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%I #18 Oct 30 2021 07:17:26
%S 0,0,0,0,1,0,0,1,1,0,0,2,2,1,0,0,2,1,2,1,0,0,3,3,3,3,1,0,0,3,3,4,3,3,
%T 1,0,0,4,3,6,6,3,4,1,0,0,4,4,8,5,8,4,4,1,0,0,5,5,10,10,10,10,5,5,1,0,
%U 0,5,4,12,10,17,10,12,4,5,1,0,0,6,6,15,15,20,20,15,15,6,6,1,0
%N 'ADPE(n,k)' triangle read by rows. ADPE(n,k) is the number of aperiodic k-double-palindromes of n up to cyclic equivalence.
%C 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, i.e., if it is not the concatenation of at least two smaller compositions. A palindrome is a word which is the same when written backwards.
%C A k-double-palindrome of n is a k-composition of n which is the concatenation of two palindromes, PP'=P|P', where both |P|, |P'|>=1.
%C For example 1123532=11|23532 is a 7-double-palindrome of 17 since both 11 and 23532 are palindromes. It is also aperiodic, and so it is an aperiodic 7-double-palindrome of 17. See sequence A181111.
%C Two k-compositions of n are cyclically equivalent if one can be obtained from the other by a cyclic permutation of its parts.
%C Let ADPE(n,k) denote the number of aperiodic k-double-palindromes of n up to cyclic equivalence, i.e., the number of cyclic equivalence classes containing at least one aperiodic k-double-palindrome.
%C This sequence is the `ADPE(n,k)' triangle read by rows.
%D John P. McSorley: Counting k-compositions of n with palindromic and related structures. Preprint, 2010.
%H Andrew Howroyd, <a href="/A181169/b181169.txt">Table of n, a(n) for n = 1..1275</a>
%F T(n, 1) = 0; T(n, k) = A180424(n, k) for k > 1. - _Andrew Howroyd_, Sep 28 2019
%e The triangle begins:
%e 0
%e 0 0
%e 0 1 0
%e 0 1 1 0
%e 0 2 2 1 0
%e 0 2 1 2 1 0
%e 0 3 3 3 3 1 0
%e 0 3 3 4 3 3 1 0
%e 0 4 3 6 6 3 4 1 0
%e 0 4 4 8 5 8 4 4 1 0
%e ...
%e For example, row 8 is: 0 3 3 4 3 3 1 0.
%e We have ADPE(8,3)=3 because the 6 aperiodic 3-double-palindromes of 8: 116, 611, 224, 422, 233, and 332 come in 3 cyclic equivalence classes: {116, 611, 161}, {224, 422, 242}, and {233, 323, 332}.
%e We have ADPE(8,4)=4 because there are 4 4-double-palindromes of 8 up to cyclic equivalence, the 4 classes are: {1115, 5111, 1511, 1151}, {1214, 4121, 1412, 2141}, {1133, 3113, 3311, 1331}, and {1232, 2123, 3212, 2321}.
%o (PARI) \\ here RE(n, k) is A119963(n, k).
%o RE(n, k) = binomial((n-k%2)\2, k\2);
%o T(n, k) = if(k<=1, 0, sumdiv(gcd(n, k), d, moebius(d)*RE(n/d, k/d))); \\ _Andrew Howroyd_, Sep 28 2019
%Y Row sums are A181314.
%Y If we remove the aperiodic requirement we get sequence A180918.
%Y If we count the aperiodic k-double-palindromes of n (not the number of classes) we get sequence A181111 which is the 'ADP(n, k)' triangle read by rows, where ADP(n, k) is the number of aperiodic k-double-palindromes of n.
%Y Cf. A180424.
%K nonn,tabl
%O 1,12
%A _John P. McSorley_, Oct 07 2010
%E Terms a(56) and beyond from _Andrew Howroyd_, Sep 27 2019
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