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A179317 'APE(n,k)' triangle read by rows. APE(n,k) is the number of aperiodic k-palindromes of n up to cyclic equivalence. 4
1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 2, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 3, 0, 3, 0, 0, 1, 0, 3, 1, 3, 1, 1, 0, 1, 0, 3, 0, 6, 0, 4, 0, 0, 1, 0, 4, 2, 5, 2, 4, 2, 1, 0, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 0, 1, 0, 4, 2, 10, 4, 10, 4, 4, 2, 1, 0 (list; table; graph; refs; listen; history; text; internal format)
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-composition is aperiodic (primitive) if its period is k, or if it is not the concatenation of a smaller composition.

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

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

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

The only possibility for two distinct aperiodic palindromes to be cyclically equivalent is with an even number of terms and with a rotation by half the number of terms. For example, 123321 is cyclically equivalent to 321123. - Andrew Howroyd, Oct 07 2017

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 (Rows n=1..50 of triangle, flattened)

FORMULA

APE(n,k) = (3-(-1)^k)/4 * A179519(n,k). - Andrew Howroyd, Oct 07 2017

EXAMPLE

The triangle begins

1

1,0

1,0,0

1,0,1,0

1,0,2,0,0

1,0,1,1,1,0

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

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

1,0,3,0,6,0,4,0,0

1,0,4,2,5,2,4,2,1,0

For example, row 8 is 1,0,3,1,3,1,1,0.

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

We have APE(8,4)=1 because there are 2 aperiodic 4-palindromes of 8, namely: 3113 and 1331, but they 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[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-Fran├žois Alcover, Sep 24 2019 *)

PROG

(PARI) \\ here p(n, k)=A051159(n-1, k-1) is number of k-palindromes of n.

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

T(n, k) = if(k%2, 1, 1/2) * 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 07 2017

CROSSREFS

The row sums of the 'APE(n, k)' triangle give sequence A056513.

If cyclic equivalence is ignored, we get sequence A179519. - John P. McSorley, Jul 26 2010

Sequence in context: A325122 A234578 A069859 * A307598 A324967 A320332

Adjacent sequences:  A179314 A179315 A179316 * A179318 A179319 A179320

KEYWORD

nonn,tabl

AUTHOR

John P. McSorley, Jul 10 2010

EXTENSIONS

Terms a(56) and beyond from Andrew Howroyd, Oct 07 2017

STATUS

approved

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Last modified October 14 01:36 EDT 2019. Contains 327994 sequences. (Running on oeis4.)