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A180322
a(n) = AR(n) is the total number of aperiodic k-reverses of n.
4
1, 1, 3, 6, 15, 21, 49, 72, 126, 195, 341, 486, 819, 1225, 1845, 2880, 4335, 6552, 9709, 14850, 21315, 33077, 47081, 72360, 102300, 158067, 220752, 341334, 475107, 732735, 1015777, 1566720, 2161599, 3333615, 4587135, 7062552, 9699291, 14922733, 20444697
OFFSET
1,3
COMMENTS
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.
See sequence A180279 for the definition of an aperiodic k-reverse of n.
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.
For example a(6)=21 because there are 21 aperiodic k-reverses of n=6 for k=1,2,3,4,5, or 6.
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}.
REFERENCES
John P. McSorley: Counting k-compositions with palindromic and related structures. Preprint, 2010.
LINKS
FORMULA
a(n) = n * A056493(n) / 2. - Andrew Howroyd, Oct 07 2017
MATHEMATICA
a[n_] := n*Sum[MoebiusMu[n/d]*If[OddQ[d], 2, 3]*2^Quotient[d-1, 2], {d, Divisors[n]}]/2;
Array[a, 40] (* Jean-François Alcover, Jul 06 2018, after Andrew Howroyd *)
PROG
(PARI)
a(n) = n * sumdiv(n, d, moebius(n/d) * if(d%2, 2, 3) * 2^((d-1)\2)) / 2; \\ Andrew Howroyd, Oct 07 2017
CROSSREFS
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.
Row sums of A180279.
Sequence in context: A093799 A087359 A253651 * A244164 A129602 A044888
KEYWORD
nonn
AUTHOR
John P. McSorley, Aug 27 2010
EXTENSIONS
Terms a(11) and beyond from Andrew Howroyd, Oct 07 2017
STATUS
approved