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

Author

Max Alekseyev, Jul 22 2019

Keywords

• nonn
• base

Comments

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

Links

• International Mathematical Olympiad, Problem 5 of IMO 2019.

Crossrefs

A variant of A326729.

Cf. A000120, A029837, A326730, A326732.

Programs

PARI

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

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.