A253556 a(1) = 0; after which, a(2n) = a(n), a(2n+1) = 1 + a(A250470(n)).

0, 0, 1, 0, 2, 1, 3, 0, 1, 2, 4, 1, 5, 3, 2, 0, 6, 1, 7, 2, 1, 4, 8, 1, 2, 5, 3, 3, 9, 2, 10, 0, 2, 6, 3, 1, 11, 7, 4, 2, 12, 1, 13, 4, 1, 8, 14, 1, 3, 2, 2, 5, 15, 3, 2, 3, 3, 9, 16, 2, 17, 10, 5, 0, 4, 2, 18, 6, 2, 3, 19, 1, 20, 11, 6, 7, 4, 4, 21, 2, 4, 12, 22, 1, 3, 13, 3, 4, 23, 1, 3, 8, 1, 14, 5, 1, 24, 3, 7, 2, 25

Offset

1,5

Comments

Consider the binary tree 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 > 1 encountered on the path (i.e., excluding the final 1 from the count but including the starting n if it was odd).

Links

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

Formula

a(1) = 0; after which, a(2n) = a(n), a(2n+1) = 1 + a(A250470(n)).

a(n) = A253555(n) - A253557(n).

a(n) = A253558(n) - 1.

a(n) = A080791(A252754(n)). [Number of nonleading 0-bits in A252754(n).]

Other identities. For all n >= 2:

a(n) = A000120(A252756(n)) - 1. [One less than the binary weight of A252756(n).]

Prog

(Scheme, with memoization-macro definec)
(definec (A253556 n) (cond ((= 1 n) 0) ((odd? n) (+ 1 (A253556 (A250470 n)))) (else (A253556 (/ n 2)))))

Crossrefs

One less than A253558.

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

Cf. A000120, A080791, A252753, A252754, A252755, A252756, A253554, A253555, A253557, A253559.

Differs from A252735 for the first time at n=21, where a(21) = 1, while A252735(21) = 3.

Sequence in context: A307744 A119709 A328167 * A252735 A120251 A333429

Adjacent sequences: A253553 A253554 A253555 * A253557 A253558 A253559

Keyword

nonn

Author

Antti Karttunen, Jan 12 2015

Status

approved