

A179317


'APE(n,k)' triangle read by rows. APE(n,k) is the number of aperiodic kpalindromes 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
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OFFSET

1,13


COMMENTS

A kcomposition of n is an ordered collection of k positive integers (parts) which sum to n.
Two kcompositions of n are cyclically equivalent if one can be obtained from the other by a cyclic permutation of its parts.
A kcomposition is aperiodic (primitive) if its period is k, or if it is not the concatenation of a smaller composition.
A kpalindrome of n is a kcomposition of n which is a palindrome.
Let APE(n,k) denote the number of aperiodic kpalindromes 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 kcompositions 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 3palindromes 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 4palindromes 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 (* JeanFrançois Alcover, Sep 24 2019 *)


PROG

(PARI) \\ here p(n, k)=A051159(n1, k1) is number of kpalindromes of n.
p(n, k) = if(n%2==1&&k%2==0, 0, binomial((n1)\2, (k1)\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



