Palindromic sequences
Problem 656
Given an irrational number , let be the sequence for .
( is the floor-function.)
( is the floor-function.)
It can be proven that for any irrational there exist infinitely many values of such that the subsequence is palindromic.
The first 20 values of that give a palindromic subsequence for are:1, 3, 5, 7, 44, 81, 118, 273, 3158, 9201, 15244, 21287, 133765, 246243, 358721, 829920, 9600319, 27971037, 46341755, 64712473.
Let be the sum of the first values of for which the corresponding subsequence is palindromic.
So .
So .
Let be the set of positive integers, not exceeding 1000, excluding perfect squares.
Calculate the sum of for . Give the last 15 digits of your answer.
Calculate the sum of for . Give the last 15 digits of your answer.