A183301 -id:A183301 - cp's OEIS frontend

A318280 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.

1, 3, 2, 10, 4, 220, 5, 235, 6, 354, 7, 497, 8, 664, 9, 1143, 11, 79117, 12, 2445932, 13, 87580535, 14, 3572000558, 15, 163703541857, 16, 8336823369072, 17, 467409009871723, 18, 28624087521132434, 19, 1901883146740912949, 20

Offset: 1

Jinyuan Wang, Aug 23 2018

nonn

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]
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.
If we always choose the smallest candidate for each a(n), we get A019444. - Jianing Song, Oct 04 2019
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]

			The first term is 1. So S(1) = 1, a(3) = 2.
This gives a(2) = 2*(3^t-1) - 1 = 3, here t = 1. So S(2) = 6, a(5) = 4.
This gives a(4) = 4*(5^t-1) - 6 = 10 > a(2), here t = 1. So S(3) = 20, a(7) = 5.
This gives a(6) = 5*(7^t-1) - 20 = 220 > a(4), here t = 2. So S(4) = 245, a(9) = 6.
...
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]

The IMO Compendium Group, 21st All-Russian Mathematical Olympiad 1995: Grade 10 Problem 3

Cf. A019444.

Incorrect definition removed by Jianing Song, Oct 04 2019

cp's OEIS Frontend

A318280 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.

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