It was an exam question. $$ f(n)= \begin{cases} 0 & \mbox{if } n \leq 1 \\ 3 f(\lfloor n/5 \rfloor) + 1 & \mbox{if } n > 1 \\ \end{cases}$$ So by calculating some I have $f(5) = 1$, $f(10) = 4$, $f(50) = 13$.
I had to solve this recurrence. So to get rid of the floor operator I said lets $n$ be $5^k$ so a multiple of $5$. So started to solve this.
\begin{equation} \begin{split} f(5^k) & = 3\cdot f(5^{k-1}) + 1\\ & = 3\cdot[ 3f(5^{k-2}) + 1 ]+ 1\\ & = 3^2f(5^{k-2}) + 3 + 1\\ & = 3^2\cdot[ 3f(5^{k-3}) + 1 ]+ 3 + 1\\ & = 3^3f(5^{k-3}) + 9 + 3 + 1 \\ & = ...\\ & = 3^jf(5^{k-j}) + \sum_{j=0}^k 3^j\\ & = 3^jf(5^{k-j}) + \frac{3^{k+1} - 1}{2}\\ \text{let $j = k$}\\ & = 3^kf(5^0) + \frac{3^{k+1} - 1}{2}\\ & = 3^kf(1) + \frac{3^{k+1} - 1}{2}\\ & = 3^k\cdot 0 + \frac{3^{k+1} - 1}{2}\\ & = \frac{3^{k+1} - 1}{2}\\ \end{split} \end{equation} And now I'm stuck here trying to express my $3^{k+1}$ as a function of $n$. I'm pretty I probably made some error up there, if I did I don't see them.
I didn't check your whole argument, but assuming it's correct, you can use the fact that if $n$ is $5^k$, then $k$ is $\log_5 n$ to express the result in terms of $n$.
(Side Question: are the $x$'s in the function definition supposed to be $n$'s?)