The initial conditions are all negative numbers give zero (similar to how $\displaystyle \binom{n}{r}$ gives zero if $\displaystyle r<0$). Can I use something like

Code:

RSolve[f(n) == n+1+f(n-2)+f(n-6)+f(n-18)+f(n-54) && RecurrenceTable[f(k)=0,{k,-54,-1}], f(n), n]

??

Wolframalpha doesn't understand that formula, though. When I tried a simpler one, it gave me a closed form in terms of a DifferenceRoot, so there may not be any general closed form. What I am hoping to evaluate is:

$\displaystyle f_k(n) = \begin{cases}0 & n<0 \\ n+1+\sum_{i=0}^k f_k(n-2\cdot 3^i) & \text{otherwise}\end{cases}$

Then for each $\displaystyle k$, I want to know if this limit converges to something nonzero:

$\displaystyle \lim_{n \to \infty} \dfrac{f_k(n-3^{k+1})}{f_k(n)}$

It is a relation I found while examining primitive roots of $\displaystyle 3^{k+1}$. It seems certain sums can fall seemingly arbitrarily in a relatively fixed range, and I want to know if it matters if I am off by a couple powers of 3 when taking the limit of that function over another function (as n approaches infinity).