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A328384
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If n is of the form p^p, a(n) = 0, otherwise a(n) gives the number of iterations of x -> A003415(x) needed to reach the first number different from n which is either a prime, or whose degree (A051903) differs from the degree of n. If the degree of the final number is <= that of n, then a(n) = -1 * iteration count.
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4
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-1, -1, -1, 0, -1, -1, -1, -1, -1, -1, -1, 1, -1, 1, 1, 1, -1, -1, -1, 1, -2, -1, -1, -1, -1, 2, 0, 1, -1, -1, -1, -1, 2, -1, 1, 3, -1, -3, 1, -1, -1, -1, -1, 1, -1, 1, -1, 3, -1, -2, 1, 1, -1, 1, 1, -1, -2, -1, -1, 2, -1, 3, -1, 2, 1, -1, -1, 1, 3, -1, -1, -1, -1, 2, -1, 1, 1, -1, -1, 5, -1, -1, -1, 2, -2, 1, 1, -1, -1, -1, 1, 1, -2, 1
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OFFSET
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1,21
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COMMENTS
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The records -1, 0, 1, 2, 3, 5, 8, 10, 11, 13, ... occur at n = 1, 4, 12, 26, 36, 80, 108, 4887, 18688, 22384, ...
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LINKS
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FORMULA
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a(1) = -1 as 0 is here considered having a smaller degree than 1.
a(p) = -1 for all primes.
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EXAMPLE
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For n = 21 = 3*7, A051903(21) = 1, A003415(21) = 10 = 2*5, is of the same degree as A051903(10) = 1, but then A003415(10) = 7, which is a prime, having degree <= of the starting value (as we have A051903(7) <= A051903(21)), thus a(21) = -1 * 2 = -2.
For n = 33 = 3*11, A051903(33) = 1, A003415(33) = 14 = 2*7, is of the same degree, but on the second iteration, A003415(14) = 9 = 3^2, with A051903(9) = 2, which is larger than the initial degree, thus a(33) = +2.
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PROG
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(PARI)
A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A051903(n) = if((n<=1), n-1, vecmax(factor(n)[, 2]));
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CROSSREFS
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Cf. A328385 (the number found in the iteration).
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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