This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A318280 #57 Oct 09 2019 12:52:39 %S A318280 1,3,2,10,4,220,5,235,6,354,7,497,8,664,9,1143,11,79117,12,2445932,13, %T A318280 87580535,14,3572000558,15,163703541857,16,8336823369072,17, %U A318280 467409009871723,18,28624087521132434,19,1901883146740912949,20 %N 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. %C A318280 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 A318280 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 A318280 If we always choose the smallest candidate for each a(n), we get A019444. - _Jianing Song_, Oct 04 2019 %C A318280 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 A318280 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 A318280 The first term is 1. So S(1) = 1, a(3) = 2. %e A318280 This gives a(2) = 2*(3^t-1) - 1 = 3, here t = 1. So S(2) = 6, a(5) = 4. %e A318280 This gives a(4) = 4*(5^t-1) - 6 = 10 > a(2), here t = 1. So S(3) = 20, a(7) = 5. %e A318280 This gives a(6) = 5*(7^t-1) - 20 = 220 > a(4), here t = 2. So S(4) = 245, a(9) = 6. %e A318280 ... %e A318280 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 A318280 Cf. A019444. %K A318280 nonn %O A318280 1,2 %A A318280 _Jinyuan Wang_, Aug 23 2018 %E A318280 Incorrect definition removed by _Jianing Song_, Oct 04 2019