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

1,5

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.

Links

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

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)).

Example

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.

Prog

(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)));

Crossrefs

Cf. A000265, A209229, A336650.

Cf. also A331410, A336361, A336363.

Sequence in context: A336361 A364260 A334204 * A336363 A308451 A308067

Adjacent sequences: A336359 A336360 A336361 * A336363 A336364 A336365

Keyword

nonn

Author

Antti Karttunen, Jul 30 2020

Status

approved