## Infinite string tour

### Problem 238

Published on Sunday, 29th March 2009, 03:00 pm; Solved by 636Create a sequence of numbers using the "Blum Blum Shub" pseudo-random number generator:

s_{0} |
= | 14025256 |

s_{n+1} |
= | s_{n}^{2} mod 20300713 |

Concatenate these numbers `s`_{0}`s`_{1}`s`_{2}… to create a string `w` of infinite length.

Then, `w` = 14025256741014958470038053646…

For a positive integer `k`, if no substring of `w` exists with a sum of digits equal to `k`, `p`(`k`) is defined to be zero. If at least one substring of `w` exists with a sum of digits equal to `k`, we define `p`(`k`) = `z`, where `z` is the starting position of the earliest such substring.

For instance:

The substrings 1, 14, 1402, …

with respective sums of digits equal to 1, 5, 7, …

start at position **1**, hence `p`(1) = `p`(5) = `p`(7) = … = **1**.

The substrings 4, 402, 4025, …

with respective sums of digits equal to 4, 6, 11, …

start at position **2**, hence `p`(4) = `p`(6) = `p`(11) = … = **2**.

The substrings 02, 0252, …

with respective sums of digits equal to 2, 9, …

start at position **3**, hence `p`(2) = `p`(9) = … = **3**.

Note that substring 025 starting at position **3**, has a sum of digits equal to 7, but there was an earlier substring (starting at position **1**) with a sum of digits equal to 7, so `p`(7) = 1, *not* 3.

We can verify that, for 0 < `k` ≤ 10^{3}, ∑ `p`(`k`) = 4742.

Find ∑ `p`(`k`), for 0 < `k` ≤ 2·10^{15}.