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%I #22 Feb 19 2021 20:10:00
%S 15,105,1120,12600,151872,1897840,24408480,320355000,4271484000,
%T 57664963104,786341441760,10812193870800,149707312950720,
%U 2085208989609360,29192926025339776,410525522071875000,5795654431511374080,82105104444274758000,1166756747396368729440,16626283650369421872480
%N Number of orbits of length n under the full 15-shift (whose periodic points are counted by A001024).
%C Number of Lyndon words (aperiodic necklaces) with n beads of 15 colors. - _Andrew Howroyd_, Dec 10 2017
%H Robert Israel, <a href="/A060218/b060218.txt">Table of n, a(n) for n = 1..851</a>
%H Y. Puri and T. Ward, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL4/WARD/short.html">Arithmetic and growth of periodic orbits</a>, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
%H Yash Puri and Thomas Ward, <a href="http://www.fq.math.ca/Scanned/39-5/puri.pdf">A dynamical property unique to the Lucas sequence</a>, Fibonacci Quarterly, Volume 39, Number 5 (November 2001), pp. 398-402.
%H T. Ward, <a href="http://www.mth.uea.ac.uk/~h720/research/files/integersequences.html">Exactly realizable sequences</a>
%F a(n) = (1/n)* Sum_{d|n} mu(d)*A001024(n/d).
%F G.f.: Sum_{k>=1} mu(k)*log(1/(1 - 15*x^k))/k. - _Ilya Gutkovskiy_, May 19 2019
%e 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.
%p f:= n -> 1/n*add(numtheory:-mobius(d)*15^(n/d), d = numtheory:-divisors(n)):
%p map(f, [$1..30]); # _Robert Israel_, Oct 28 2018
%o (PARI) a001024(n) = 15^n;
%o a(n) = (1/n)*sumdiv(n, d, moebius(d)*a001024(n/d)); \\ _Michel Marcus_, Sep 11 2017
%Y Column 15 of A074650.
%Y Cf. A001024.
%K easy,nonn
%O 1,1
%A _Thomas Ward_, Mar 21 2001
%E More terms from _Michel Marcus_, Sep 11 2017
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