OFFSET
1,2
COMMENTS
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]
LINKS
The IMO Compendium Group, 21st All-Russian Mathematical Olympiad 1995: Grade 10 Problem 3
EXAMPLE
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]
CROSSREFS
KEYWORD
nonn
AUTHOR
Jinyuan Wang, Aug 23 2018
EXTENSIONS
Incorrect definition removed by Jianing Song, Oct 04 2019
STATUS
approved