Tessellate S.T.E.M.S. (2019) - Mathematics - Category A (School) - Set 4 - Objective Problem 3

Number Theory Level 2

Let ( a n ) n ∈ N (a_n)_{n \in \mathbb{N} } be an infinite sequence of positive integers satisfying a 1 = 1 a_1=1 , a n ∣ ∑ i = 0 n − 1 a k + i a_n| \sum_{i=0}^{n-1} a_{k+i} for all k , n ∈ N k, n \in \mathbb{N} . Compute the maximum possible value of a 2018 a_{2018} .

This problem is a part of Tessellate S.T.E.M.S. (2019)

2 2018 − 1 2^{2018}-1 2 2019 − 1 2^{2019}-1 2 1010 − 1 2^{1010}-1 2 1009 − 1 2^{1009}-1

Tessellate S.T.E.M.S. Mathematics by Tessellate S.T.E.M.S. Ma… , 26

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1 solution

Patrick Corn
Jan 30, 2019

Well, it's not hard to verify that the sequence a k = { 1 if k is odd 2 k / 2 − 1 if k is even a_k = \begin{cases} 1 & \text{if } k \text{ is odd} \\ 2^{k/2}-1 & \text{if } k \text{ is even} \end{cases} satisfies the conditions. (The sequence goes 1 , 1 , 1 , 3 , 1 , 7 , 1 , 15 , 1 , 31 , … . ) 1,1,1,3,1,7,1,15,1,31,\ldots.) If n is odd, the condition is vacuous (since 1 1 divides everything), and if n n is even, one checks that the sum of any n n consecutive terms of the sequence equals a power of 2 2 times a n . a_n.

I'm not sure how to show this is maximal...but assuming it is, this gives a 2018 = 2 1009 − 1. a_{2018} = 2^{1009}-1.

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