A253887 -id:A253887 - cp's OEIS frontend

A253893 a(1) = 0, for n > 1, a(n) = 1 + a(A253889(n)).

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

Offset: 1

Antti Karttunen, Jan 22 2015

nonn

When A048673 is represented as a binary tree, then a(n) gives the distance of node containing n from 1 at top.

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

One less than A253894.

Cf. A000523, A007051, A064216, A253786, A253887, A253889.

a(1) = 0, for n > 1, a(n) = 1 + A253893(A253889(n)).
a(n) = A000523(A064216(n)).
a(n) = A253894(n) - 1.
Other identities:
a(A007051(n)) = n for all n >= 0.

A254048 a(n) = A126760(A007494(n)).

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 4, 1, 3, 2, 6, 1, 2, 3, 8, 1, 5, 1, 10, 2, 1, 4, 12, 1, 7, 5, 14, 3, 4, 2, 16, 1, 9, 6, 18, 1, 3, 7, 20, 2, 11, 3, 22, 4, 6, 8, 24, 1, 13, 9, 26, 5, 2, 1, 28, 3, 15, 10, 30, 2, 8, 11, 32, 1, 17, 4, 34, 6, 5, 12, 36, 1, 19, 13, 38, 7, 10, 5, 40, 2, 21, 14, 42, 3, 1, 15, 44, 4, 23, 2, 46, 8, 12, 16, 48, 1, 25, 17, 50, 9, 7, 6, 52, 5, 27, 18, 54, 1, 14, 19, 56, 3, 29, 7, 58, 10, 4, 20, 60, 2

Offset: 0

Antti Karttunen, Jan 28 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..19683

Cf. A006370, A007494, A126760, A139391, A253887, A254102.

(define (A254048 n) (A126760 (A007494 n)))

a(n) = A126760(A007494(n)).
Other identities:
a(4n) = A126760(n).
a(4n+1) = A126760(3n+1).
a(4n+2) = A126760(2n+1) = A253887(n+1).
a(4n+3) = 2n+2.
For all n >= 1, a(n) = A126760(A139391(n)). [Conjecture. The proof should be easy. Holds at least up to n = 2^25 = 33554432.]

cp's OEIS Frontend

A253893 a(1) = 0, for n > 1, a(n) = 1 + a(A253889(n)).

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