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%I #57 Oct 09 2019 12:52:39
%S 1,3,2,10,4,220,5,235,6,354,7,497,8,664,9,1143,11,79117,12,2445932,13,
%T 87580535,14,3572000558,15,163703541857,16,8336823369072,17,
%U 467409009871723,18,28624087521132434,19,1901883146740912949,20
%N A permutation of the positive integers defined in the comment section such that the sum of the first n terms of the sequence is divisible by n.
%C Start the sequence at a(1) = 1. For each n, if the sum of the first 2n-1 terms is S(n), then define a(2n+1) to be the smallest positive integer that has not appeared in {a(1), a(2), ..., a(2n-1)}, and a(2n) = a(2n+1)*[(2n+1)^t-1] - S(n), where t is the smallest positive integer that makes a(2n) > a(2n-2) (if n = 1, choose t = 1). [Simplified and corrected by _Jianing Song_, Oct 04 2019]
%C This is a sequence of positive integers in which each number occurs exactly once such that for each n = 1,2,3,... the sum of the first n terms of the sequence is divisible by n.
%C If we always choose the smallest candidate for each a(n), we get A019444. - _Jianing Song_, Oct 04 2019
%C Is a(2n-1) = A183301(n)? - _Dmitry Kamenetsky_, Jul 11 2019 [Answer: No, because a(37) = 21 while A183301(19) = 22. - _Jianing Song_, Oct 04 2019]
%H The IMO Compendium Group, <a href="https://imomath.com/othercomp/Rus/RusMO95.pdf">21st All-Russian Mathematical Olympiad 1995: Grade 10 Problem 3</a>
%e The first term is 1. So S(1) = 1, a(3) = 2.
%e This gives a(2) = 2*(3^t-1) - 1 = 3, here t = 1. So S(2) = 6, a(5) = 4.
%e This gives a(4) = 4*(5^t-1) - 6 = 10 > a(2), here t = 1. So S(3) = 20, a(7) = 5.
%e This gives a(6) = 5*(7^t-1) - 20 = 220 > a(4), here t = 2. So S(4) = 245, a(9) = 6.
%e ...
%e S(7) = 2025, a(17) = 11, so a(16) = 11*(17^t-1) - 2025 = 1143 > a(14) = 664, here t = 2. [Rewritten by _Jianing Song_, Oct 04 2019]
%Y Cf. A019444.
%K nonn
%O 1,2
%A _Jinyuan Wang_, Aug 23 2018
%E Incorrect definition removed by _Jianing Song_, Oct 04 2019
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