cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A336363 Number of iterations of map k -> k*sigma(p^e)/p^e needed to reach a power of 2, where p is the largest 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.

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 1, 0, 3, 2, 2, 1, 2, 1, 4, 0, 4, 3, 3, 2, 2, 2, 2, 1, 2, 2, 3, 1, 5, 4, 1, 0, 4, 4, 3, 3, 4, 3, 3, 2, 3, 2, 3, 2, 4, 2, 2, 1, 6, 2, 4, 2, 4, 3, 5, 1, 5, 5, 5, 4, 2, 1, 4, 0, 4, 4, 5, 4, 4, 3, 4, 3, 5, 4, 3, 3, 3, 3, 3, 2, 6, 3, 3, 2, 5, 3, 5, 2, 5, 4, 7, 2, 2, 2, 3, 1, 7, 6, 4, 2, 5, 4, 3, 2, 5
Offset: 1

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Author

Antti Karttunen, Jul 30 2020

Keywords

Comments

Informally: starting from k=n, keep on replacing p^e, the maximal power of the largest 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 largest 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 -> 3*(5+1) = 18 = 2^1 * 3^2 -> 2 * (1+3+9) = 26 = 2 * 13 -> 2 * (13+1) = 28 = 2^2 * 7 -> 4*(7+1) = 2^5, thus it took four iterations to reach a power of two, and a(15) = 4.
		

Crossrefs

Programs

  • PARI
    A053585(n) = if(1==n, 1, my(f=factor(n)); f[#f~, 1]^f[#f~, 2]);
    A336363(n) = if(!bitand(n,n-1),0,my(pe=A053585(n)); 1+A336363((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(A053585(n))*(n/A053585(n))).
a(n) = a(2n) = a(A000265(n)).