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

Author

Antti Karttunen, Jan 22 2015

Keywords

• nonn

Comments

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

Links

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

Crossrefs

One less than A253894.

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

Formula

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.