A332894 a(1) = 0, a(2n) = 1 + a(n), a(2n+1) = 1 + a(A332819(2n+1)); also binary width of terms of A332816.

0, 1, 2, 2, 3, 3, 4, 3, 3, 4, 6, 4, 5, 5, 4, 4, 7, 4, 8, 5, 5, 7, 10, 5, 4, 6, 4, 6, 9, 5, 12, 5, 7, 8, 5, 5, 11, 9, 6, 6, 13, 6, 14, 8, 5, 11, 16, 6, 5, 5, 8, 7, 15, 5, 7, 7, 9, 10, 18, 6, 17, 13, 6, 6, 6, 8, 20, 9, 11, 6, 22, 6, 19, 12, 5, 10, 7, 7, 24, 7, 5, 14, 26, 7, 8, 15, 10, 9, 21, 6, 6, 12, 13, 17, 9, 7, 23, 6, 8, 6, 25, 9, 28, 8, 6

Offset: 1

Author

Antti Karttunen, Mar 04 2020

Keywords

• nonn

Comments

a(n) tells how many iterations of A332893 are needed before 1 is reached, i.e., the distance of n from 1 in binary trees like A332815.

Each n > 0 occurs 2^(n-1) times in total.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

A left inverse of A000079 and A108546.

Cf. A070939, A252464, A332808, A332815, A332816, A332819, A332893, A332897, A332898, A332899.

Programs

PARI

A332894(n) = if(1==n,0,if(!(n%2),1+A332894(n/2),1+A332894(A332819(n))));

PARI

A332894(n) = if(1==n,0,1+A332894(A332893(n)));

Formula

a(n) = A252464(A332808(n)).

a(1) = 0, and for n > 1, a(n) = 1 + a(A332893(n)).

For n >= 1, a(A108546(n)) = n; for all n >= 0, a(2^n) = n.

For n > 1: (Start)

a(n) = 1 + a(n/2) if n is even, and a(n) = 1 + a(A332819(n)), if n is odd.

a(n) = A070939(A332816(n)).

a(n) >= A332899(n).

(End)