A336362 Number of iterations of map k -> k*sigma(p^e)/p^e needed to reach a power of 2, where p is the smallest odd prime factor of k and e is its exponent, when starting from k = n. a(n) = -1 if number of the form 2^k is never reached.

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

Offset: 1

Author

Antti Karttunen, Jul 30 2020

Keywords

• nonn

Comments

Informally: starting from k=n, keep on replacing p^e, the maximal power of the smallest odd prime factor in k, with (1 + p + p^2 + ... + p^e), until a power of 2 is reached. Sequence counts the steps needed.

Examples

			For n = 15 = 3*5, we obtain the following path, when starting from k = n, and when we always replace the maximal power of the lowest odd prime factor, p^e of k with sigma(p^e) = (1 + p + p^2 + ... + p^e) in the prime factorization k: 3^1 * 5^1 -> (1+3)*5 = 20 = 2^2 * 5 -> 4 * (1+5) = 24 = 2^3 * 3^1 -> 2^3 * 2^2 = 2^5, thus it took three iterations to reach a power of two, and a(15) = 3.
		

Links

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

Crossrefs

Cf. A000265, A209229, A336650.

Cf. also A331410, A336361, A336363.

Programs

PARI

A336650(n) = if(!bitand(n,n-1),1,my(f=factor(n>>valuation(n,2))); f[1, 1]^f[1, 2]);
A336362(n) = if(!bitand(n,n-1),0,my(pe=A336650(n)); 1+A336362((n/pe)*sigma(pe)));

Formula

If A209229(n) = 1 [when n is a power of 2], a(n) = 0, otherwise a(n) = 1 + a(sigma(A336650(n))*(n/A336650(n))).

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