Solving the recurrence: $h(i) = h\left(\left\lfloor\frac{i+1}{d}\right\rfloor\right)+1 $ I want to solve the following recurrence:
\begin{equation}
h(1) = 0\\
h(i) = h\left(\left\lfloor\frac{i+1}{d}\right\rfloor\right)+1 
\end{equation}
What are some basic "methods" I can use to guess a closed form of this recurrence? I see that I am dividing by $d$ and this would mean that I am doing repeated division which would imply some sort of $\log_d$ term. Is this correct? 
What is the closed form of this recurrence?
 A: If it's
$$\begin{equation}
h(0) = 0\\
h(i) = h\left(\left\lfloor\frac{i+1}{d}\right\rfloor\right)+1 
\end{equation}$$
then
$$h(i)=\left\lfloor \log_{d}{\frac{d-1}{d-2}i}\right\rfloor+1$$
Solution (I believe that it contains mistake I couldn't find because it gives different result):
Will use this later $$a\leq\left\lfloor\frac{i+1}{d} \right\rfloor < b \Longleftrightarrow ad-1\leq i < bd-1$$
Also such definition will be useful
$$a_n=\min \left\{ i \vert f(i)=n\right\}$$  
Then (without proof here) 
$$f(i)=n \Longleftrightarrow a_n \leq i<a_{n+1}$$  
Because $f(i)=n+1$ iff $f\left(\left\lfloor\frac{i+1}{d}\right\rfloor\right)=n$
$$f(i)=n+1 \Longleftrightarrow a_n \leq \left\lfloor\frac{i+1}{d}\right\rfloor<a_{n+1}\\
f(i)=n+1 \Longleftrightarrow da_n-1 \leq i < da_{n+1}-1$$
So
$$a_{n+1}=da_n-1, \ a_1=1, a_2=d-1$$
gives
$$a_n=d^{n-1}-d^{n-2}-...-d-1=2d^{n-1}-\frac{d^n-1}{d-1}=\frac{(d-2)d^{n-1}+1}{d-1}$$
Next step is to get $n$ from inequalities $\frac{(d-2)d^{n-1}+1}{d-1}\leq i < \frac{(d-2)d^{n}+1}{d-1}$:
$$\frac{(d-2)d^{n-1}+1}{d-1}\leq i < \frac{(d-2)d^n+1}{d-1}\\
(d-2)d^{n-1}+1\leq (d-1)i < (d-2)d^n+1 \\
(d-2)d^{n-1}\leq (d-1)i-1 < (d-2)d^n \\
d^n\leq d\frac{(d-1)i-1}{d-2} < d^{n+1} \\
n=\left\lfloor \log_{d}{\frac{(d-1)i-1}{d-2}}\right\rfloor+1
$$
And $h(i)$ is equal to this value of $n$.
