A180322 a(n) = AR(n) is the total number of aperiodic k-reverses of n.

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

Author

John P. McSorley, Aug 27 2010

Keywords

• nonn

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

• Andrew Howroyd, Table of n, a(n) for n = 1..200

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.

Programs

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 *)

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

Formula

a(n) = n * A056493(n) / 2. - Andrew Howroyd, Oct 07 2017

Extensions

Terms a(11) and beyond from Andrew Howroyd, Oct 07 2017