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Search: id:A067631
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| A067631 |
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If n is composite then a(n) is the standard deviation of the prime factors of n, rounded off to the nearest integer (rounding up if there's a choice), with each factor counted according to its frequency of occurrence in the prime factorization. If n is 1 or prime then a(n)=0. |
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+0
2
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| 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 1, 0, 4, 1, 0, 0, 1, 0, 2, 3, 6, 0, 1, 0, 8, 0, 3, 0, 2, 0, 0, 6, 11, 1, 1, 0, 12, 7, 2, 0, 3, 0, 5, 1, 15, 0, 0, 0, 2, 10, 6, 0, 1, 4, 3, 11, 19, 0, 1, 0, 21, 2, 0, 6, 5, 0, 9, 14, 3, 0, 1, 0, 25, 1, 10, 3, 6, 0, 1, 0, 28, 0, 2, 8, 29, 18, 5, 0, 1, 4, 12, 20, 32, 10, 0, 0
(list; graph; listen)
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OFFSET
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2,9
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COMMENT
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The (sample) standard deviation sigma of {x_1,...,x_n} is calculated from sigma^2 = 1/(n-1) * sum_{1,...,n}(x_i - mu)^2, where mu denotes the average of {x_1,...,x_n}.
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EXAMPLE
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24 = 2^3 * 3^1, so the corresponding average = (2 + 2 + 2 + 3)/ 4 = 2.25 and the standard deviation is [(1/3){3 * (2-2.25)^2 + (3-2.25)^2}]^0.5 = 0.5, which rounds to 1. So a(24) = 1.
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MATHEMATICA
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<<Statistics`NormalDistribution` f[n_] := Flatten[Table[ #[[1]], {#[[2]]}]&/@FactorInteger[n]]; a[n_] := If[PrimeQ[n]||n==1, 0, Floor[StandardDeviation[f[n]]+1/2]]
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CROSSREFS
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Sequence in context: A131321 A111959 A110109 this_sequence A123641 A134317 A132277
Adjacent sequences: A067628 A067629 A067630 this_sequence A067632 A067633 A067634
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KEYWORD
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easy,nonn
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AUTHOR
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Joseph L. Pe (joseph_l_pe(AT)hotmail.com), Feb 02 2002
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EXTENSIONS
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Edited and extended by Robert G. Wilson v (rgwv(AT)rgwv.com), Feb 05 2002 and edited by Dean Hickerson (dean(AT)math.ucdavis.edu), Feb 12 2002
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