A256993 a(1) = 0; for n > 1, a(n) = 1 + a(A256992(n)).

0, 1, 2, 3, 2, 3, 4, 3, 4, 4, 5, 3, 4, 5, 4, 5, 4, 5, 6, 5, 5, 4, 5, 6, 6, 5, 4, 5, 6, 5, 6, 5, 6, 6, 7, 5, 6, 6, 6, 7, 5, 6, 6, 6, 5, 7, 7, 6, 6, 5, 7, 7, 6, 7, 6, 6, 7, 5, 6, 7, 6, 7, 6, 7, 6, 7, 8, 7, 7, 6, 7, 8, 7, 7, 6, 7, 7, 8, 6, 7, 7, 7, 8, 6, 7, 6, 7, 8, 8, 7, 7, 6, 8, 7, 7, 8, 6, 8, 7, 7, 8, 7, 6, 8, 7, 8, 8, 7, 7, 8, 8, 6, 7, 7, 7, 8, 7, 8, 8, 7, 6, 7, 8, 7, 8, 7, 8, 7

Offset

1,3

Comments

Number of iterations of A256992 needed to reach one when starting from n.

Links

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

Formula

a(1) = 0; for n > 1, a(n) = 1 + a(A256992(n)).

Other observations. For all n >= 1 it holds that:

a(n) >= A254110(n).

a(n) >= A256989(n).

a(n) >= A255559(n)-1.

Also it seems that a(n) - A070939(n) = -1, 0 or +1 for all n >= 1. [Compare A256991 and A256992 to see the connection.]

It is also very likely that a(n) <= A071542(n) for all n.

From Antti Karttunen, Dec 10 2016: (Start)

For all n >= 2, a(n) = A070939(A279341(n)) = A070939(A279343(n)).

For all n >= 2, a(n) = A279345(n) + A279346(n) - 1.

(End)

Prog

(Scheme, with memoization macro definec)
(definec (A256993 n) (if (= 1 n) 0 (+ 1 (A256993 (A256992 n)))))

Crossrefs

Cf. A070939, A071542, A254110, A255559, A256991, A256992, A257264, A257265, A279341, A279343, A279345, A279346.

Sequence in context: A069464 A156723 A240834 * A348192 A173523 A199323

Adjacent sequences: A256990 A256991 A256992 * A256994 A256995 A256996

Keyword

nonn

Author

Antti Karttunen, Apr 15 2015

Status

approved