A326731 a(0) = 0; for n >= 1, a(n) = result of inverting s-th bit (from left) in n, where s is the number of ones in the binary representation of n.

0, 0, 0, 2, 0, 7, 4, 6, 0, 13, 14, 9, 8, 15, 12, 14, 0, 25, 26, 23, 28, 17, 18, 21, 16, 29, 30, 25, 24, 31, 28, 30, 0, 49, 50, 43, 52, 45, 46, 35, 56, 33, 34, 47, 36, 41, 42, 45, 32, 57, 58, 55, 60, 49, 50, 53, 48, 61, 62, 57, 56, 63, 60, 62, 0, 97, 98, 83, 100, 85, 86, 79, 104, 89, 90, 67, 92, 69, 70, 75, 112, 65, 66, 91, 68, 93, 94, 83, 72, 81, 82, 95, 84, 89, 90, 93, 64, 113, 114, 107, 116

Offset

0,4

Comments

Iterations of a(n) always reach 0 (cf. A326732), see Problem 5 of IMO 2019.

Links

Table of n, a(n) for n=0..100.

International Mathematical Olympiad, Problem 5 of IMO 2019.

Formula

For n>=1, a(n) = n XOR 2^(A029837(n)-A000120(n)).

For n>=1, a(n) = 0 iff n is a power of 2.

Prog

(PARI) A326731(n) = if(n==0, return(0)); my(b=binary(n)); bitxor(n, 2^(#b-vecsum(b)));

Crossrefs

A variant of A326729.

Cf. A000120, A029837, A326730, A326732.

Sequence in context: A196767 A307216 A011343 * A021486 A258990 A351727

Adjacent sequences: A326728 A326729 A326730 * A326732 A326733 A326734

Keyword

nonn,base

Author

Max Alekseyev, Jul 22 2019

Status

approved