x^3)$$ if i understand it correctly.
I searched the internet for that and even Wilf's gfology book, but i couldn't find explaination for this case. Maybe we can't solve this recurence easily and in effect, can't have closed form for this recurence? Are there other tools to solve it? Thanks in advance.
Intersting question, but I think, only constant sequences verify this. take $(a_n)$ a sequence verifying a recurrence equation :
$$a_n=\sum_{k=2}^r\lambda_ka_{\lfloor \frac{n}{k} \rfloor}$$
Then, evaluating it for $n=0$ we get :
$$a_0=\sum_{k=2}^r\lambda_ka_0 $$
That is either $a_0=0$ or :
$$\sum_{k=2}^r\lambda_k=1 $$
First, if $a_0=0$ then $a_1=a_0*=0$ and with a trivial induction you have that $a_n=0$ for all $n$.
Suppose now that $a_0\neq 0$ then we will show by induction that $a_n=a_0$. The case $n=0$ is trivial. Assume that $a_l=a_0$ for $0\leq l<n$ then for all $k\geq 2$ :
$$\lfloor \frac{n}{k} \rfloor <n$$
So by induction hypothesis :
$$a_{\lfloor \frac{n}{k} \rfloor}=a_0 $$
Finally :
$$a_n=\sum_{k=2}^r\lambda_ka_{\lfloor \frac{n}{k} \rfloor}=\sum_{k=2}^r\lambda_ka_0=a_0$$
If we add together all the pieces this shows that any function verifying a recurrence equation involving a division should be constant.