Given an irrational number $\alpha$, let $S_{\alpha }(n)$ be the sequence $S_{\alpha }(n)=\lfloor \alpha \cdot n\rfloor -\lfloor \alpha \cdot (n-1)\rfloor$ for $n\ge 1$.
($\lfloor \cdots \rfloor$ is the floor-function.)

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

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

Let $H_g(\alpha )$ be the sum of the first $g$ values of $n$ for which the corresponding subsequence is palindromic.
So $H_{20}(\sqrt{31})=150243655$.

Let $T=\{2,3,5,6,7,8,10,\dots ,1000\}$ be the set of positive integers, not exceeding $1000$, excluding perfect squares.
Calculate the sum of $H_{100}(\sqrt{\beta })$ for $\beta \in T$. Give the last $15$ digits of your answer.