Solving a recurrence relation when $n$ is a power of $7$ I'm working through some recurrence relation examples and am struggling with this question that assumes inputs of $n$ are powers of $7$.
Essentially we have
$T(n) = T({\lfloor}n/7{\rfloor}) + \log_{3}(n)$
$T(1) = 0$
Using the bottom up method I have found the following
$T(7) = T(1) + \log_{3}(7) = \log_{3}(7)$
$T(49) = T(7) + \log_{3}(49) = \log_{3}(7) + \log_{3}(49)$
$T(343) = T(49) + \log_{3}(343) = \log_{3}(7) + \log_{3}(49) 
+ log_{3}(343)$
I'm struggling to turn this into an equation though. I can see that essentially we have
$T(n) = \log_{3}(7^1) + \log_{3}(7^2) + ... + \log_{3}(7^k)$ where $7^k = n$.
How would I transform that logic into an equation?

It seems more helpful to consider the function
$f(0) = 1$
$f(n + 1) = f(n) + \log_3(7^{n + 1}) = f(n) + (n + 1) \log_3(7)$
Here, we have $f(n) = T(7^n)$.
Then it is immediate that $f(n) = 1 + \sum\limits_{i = 1}^n n \log_3(7) = 1 + \log_3(7) \sum\limits_{i= 1}^n n$. Using the identity $\sum\limits_{i = 1}^n n = \frac{n (n + 1)}{2}$, we have
$f(n) = 1 + \frac{n(n+1) \log_3(7)}{2}$
hence
$T(n) = 1 + \frac{\log_7(n) (\log_7(n) + 1) \log_3(7)}{2} = 1 + \frac{\log_3(n) (\log_7(n) + 1)}{2}$
whenever $n$ is a power of 7.
