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A180279
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Triangle read by rows: AR(n,k) is the number of aperiodic k-reverses of n (n >= 1, 1 <= k <= n).
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6
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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
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OFFSET
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1,5
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COMMENTS
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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.
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REFERENCES
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John P. McSorley: Counting k-compositions of n with palindromic and related structures. Preprint, 2010.
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LINKS
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FORMULA
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EXAMPLE
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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]
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MATHEMATICA
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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 *)
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PROG
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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
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CROSSREFS
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If we ignore the aperiodic requirement, we get the sequence A180171.
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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