A253555 a(1) = 0, a(2n) = 1 + a(n), a(2n+1) = 1 + a(A250470(2n+1)); also binary width of terms of A252754 and A252756.

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

Offset

1,3

Comments

a(n) tells how many iterations of A253554 are needed before 1 is reached, i.e., the distance of n from 1 in binary trees like A252753 and A252755.

Links

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

Formula

a(1) = 0; for n > 1: a(n) = 1 + a(A253554(n)).

a(n) = A029837(1+A252754(n)) = A029837(1+A252756(n)).

a(n) = A253556(n) + A253557(n).

Other identities.

For all n >= 1:

a(A000079(n)) = n. [I.e., a(2^n) = n.]

a(A000040(n)) = n.

a(A001248(n)) = n+1.

For n >= 2, a(n) = A253558(n) + A253559(n).

Prog

(Scheme, three versions, first one using memoization macro definec)
(definec (A253555 n) (if (<= n 1) 0 (+ 1 (A253555 (A253554 n)))))
(define (A253555 n) (A029837 (+ 1 (A252754 n))))
(define (A253555 n) (A029837 (+ 1 (A252756 n))))

Crossrefs

Cf. A000040, A000079, A001248, A253554.

Cf. also A252753, A252754, A252755, A252756, A253557, A253558, A253559.

Differs from A252464 for the first time at n=21, where a(21) = 4, while A252463(21) = 5.

Sequence in context: A230697 A322163 A075167 * A252464 A324861 A324863

Adjacent sequences: A253552 A253553 A253554 * A253556 A253557 A253558

Keyword

nonn

Author

Antti Karttunen, Jan 12 2015

Status

approved