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%I #14 Jun 09 2021 02:30:56
%S 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,
%T 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,
%U 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
%N 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.
%C 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.
%H Antti Karttunen, <a href="/A336362/b336362.txt">Table of n, a(n) for n = 1..65537</a>
%F 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))).
%F a(n) = a(2n) = a(A000265(n)).
%e 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.
%o (PARI)
%o A336650(n) = if(!bitand(n,n-1),1,my(f=factor(n>>valuation(n,2))); f[1, 1]^f[1, 2]);
%o A336362(n) = if(!bitand(n,n-1),0,my(pe=A336650(n)); 1+A336362((n/pe)*sigma(pe)));
%Y Cf. A000265, A209229, A336650.
%Y Cf. also A331410, A336361, A336363.
%K nonn
%O 1,5
%A _Antti Karttunen_, Jul 30 2020
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