A252736 a(1) = a(2) = 0; for n > 2: a(2n) = 1 + a(n), a(2n+1) = a(A064989(2n+1)).

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

Offset: 1

Author

Antti Karttunen, Dec 21 2014

Keywords

• nonn

Comments

Consider the binary tree illustrated in A005940: If we start from any n, computing successive iterations of A252463 until 1 is reached (i.e., we are traversing level by level towards the root of the tree, starting from that vertex of the tree where n is located), a(n) gives the number of even numbers > 2 encountered on the path (i.e., excluding the 2 from the count but including the starting n if it was even).

The number of pairs in any factorization tree of n. For example, a possible factorization tree of 12 is 12 -> (4*3) -> (2*2)*3. There are 2 pairs in this factor tree: (4*3) and (2*2). Thus, a(12) - 1 = 3 - 1 = 2. - Melvin Peralta, Aug 29 2016

Links

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

Crossrefs

Essentially one less than A001222.

Cf. A000120, A005940, A064989, A080791, A156552, A243071, A252464, A252735.

Cf. also A246370.

Programs

Mathematica

a[1] = a[2] = 0; a[n_] := a[n] = If[EvenQ@ n, 1 + a[n/2], a[Times @@ Power[Which[# == 1, 1, # == 2, 1, True, NextPrime[#, -1]] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ n]]; Array[a, 120] (* Michael De Vlieger, Aug 30 2016 *)

Formula

a(1) = a(2) = 0; for n > 2: a(2n) = 1 + a(n), a(2n+1) = a(A064989(2n+1)).

a(n) = A080791(A243071(n)). [Number of nonleading 0-bits in A243071(n).]

Other identities. For all n >= 2:

a(n) = A000120(A156552(n)) - 1. [One less than the binary weight of A156552(n).]

a(n) = A252464(n) - A252735(n) - 1.

a(n) = A001222(n) - 1.