Note that $$\frac{1}{3} = \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \cdots$$ In fact, we have

\begin{aligned} \frac{1}{3} &= \sum_{k=1}^{\infty} \frac{1}{4^k} \frac{1}{4} &= \sum_{k=1}^{\infty} \frac{1}{5^k} \frac{1}{n-1} &= \sum_{k=1}^{\infty} \frac{1}{n^k} \end{aligned}

This fact is useful in performing integer division. We can compute $$m$$ divided by $$n$$ by recursion (note the summation sign above). Consider $$m/(n-1)$$, we have

\begin{aligned} m &= nA + B & \exists A,B\in\mathbb{N},\; 0\le B< n \textrm{therefore, }\frac{m}{n-1} &= A + \frac{A+B}{n-1}. \end{aligned}

Since $$A = m \textrm{ div } n$$, $$B = m \textrm{ mod } n$$, if $$n$$ is a power of 2, we can perform the div and mod operations by bit shift.

def div(m, n):
# n+1 is 2^k assumed here
result = 0
A = m
while (True):
A = A >> k
B = A & n
result = add(result, A)
if A <= n:
break
if A == n: result = result + 1
return result

def add(A, B):
result = A XOR B
carry = (A AND B) << 1
return add(result, carry) if carry else result