

A180653


'DP(n,k)' triangle read by rows. DP(n,k) is the number of kdoublepalindromes of n.


5



0, 0, 1, 0, 2, 1, 0, 3, 2, 1, 0, 4, 4, 4, 1, 0, 5, 3, 8, 4, 1, 0, 6, 6, 12, 12, 6, 1, 0, 7, 6, 17, 12, 19, 6, 1, 0, 8, 7, 24, 24, 20, 24, 8, 1, 0, 9, 8, 32, 21, 50, 24, 32, 8, 1, 0, 10, 10, 40, 40, 60, 60, 40, 40, 10, 1, 0, 11, 9, 49, 40, 100, 60, 98, 35, 51, 10, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,5


COMMENTS

A kcomposition of n is an ordered collection of k positive integers (parts) which sum to n. 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.
For example 1123532=1123532 is a 7doublepalindrome of 17 since both 11 and 23532 are palindromes.
Let DP(n,k) denote the number of kdoublepalindromes of n.
This sequence is the 'DP(n,k)' triangle read by rows.


REFERENCES

John P. McSorley: Counting kcompositions of n with palindromic and related structures. Preprint, 2010.


LINKS



FORMULA



EXAMPLE

The triangle begins
0
0 1
0 2 1
0 3 2 1
0 4 4 4 1
0 5 3 8 4 1
0 6 6 12 12 6 1
0 7 6 17 12 19 6 1
0 8 7 24 24 20 24 8 1
0 9 8 32 21 50 24 32 8 1
...
For example, row 8 is: 0 7 6 17 12 19 6 1.
We have DP(8,3)=6 because there are 6 3doublepalindromes of 8: 116, 611, 224, 422, 233, and 332.
We have DP(8,4)=17 because there are 17 4doublepalindromes of 8: 1115, 5111, 1511, 1151, 1214, 4121, 1412, 2141, 1133, 3311, 1313, 3131, 1232, 2123, 3212, 2321, and 2222.


PROG

p(n, k) = {k*binomial((nk%2)\2, k\2)}
q(n, k) = {if(n%2==1&&k%2==0, 0, binomial((n1)\2, (k1)\2))}
invphi(n) = {sumdiv(n, d, d*moebius(d))}
T(n, k) = sumdiv(gcd(n, k), d, invphi(d) * p(n/d, k/d)  moebius(d) * q(n/d, k/d)); \\ Andrew Howroyd, Sep 27 2019


CROSSREFS

See sequence A051159 for the triangle whose (n, k) term gives the number of kpalindromes (singlepalindromes) of n.


KEYWORD



AUTHOR



EXTENSIONS



STATUS

approved



