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

0,3

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.

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)

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

Prog

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

Crossrefs

Cf. A000120, A326730, A326731, A326732.

Sequence in context: A077020 A107920 A169998 * A171998 A343615 A159285

Adjacent sequences: A326726 A326727 A326728 * A326730 A326731 A326732

Keyword

base,nonn,look

Author

Max Alekseyev, Jul 22 2019

Status

approved