A336361 Number of iterations of A000593 (sum of divisors of odd part of n) needed to reach a power of 2, or -1 if never reached.

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

Offset: 1

Author

Antti Karttunen, Jul 30 2020

Keywords

• nonn

Comments

Also, for n > 1, one less than the number of iterations of A000593 to reach 1.

If there exists any hypothetical odd perfect numbers w, then the iteration will get stuck into a fixed point after encountering them, and we will have a(w) = a(2^k * w) = -1 by the escape clause.

Links

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

Crossrefs

Cf. A000265, A000593, A161942, A209229, A336362, A336363, A347249, A347250.

Cf. A054784 (positions of 0's and 1's in this sequence).

Cf. also A347240, A347241, A347242, A347243, A347244, A347245.

Programs

PARI

A336361(n) = if(!bitand(n,n-1),0,1+A336361(sigma(n>>valuation(n,2))));

Formula

If A209229(n) = 1 [when n is a power of 2], a(n) = 0, otherwise a(n) = 1+a(A000593(n)).

a(n) = a(2n) = a(A000265(n)).