Problem 316
Numbers in decimal expansions
Let p = p1 p2 p3 … be an infinite sequence of random digits, selected from {0,1,2,3,4,5,6,7,8,9} with equal probability.
It can be seen that p corresponds to the real number 0.p1 p2 p3 ….
It can also be seen that choosing a random real number from the interval [0,1) is equivalent to choosing an infinite sequence of random digits selected from {0,1,2,3,4,5,6,7,8,9} with equal probability.
For any positive integer n with d decimal digits, let k be the smallest index such that pk, pk+1, …pk+d-1 are the decimal digits of n, in the same order.
Also, let g(n) be the expected value of k; it can be proven that g(n) is always finite and, interestingly, always an integer number.
For example, if n = 535, then
for p = 31415926535897…., we get k = 9
for p = 355287143650049560000490848764084685354…, we get k = 36
etc and we find that g(535) = 1008.
Given that $\sum^{999}_{n=2}g(\lfloor \frac{10^6}{n} \rfloor)=27280188$, find $\sum^{999999}_{n=2}g(\lfloor \frac{10^{16}}{n} \rfloor)$。
Note : $\lfloor x \rfloor$ represents the floor function.
十进制展开中的数
记p = p1 p2 p3 … 为无限随机数字序列，每一位数字都从{0,1,2,3,4,5,6,7,8,9}中等概率地选取。
可以看出，p对应了一个实数0.p1 p2 p3 ….
同样可以看出，从区间[0,1)中随机选择一个实数等价于从{0,1,2,3,4,5,6,7,8,9}中等概率地选择数字构成一个无限随机数字序列。
对于任意d位十进制正整数n，存在最小的k，使得pk、pk+1、……pk+d-1恰好是n按顺序排列的各位数字。
此外，记g(n)为k的期望值；可以证明，g(n)总是有限的，而且有趣的是，g(n)总是整数。
例如，取n = 535，那么
若p = 31415926535897….，我们有k = 9
若p = 355287143650049560000490848764084685354…，我们有k = 36
依此类推，我们可以得到g(535) = 1008.
已知$\sum^{999}_{n=2}g(\lfloor \frac{10^6}{n} \rfloor)=27280188$，求$\sum^{999999}_{n=2}g(\lfloor \frac{10^{16}}{n} \rfloor)$。
注意：$\lfloor x \rfloor$表示下取整函数。