Solving recursive function with floor The recursive function is this:
$$ T(n) = 
\begin{cases}
2 & \text{ for }n=1;\\
T \left( \lfloor \frac{n}{2} \rfloor \right) + 7 &,\text{ otherwise}
\end{cases}
$$
Based on the definition of the function, the right side becomes: 
$T(n) = T( \lfloor \frac{n}{2^i} \rfloor) + 7 * i;$
The procedure stops when $\lfloor \frac{n}{2^i} \rfloor == 1$
The problem is how do I continue it from there?

We list first some terms:
$$\underbrace{2}_{1}, \underbrace{9, 9}_{2}, \underbrace{16, 16, 16, 16}_{4}, 
\underbrace{23, 23, 23, 23, 23, 23, 23, 23}_{8}, 30, 30, \cdots$$
Conjecture: 
$$T(n) = 2 + 7\lfloor \log_2 n\rfloor, \quad n = 1, 2, 3, \cdots.$$
We need to prove it. To this end, let
$$S(n) = 2 + 7\lfloor \log_2 n\rfloor, \quad n = 1, 2, 3, \cdots.$$
Let us prove that $S(n) = T(n)$ for $n = 1, 2, 3, \cdots$.
We use mathematical induction.
First, $S(1) = T(1) = 2$, and $S(2) = T(2) = 9$. 
Assume that $S(k) = T(k)$ for $k = 1, 2, \cdots, n$ ($n\ge 2$).
We need to prove that $S(n+1) = T(n+1)$. 
There exist integer $m\ge 1$ and integer $r$ with $0\le r < 2^m$
such that $n = 2^m + r$. We split into two cases:
1) $r = 2^m - 1$: We have $S(n+1) = 2 + 7(m+1)$
and $S(\lfloor \frac{n+1}{2}\rfloor ) = 2 + 7m$
which results in
$S(n+1) = S(\lfloor \frac{n+1}{2}\rfloor ) + 7 = T(\lfloor \frac{n+1}{2}\rfloor ) + 7 = T(n+1)$.
2) $r < 2^m - 1$: We have $S(n+1) = 2 + 7m$
and $S(\lfloor \frac{n+1}{2}\rfloor ) = 2 + 7(m-1)$
which results in
$S(n+1) = S(\lfloor \frac{n+1}{2}\rfloor ) + 7 = T(\lfloor \frac{n+1}{2}\rfloor ) + 7 = T(n+1)$.  $\quad$ Q.E.D.
Thus, $T(n) = 2 + 7\lfloor \log_2 n\rfloor, \quad n = 1, 2, 3, \cdots.$
