A253558 a(n) = A253556(n) + 1.

1, 1, 2, 1, 3, 2, 4, 1, 2, 3, 5, 2, 6, 4, 3, 1, 7, 2, 8, 3, 2, 5, 9, 2, 3, 6, 4, 4, 10, 3, 11, 1, 3, 7, 4, 2, 12, 8, 5, 3, 13, 2, 14, 5, 2, 9, 15, 2, 4, 3, 3, 6, 16, 4, 3, 4, 4, 10, 17, 3, 18, 11, 6, 1, 5, 3, 19, 7, 3, 4, 20, 2, 21, 12, 7, 8, 5, 5, 22, 3, 5, 13, 23, 2, 4, 14, 4, 5, 24, 2, 4, 9, 2, 15, 6, 2, 25, 4, 8, 3, 26, 3

Offset: 1

Author

Antti Karttunen, Jan 12 2015

Keywords

• nonn

Comments

Consider the binary trees illustrated in A252753 and A252755: If we start from any n, computing successive iterations of A253554 until 1 is reached (i.e., we are traversing level by level towards the root of the tree, starting from that vertex of the tree where n is located at), a(n) gives the number of odd numbers encountered on the path (i.e., including both the final 1 and the starting n if it was odd).

Links

• Antti Karttunen, Table of n, a(n) for n = 1..8192

Crossrefs

One more than A253556.

Powers of two, A000079, gives the positions of ones.

Cf. A000040, A253555, A253557, A253559.

After n=1, differs from A061395 for the first time at n=21, where a(21) = 2, while A061395(21) = 4.

Programs

Scheme

(define (A253558 n) (+ 1 (A253556 n)))

Formula

a(n) = A253556(n) + 1.

a(n) = A080791(A252754(n)) + 1. [One more than the number of nonleading 0-bits in A252754(n).]

Other identities.

For all n >= 1:

a(A000040(n)) = n.

For all n >= 2:

a(n) = A000120(A252756(n)). [Binary weight of A252756(n).]

a(n) = A253555(n) - A253559(n).