

A180424


'ARE(n,k)' triangle read by rows. ARE(n,k) is the number of aperiodic kreverses of n up to cyclic equivalence.


2



1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 3, 3, 3, 3, 1, 0, 1, 3, 3, 4, 3, 3, 1, 0, 1, 4, 3, 6, 6, 3, 4, 1, 0, 1, 4, 4, 8, 5, 8, 4, 4, 1, 0, 1, 5, 5, 10, 10, 10, 10, 5, 5, 1, 0, 1, 5, 4, 12, 10, 17, 10, 12, 4, 5, 1, 0
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OFFSET

1,12


COMMENTS

A kcomposition of n is an ordered collection of k positive integers (parts) which sum to n.
A kcomposition is aperiodic (primitive) if its period is k, or if it is not the concatenation of at least two smaller compositions.
Two kcompositions are cyclically equivalent if one can be obtained from the other by a cyclic permutation of its parts.
The reverse of a kcomposition is the kcomposition obtained by writing its parts in reverse. For example the reverse of 123 is 321.
A kreverse of n is a kcomposition of n which is cyclically equivalent to its reverse. And an aperiodic kreverse of n is a kreverse of n which is aperiodic.
For example 114 is an aperiodic 3reverse 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 3reverse 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 kreverses of n, then sequence A180279 is the 'AR(n,k)' triangle read by rows.
For the above sequence we count the aperiodic kreverses of n up to cyclic equivalence, ARE(n,k), in other words the number of equivalence classes under cyclic permutation which contain at least one aperiodic kreverse of n.
The triangle begins
1
1 0
1 1 0
1 1 1 0
1 2 2 1 0
1 2 1 2 1 0
1 3 3 3 3 1 0
1 3 3 4 3 3 1 0
1 4 3 6 6 3 4 1 0
1 4 4 8 5 8 4 4 1 0
For example row 8 is 1 3 3 4 3 3 1 0.
We have ARE(8,3)=3 because there are 9 aperiodic 3reverses of 8.
In the 3 classes: {116,611,161}, {224,422,242}, and {233,323,332}, and so there are ARE(8,3)=3 aperiodic 3reverses of 8 up to cyclic equivalence.
We have ARE(8,6)=3 because there are 3 aperiodic 6reverses of 8 up to cyclic equivalence. The representatives of the 3 classes are: 111113, 111122, and, 111212.


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


FORMULA

ARE(n,k) = Sum_{dgcd(n,k)} mu(d) * A119963(n/d,k/d).  Andrew Howroyd, Oct 07 2017


PROG

(PARI) \\ here RE(n, k) is A119963(n, k).
RE(n, k) = binomial((nk%2)\2, k\2);
T(n, k) = sumdiv(gcd(n, k), d, moebius(d)*RE(n/d, k/d));
for(n=1, 10, for(k=1, n, print1(T(n, k), ", ")); print) \\ Andrew Howroyd, Oct 07 2017


CROSSREFS

As mentioned above if we don't count the classes, but rather the elements in the classes, we get sequence A180279.
If we remove the aperiodic requirement, see sequence A119963.
The row sums of the 'ARE(n, k)' triangle above give sequence A056493 (except for the first term). See also sequence A056498.
Sequence in context: A108423 A208568 A193804 * A092339 A079693 A117444
Adjacent sequences: A180421 A180422 A180423 * A180425 A180426 A180427


KEYWORD

nonn,tabl


AUTHOR

John P. McSorley, Sep 03 2010


EXTENSIONS

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


STATUS

approved



