A180653 'DP(n,k)' triangle read by rows. DP(n,k) is the number of k-double-palindromes of n.
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
Offset: 1
Examples
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 3-double-palindromes of 8: 116, 611, 224, 422, 233, and 332. We have DP(8,4)=17 because there are 17 4-double-palindromes of 8: 1115, 5111, 1511, 1151, 1214, 4121, 1412, 2141, 1133, 3311, 1313, 3131, 1232, 2123, 3212, 2321, and 2222.
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
Crossrefs
Programs
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PARI
\\ p(n,k) is k*A119963(n,k); q(n,k) is A051159(n-1, k-1). p(n, k) = {k*binomial((n-k%2)\2, k\2)} q(n, k) = {if(n%2==1&&k%2==0, 0, binomial((n-1)\2, (k-1)\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
Formula
Extensions
Terms a(56) and beyond from Andrew Howroyd, Sep 27 2019
Comments