A326729 a(0) = 0; for n >= 1, a(n) is the result of inverting s-th bit (from right) in n, where s is the number of ones in the binary representation of n.

0, 0, 3, 1, 5, 7, 4, 3, 9, 11, 8, 15, 14, 9, 10, 7, 17, 19, 16, 23, 22, 17, 18, 31, 26, 29, 30, 19, 24, 21, 22, 15, 33, 35, 32, 39, 38, 33, 34, 47, 42, 45, 46, 35, 40, 37, 38, 63, 50, 53, 54, 59, 48, 61, 62, 39, 60, 49, 50, 43, 52, 45, 46, 31, 65, 67, 64, 71, 70, 65, 66, 79, 74, 77, 78, 67, 72, 69, 70, 95, 82, 85, 86, 91, 80, 93, 94, 71, 92, 81, 82, 75, 84, 77, 78, 127, 98, 101, 102, 107, 96

Offset: 0

Author

Max Alekseyev, Jul 22 2019

Keywords

• base
• nonn
• look

Comments

Iterations of a(n) always reach 0 (cf. A326730), see Problem 5 of IMO 2019.

Links

• Robert Israel, Table of n, a(n) for n = 0..10000
• International Mathematical Olympiad, Problem 5 of IMO 2019.
• Index to sequences related to Olympiads.

Crossrefs

Cf. A000120, A326730, A326731, A326732.

Programs

Maple

f:= proc(n) local s;
  s:= convert(convert(n,base,2),`+`);
  Bits:-Xor(n,2^(s-1))
end proc:
f(0):= 0:
map(f, [$0..100]); # Robert Israel, Oct 01 2020

PARI

A326729(n) = if(n==0,return(0)); bitxor(n,2^(hammingweight(n)-1));

Formula

For n>=1, a(n) = n XOR 2^(A000120(n)-1).

From Robert Israel, Oct 01 2020: (Start)

a(2*n+1) = 2*a(n).

a(2*n + 2^k) = 2*a(n)+2^k if 2^k > 2*n. (End)