A144484 Triangle read by rows: T(n, k) = binomial(3*n+1-k, n-k) for n, k >= 0.
1, 4, 1, 21, 6, 1, 120, 36, 8, 1, 715, 220, 55, 10, 1, 4368, 1365, 364, 78, 12, 1, 27132, 8568, 2380, 560, 105, 14, 1, 170544, 54264, 15504, 3876, 816, 136, 16, 1, 1081575, 346104, 100947, 26334, 5985, 1140, 171, 18, 1, 6906900, 2220075, 657800, 177100
Offset: 0
Examples
{1}, {4, 1}, {21, 6, 1}, {120, 36, 8, 1}, {715, 220, 55, 10, 1}, {4368, 1365, 364, 78, 12, 1}, {27132, 8568, 2380, 560, 105, 14, 1}, {170544, 54264, 15504, 3876, 816, 136, 16, 1}, {1081575, 346104, 100947, 26334, 5985, 1140, 171, 18, 1}, {6906900, 2220075, 657800, 177100, 42504, 8855, 1540, 210, 20, 1}, {44352165, 14307150, 4292145, 1184040, 296010, 65780, 12650, 2024, 253, 22, 1}
References
- M. Jones, Further remarks on the enumeration of graphs, preprint, 2001.
Links
- Omer Egecioglu, Timothy Redmond, Charles Ryavec, Almost Product Evaluation of Hankel Determinants, arXiv:0704.3398 [math.CO], 2007.
Crossrefs
Cf. A025174 (row sums).
Programs
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Mathematica
p[x_, n_] = Sum[Binomial[3*n + 1 - m, n - m]*x^m, {m, 0, n}]; Table[CoefficientList[p[x, n], x], {n, 0, 10}]; Flatten[%]
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PARI
T(n, k) = binomial(3*n+1-k, n-k); tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, May 13 2018
Formula
p(x,n)=Sum[Binomial[3*n + 1 - m, n - m]*x^m, {m, 0, n}];
p(x,n)=Gamma[2*n+3]*Hypergeometric2F1[1,-n-1-3*n,x]/(Gamma[1+n]*Gamma[2+2*n});
Extensions
New name from Michel Marcus, May 13 2018
Comments