Problem 601

Divisibility streaks

For every positive number $n$ we define the function $s t r e a k (n) = k$ as the smallest positive integer $k$ such that $n + k$ is not divisible by $k + 1$ .
E.g:
13 is divisible by 1
14 is divisible by 2
15 is divisible by 3
16 is divisible by 4
17 is NOT divisible by 5
So $s t r e a k (13) = 4$ .
Similarly:
120 is divisible by 1
121 is NOT divisible by 2
So $s t r e a k (120) = 1$ .

Define $P (s, N)$ to be the number of integers $n$ , $1 < n < N$ , for which $s t r e a k (n) = s$ .
So $P (3, 14) = 1$ and $P (6, 10^{6}) = 14286$ .

Find the sum, as $i$ ranges from 1 to 31, of $P (i, 4^{i})$ .

连续整除

对于任意正整数 $n$ ，我们定义函数 $s t r e a k (n) = k$ 为使 $n + k$ 不能被 $k + 1$ 整除的最小正整数 $k$ 。
例如：
13能被1整除
14能被2整除
15能被3整除
16能被4整除
17不能被5整除
因此 $s t r e a k (13) = 4$ 。
类似地：
120能被1整除
121不能被2整除
因此 $s t r e a k (120) = 1$ 。

定义 $P (s, N)$ 为 $1 < n < N$ 中满足 $s t r e a k (n) = s$ 的整数 $n$ 的数量。
因此 $P (3, 14) = 1$ ，而 $P (6, 10^{6}) = 14286$ 。

令 $i$ 取值从1到31，求对应的 $P (i, 4^{i})$ 之和。