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%I #13 Jun 09 2021 02:31:03
%S 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,
%T 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,
%U 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
%N 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.
%C 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.
%H Antti Karttunen, <a href="/A336363/b336363.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(A053585(n))*(n/A053585(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 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.
%o (PARI)
%o A053585(n) = if(1==n, 1, my(f=factor(n)); f[#f~, 1]^f[#f~, 2]);
%o A336363(n) = if(!bitand(n,n-1),0,my(pe=A053585(n)); 1+A336363((n/pe)*sigma(pe)));
%Y Cf. A000203, A000265, A000593, A209229, A053585.
%Y Cf. also A331410, A336361, A336362.
%K nonn
%O 1,5
%A _Antti Karttunen_, Jul 30 2020
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