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a_{2n-1} = a_{2n-2} + n$ , $a_{2n} = a_{2n-1} + n , perfect square

Source: Dutch NMO 2009 p2

September 6, 2019
number theoryPerfect Squarerecurrence relation

Problem Statement

Consider the sequence of integers 0,1,2,4,6,9,12,...0, 1, 2, 4, 6, 9, 12,... obtained by starting with zero, adding 11, then adding 11 again, then adding 22, and adding 22 again, then adding 33, and adding 33 again, and so on. If we call the subsequent terms of this sequence a0,a1,a2,...a_0, a_1, a_2, ..., then we have a0=0a_0 = 0, and a2n−1=a2n−2+na_{2n-1} = a_{2n-2} + n , a2n=a2n−1+na_{2n} = a_{2n-1} + n for all integers n≥1n \ge 1. Find all integers k≥0k \ge 0 for which aka_k is the square of an integer.
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