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%I #16 Jul 06 2018 17:10:46
%S 1,1,3,6,15,21,49,72,126,195,341,486,819,1225,1845,2880,4335,6552,
%T 9709,14850,21315,33077,47081,72360,102300,158067,220752,341334,
%U 475107,732735,1015777,1566720,2161599,3333615,4587135,7062552,9699291,14922733,20444697
%N a(n) = AR(n) is the total number of aperiodic k-reverses of n.
%C The n-th term of this sequence a(n) = AR(n) gives the total number of aperiodic k-reverses of n for k=1,2,...,n. It is the sum of the n-th row of the 'AR(n,k)' triangle from sequence A180279.
%C See sequence A180279 for the definition of an aperiodic k-reverse of n.
%C Briefly, a k-reverse of n is a k-composition of n whose reverse is cyclically equivalent to itself, and an aperiodic k-reverse of n is a k-reverse of n which is also aperiodic.
%C For example a(6)=21 because there are 21 aperiodic k-reverses of n=6 for k=1,2,3,4,5, or 6.
%C They are, in cyclically equivalent classes: {6}, {15,51}, {24,42}, {114,411,141}, {1113,3111,1311,1131}, {1122,2112,2211,1221},{11112,21111,12111,11211,11121}.
%D John P. McSorley: Counting k-compositions with palindromic and related structures. Preprint, 2010.
%H Andrew Howroyd, <a href="/A180322/b180322.txt">Table of n, a(n) for n = 1..200</a>
%F a(n) = n * A056493(n) / 2. - _Andrew Howroyd_, Oct 07 2017
%t a[n_] := n*Sum[MoebiusMu[n/d]*If[OddQ[d], 2, 3]*2^Quotient[d-1, 2], {d, Divisors[n]}]/2;
%t Array[a, 40] (* _Jean-François Alcover_, Jul 06 2018, after _Andrew Howroyd_ *)
%o (PARI)
%o a(n) = n * sumdiv(n, d, moebius(n/d) * if(d%2,2,3) * 2^((d-1)\2)) / 2; \\ _Andrew Howroyd_, Oct 07 2017
%Y If we ask for the number of cyclically equivalent classes we get sequence A056493 (except for the first term). For example, the 6th term of A056493 is 7, corresponding to the 7 classes in the example above.
%Y Row sums of A180279.
%K nonn
%O 1,3
%A _John P. McSorley_, Aug 27 2010
%E Terms a(11) and beyond from _Andrew Howroyd_, Oct 07 2017
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