

A180918


'DPE(n,k)' triangle read by rows. DPE(n,k) is the number of kdoublepalindromes of n up to cyclic equivalence.


5



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, 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, 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
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OFFSET

1,8


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 palindrome is a word which is the same when written backwards.
A kdoublepalindrome of n is a kcomposition of n which is the concatenation of two palindromes, PP'=PP', where both P, P'>=1.
See sequence A180653. For example 1123532=1123532 is a 7doublepalindrome of 17 since both 11 and 23532 are palindromes.
Let DPE(n,k) denote the number of kdoublepalindromes of n up to cyclic equivalence.
This sequence is the 'DPE(n,k)' triangle read by rows.
The triangle begins:
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 1 1
0 4 4 6 6 4 4 1 1
0 5 4 10 6 10 4 5 1 1
...
For example, row 8 is: 0 4 3 6 3 4 1 1.
We have DPE(8,3)=3 because there are 3 3doublepalindromes of 8 up to cyclic equivalence: {116, 611}, {224, 422}, and {233, 332}.
We have DPE(8,4)=6 because there are 6 4doublepalindromes of 8: up to cyclic equivalence: {1115, 5111, 1511, 1151}, {1214, 4121, 1412, 2141}, {1133, 3311}, {1313, 3131}, {1232, 2123, 3212, 2321}, and {2222}.


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

T(n, 1) = 0; T(n, k) = A119963(n,k) for k > 1.


PROG

(PARI) T(n, k) = {if(k<=1, 0, binomial((nk%2)\2, k\2))} \\ Andrew Howroyd, Sep 27 2019


CROSSREFS

Row sums are A027383(n1).
If we remove the cyclic equivalence requirement, and just count kdoublepalindromes of n, then we get sequence A180653.
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 kreverses of n up to cyclic equivalence, see the McSorley reference above for more details and also sequence A119963.
See sequence A179181 for the triangle whose (n, k) term gives the number of kpalindromes (singlepalindromes) of n up to cyclic equivalence.
Sequence in context: A308061 A060582 A060450 * A152146 A025860 A322285
Adjacent sequences: A180915 A180916 A180917 * A180919 A180920 A180921


KEYWORD

nonn,tabl


AUTHOR

John P. McSorley, Sep 23 2010


EXTENSIONS

Terms a(56) and beyond from Andrew Howroyd, Sep 27 2019


STATUS

approved



