A180279 Triangle read by rows: AR(n,k) is the number of aperiodic k-reverses of n (n >= 1, 1 <= k <= n).

1, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 4, 6, 4, 0, 1, 4, 3, 8, 5, 0, 1, 6, 9, 12, 15, 6, 0, 1, 6, 9, 16, 15, 18, 7, 0, 1, 8, 9, 24, 30, 18, 28, 8, 0, 1, 8, 12, 32, 25, 48, 28, 32, 9, 0, 1, 10, 15, 40, 50, 60, 70, 40, 45, 10, 0, 1, 10, 12, 48, 50, 102, 70, 96, 36, 50, 11, 0

Offset: 1

Author

John P. McSorley, Aug 23 2010

Keywords

• nonn
• tabl

Comments

A k-composition of n is an ordered collection of k positive integers (parts) which sum to n.

A k-composition is aperiodic (primitive) if its period is k, or if it is not the concatenation of at least two smaller [identical] compositions.

Two k-compositions are cyclically equivalent if one can be obtained from the other by a cyclic permutation of its parts.

The reverse of a k-composition is the k-composition obtained by writing its parts in reverse. For example the reverse of 123 is 321.

A k-reverse of n is a k-composition of n which is cyclically equivalent to its reverse. And an aperiodic k-reverse of n is a k-reverse of n which is aperiodic.

For example, 114 is an aperiodic 3-reverse of 6 since it is aperiodic and its set of cyclic equivalents {114, 411, 141} contains its reverse 411.

But 123 is not an aperiodic 3-reverse of 6 since, even though it is aperiodic, its set of cyclic equivalents {123, 312, 231} does not contain its reverse 321.

Let AR(n,k) denote the number of aperiodic k-reverses of n.

This sequence is the 'AR(n,k)' triangle read by rows.

Examples

			Triangle AR(n,k) (with n >= 1 and 1 <= k <= n) begins as follows:
  1
  1 0
  1 2  0
  1 2  3  0
  1 4  6  4  0
  1 4  3  8  5  0
  1 6  9 12 15  6  0
  1 6  9 16 15 18  7  0
  1 8  9 24 30 18 28  8 0
  1 8 12 32 25 48 28 32 9 0
  ...
For example, row 8 is 1 6 9 16 15 18 7 0.
We have AR(8,3) = 9 because there are 9 aperiodic 3-reverses of 8.
These are in the classes {116, 611, 161}, {224, 422, 242}, and {233, 323, 332}.
We have AR(8,6) = 18 because all, except 3, of the 21 6-compositions of 8 are aperiodic 6-reverses of 8. The missing 3 form one class, {112112, 211211, 121121}, and they are each 6-reverses of 8, but they are each periodic of period 3; so, they are not aperiodic. [Edited by _Petros Hadjicostas_, Apr 27 2020]
		

References

• John P. McSorley: Counting k-compositions of n with palindromic and related structures. Preprint, 2010.

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1275

Crossrefs

If we ignore the aperiodic requirement, we get the sequence A180171.

Row sums are A180322.

Cf. A119963.

Programs

Mathematica

Table[k DivisorSum[GCD[n, k], MoebiusMu[#] Apply[Binomial[Floor[(#1 - Boole[OddQ@ #2])/2], Floor[#2/2]] &, {n/#, k/#}] &], {n, 12}, {k, n}] // Flatten (* Michael De Vlieger, Oct 11 2017 *)

PARI

\\ here p(n,k) is A119963.
p(n,k) = binomial((n-k%2)\2, k\2);
T(n, k) = k*sumdiv(gcd(n,k), d, moebius(d) * p(n/d, k/d));
for(n=1, 10, for(k=1, n, print1(T(n,k), ", ")); print) \\ Andrew Howroyd, Oct 08 2017

Formula

AR(n, k) = k * Sum_{d|gcd(n,k)} mu(d) * A119963(n/d, k/d). - Andrew Howroyd, Oct 08 2017 (Corrected by Petros Hadjicostas, Oct 11 2017.)

Extensions

Terms a(56) and beyond from Andrew Howroyd, Oct 08 2017

Name edited by Petros Hadjicostas, Apr 28 2020