Palindromic sequences (noch nicht übersetzt)

Problem 656

Given an irrational number α, let Sα(n) be the sequence Sα(n)=αnα(n1) for n1.
(... is the floor-function.)

It can be proven that for any irrational α there exist infinitely many values of n such that the subsequence {Sα(1),Sα(2)...Sα(n)} is palindromic.

The first 20 values of n that give a palindromic subsequence for α=31 are: 1, 3, 5, 7, 44, 81, 118, 273, 3158, 9201, 15244, 21287, 133765, 246243, 358721, 829920, 9600319, 27971037, 46341755, 64712473.

Let Hg(α) be the sum of the first g values of n for which the corresponding subsequence is palindromic.
So H20(31)=150243655.

Let T={2,3,5,6,7,8,10,...,1000} be the set of positive integers, not exceeding 1000, excluding perfect squares.
Calculate the sum of H100(β) for βT. Give the last 15 digits of your answer.