Let $m\geq 2$ be a fixed integer.

Let $$f(n):=\begin{cases} mf\left(\frac{n}{m}\right),&\text{if $n\mod m = 0$;}\\ 1,&\text{otherwise} \end{cases}$$ then if we have $$a(n):=\begin{cases} 1,&\text{if $n=0$;}\\ a\left(\frac{n}{m}\right)+a\left(n-f\left(\frac{n}{m}\right)\right),&\text{if $n\mod m = 0$;}\\ a\left(\left\lfloor\frac{n}{m}\right\rfloor\right),&\text{otherwise} \end{cases}$$ and also $$s(n):=\sum\limits_{k=0}^{m^n-1}a(k).$$ In particular, $s(0) = 1$ and $s(1) = m$.

I conjecture that for $m > 2$ $$s(n)=(m+3)s(n-1)-(2m+1)s(n-2).$$ For a case $m=2$ we get Bell numbers.

Is there a way to prove it?