## Infinite string tour

### Problem 238

Published on Sunday, 29th March 2009, 03:00 pm; Solved by 606

Create a sequence of numbers using the "Blum Blum Shub" pseudo-random number generator:

s0 | = | 14025256
sn+1 | = | sn2 mod 20300713

Concatenate these numberss0s1s2… 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 103, p(k) = 4742.

Find p(k), for 0< k 2·1015.