A180918 'DPE(n,k)' triangle read by rows. DPE(n,k) is the number of k-double-palindromes of n up to cyclic equivalence.

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

Offset

1,8

Comments

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.

A palindrome is a word which is the same when written backwards.

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.

See sequence A180653. For example 1123532=11|23532 is a 7-double-palindrome of 17 since both 11 and 23532 are palindromes.

Let DPE(n,k) denote the number of k-double-palindromes of n up to cyclic equivalence.

This sequence is the 'DPE(n,k)' triangle read by rows.

References

John P. McSorley: Counting k-compositions 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.

Example

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 3-double-palindromes of 8 up to cyclic equivalence: {116, 611}, {224, 422}, and {233, 332}.
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}.

Prog

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

Crossrefs

Row sums are A027383(n-1).

If we remove the cyclic equivalence requirement, and just count k-double-palindromes 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 k-reverses 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 k-palindromes (single-palindromes) of n up to cyclic equivalence.

Sequence in context: A339448 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