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Problem 601

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Divisibility streaks

For every positive number $n$ we define the function $streak(n)=k$ 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 $streak(13) = 4$.
Similarly:
120 is divisible by 1
121 is NOT divisible by 2
So $streak(120) = 1$.

Define $P(s, N)$ to be the number of integers $n$, $1 < n < N$, for which $streak(n) = s$.
So $P(3, 14) = 1$ and $P(6, 10^6) = 14286$.

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

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连续整除

对于任意正整数$n$，我们定义函数$streak(n)=k$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$streak(13) = 4$。
类似地：
120能被1整除
121不能被2整除
因此$streak(120) = 1$。

定义$P(s, N)$为$1 < n < N$中满足$streak(n) = s$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。

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