A181111 'ADP(n,k)' triangle read by rows. ADP(n,k) is the number of aperiodic k-double-palindromes of n.
0, 0, 0, 0, 2, 0, 0, 2, 2, 0, 0, 4, 4, 4, 0, 0, 4, 2, 6, 4, 0, 0, 6, 6, 12, 12, 6, 0, 0, 6, 6, 14, 12, 16, 6, 0, 0, 8, 6, 24, 24, 18, 24, 8, 0, 0, 8, 8, 28, 20, 44, 24, 28, 8, 0, 0, 10, 10, 40, 40, 60, 60, 40, 40, 10, 0, 0, 10, 8, 44, 40, 94, 60, 88, 32, 46, 10, 0
Offset: 1
Examples
The triangle begins: 0 0 0 0 2 0 0 2 2 0 0 4 4 4 0 0 4 2 6 4 0 0 6 6 12 12 6 0 0 6 6 14 12 16 6 0 0 8 6 24 24 18 24 8 0 0 8 8 28 20 44 24 28 8 0 ... For example, row 8 is: 0 6 6 14 12 16 6 0. We have ADP(8,3)=6 because there are 6 aperiodic 3-double-palindromes of 8: 116, 611, 224, 422, 233, and 332. We have ADP(8,4)=14 because there are 14 4-double-palindromes of 8: 1115, 5111, 1511, 1151, 1214, 4121, 1412, 2141, 1133, 3311, 1232, 2123, 3212, and 2321.
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
\\ here p(n,k) is A119963(n,k), q(n,k) is A051159(n-1, k-1). p(n, 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)) } T(n, k) = sumdiv(gcd(n, k), d, moebius(d) * (k*p(n/d, k/d) - q(n/d, k/d))); \\ Andrew Howroyd, Sep 27 2019
Formula
Extensions
a(37) corrected and terms a(56) and beyond from Andrew Howroyd, Sep 27 2019
Comments