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%I #18 Oct 30 2021 07:17:44
%S 0,0,1,0,1,1,0,2,1,1,0,2,2,1,1,0,3,2,3,1,1,0,3,3,3,3,1,1,0,4,3,6,3,4,
%T 1,1,0,4,4,6,6,4,4,1,1,0,5,4,10,6,10,4,5,1,1,0,5,5,10,10,10,10,5,5,1,
%U 1,0,6,5,15,10,20,10,15,5,6,1,1,0,6,6,15,15,20,20,15,15,6,6,1,1
%N 'DPE(n,k)' triangle read by rows. DPE(n,k) is the number of 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. Two k-compositions of n are cyclically equivalent if one can be obtained from the other by a cyclic permutation of its parts.
%C 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 See sequence A180653. For example 1123532=11|23532 is a 7-double-palindrome of 17 since both 11 and 23532 are palindromes.
%C Let DPE(n,k) denote the number of k-double-palindromes of n up to cyclic equivalence.
%C This sequence is the 'DPE(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="/A180918/b180918.txt">Table of n, a(n) for n = 1..1275</a>
%F T(n, 1) = 0; T(n, k) = A119963(n,k) for k > 1.
%e The triangle begins:
%e 0
%e 0 1
%e 0 1 1
%e 0 2 1 1
%e 0 2 2 1 1
%e 0 3 2 3 1 1
%e 0 3 3 3 3 1 1
%e 0 4 3 6 3 4 1 1
%e 0 4 4 6 6 4 4 1 1
%e 0 5 4 10 6 10 4 5 1 1
%e ...
%e For example, row 8 is: 0 4 3 6 3 4 1 1.
%e We have DPE(8,3)=3 because there are 3 3-double-palindromes of 8 up to cyclic equivalence: {116, 611}, {224, 422}, and {233, 332}.
%e We have DPE(8,4)=6 because there are 6 4-double-palindromes of 8: up to cyclic equivalence: {1115, 5111, 1511, 1151}, {1214, 4121, 1412, 2141}, {1133, 3311}, {1313, 3131}, {1232, 2123, 3212, 2321}, and {2222}.
%o (PARI) T(n, k) = {if(k<=1, 0, binomial((n-k%2)\2, k\2))} \\ _Andrew Howroyd_, Sep 27 2019
%Y Row sums are A027383(n-1).
%Y If we remove the cyclic equivalence requirement, and just count k-double-palindromes of n, then we get sequence A180653.
%Y If we replace the left hand column of 0's by 1's in the triangle above, we get the triangle 'RE(n, k)' where RE(n, k) is the number of k-reverses of n up to cyclic equivalence, see the McSorley reference above for more details and also sequence A119963.
%Y See sequence A179181 for the triangle whose (n, k) term gives the number of k-palindromes (single-palindromes) of n up to cyclic equivalence.
%K nonn,tabl
%O 1,8
%A _John P. McSorley_, Sep 23 2010
%E Terms a(56) and beyond from _Andrew Howroyd_, Sep 27 2019
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