A179181 - OEIS

A179181

'PE(n,k)' triangle read by rows. PE(n,k) is the number of k-palindromes of n up to cyclic equivalence.

1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 1, 1, 2, 1, 1, 1, 1, 0, 3, 0, 3, 0, 1, 1, 1, 3, 2, 3, 2, 1, 1, 1, 0, 4, 0, 6, 0, 4, 0, 1, 1, 1, 4, 2, 6, 4, 4, 2, 1, 1, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 1, 1, 1, 5, 3, 10, 6, 10, 5, 5, 3, 1, 1

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OFFSET

1,13

COMMENTS

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

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

A k-palindrome of n is a k-composition of n which is a palindrome.

Let PE(n,k) denote the number of k-palindromes of n up to cyclic equivalence.

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

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..1275

FORMULA

PE(n, k) = Sum_{d | gcd(n,k)} A179317(n/d, k/d). - Andrew Howroyd, Oct 07 2017

EXAMPLE

The triangle begins

1,1

1,0,1

1,1,1,1

1,0,2,0,1

1,1,2,1,1,1

1,0,3,0,3,0,1

1,1,3,2,3,2,1,1

1,0,4,0,6,0,4,0,1

1,1,4,2,6,4,4,2,1,1

For example, row 8 is 1,1,3,2,3,2,1,1.

We have PE(8,3)=3 because there are 3 3-palindromes of 8, namely: 161, 242, and 323, and none are cyclically equivalent to the others.

We have PE(8,4)=2 because there are 3 4-palindromes of 8, namely: 3113, 1331, and 2222, but 3113 and 1331 are cyclically equivalent.

MATHEMATICA

T[n_, k_] := (3 - (-1)^k)/4*Sum[MoebiusMu[d]*QBinomial[n/d - 1, k/d - 1, -1], {d, Divisors@ GCD[n, k]}]; Table[DivisorSum[GCD[n, k], T[n/#, k/#] &], {n, 12}, {k, n}] // Flatten (* Michael De Vlieger, Oct 31 2021, after Jean-François Alcover at A179317 *)

PROG

(PARI)

p(n, k) = if(n%2==1&&k%2==0, 0, binomial((n-1)\2, (k-1)\2));

APE(n, k) = if(k%2, 1, 1/2) * sumdiv(gcd(n, k), d, moebius(d) * p(n/d, k/d));

T(n, k) = sumdiv(gcd(n, k), d, APE(n/d, k/d));

for(n=1, 10, for(k=1, n, print1(T(n, k), ", ")); print) \\ Andrew Howroyd, Oct 07 2017

CROSSREFS

If we ignore cyclic equivalence then we have sequence A051159.

The row sums of the 'PE(n, k)' triangle give sequence A056503.

Cf. A179317, A179519.

Sequence in context: A258832 A121444 A118230 * A153246 A358006 A025889

Adjacent sequences: A179178 A179179 A179180 * A179182 A179183 A179184