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A336362
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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.
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5
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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
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OFFSET
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1,5
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COMMENTS
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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.
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LINKS
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FORMULA
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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))).
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EXAMPLE
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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.
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PROG
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(PARI)
A336650(n) = if(!bitand(n, n-1), 1, my(f=factor(n>>valuation(n, 2))); f[1, 1]^f[1, 2]);
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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