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A060218
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Number of orbits of length n under the full 15-shift (whose periodic points are counted by A001024).
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2
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15, 105, 1120, 12600, 151872, 1897840, 24408480, 320355000, 4271484000, 57664963104, 786341441760, 10812193870800, 149707312950720, 2085208989609360, 29192926025339776, 410525522071875000, 5795654431511374080, 82105104444274758000, 1166756747396368729440, 16626283650369421872480
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OFFSET
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1,1
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COMMENTS
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Number of Lyndon words (aperiodic necklaces) with n beads of 15 colors. - Andrew Howroyd, Dec 10 2017
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LINKS
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FORMULA
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a(n) = (1/n)* Sum_{d|n} mu(d)*A001024(n/d).
G.f.: Sum_{k>=1} mu(k)*log(1/(1 - 15*x^k))/k. - Ilya Gutkovskiy, May 19 2019
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EXAMPLE
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a(2)=105 since there are 225 points of period 2 in the full 15-shift and 15 fixed points, so there must be (225-15)/2 = 105 orbits of length 2.
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MAPLE
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f:= n -> 1/n*add(numtheory:-mobius(d)*15^(n/d), d = numtheory:-divisors(n)):
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PROG
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(PARI) a001024(n) = 15^n;
a(n) = (1/n)*sumdiv(n, d, moebius(d)*a001024(n/d)); \\ Michel Marcus, Sep 11 2017
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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