A142467 Triangle T(n,m) read by rows: T(n,m) = Product_{i=0..6} binomial(n+i,m)/binomial(m+i,m).
1, 1, 1, 1, 8, 1, 1, 36, 36, 1, 1, 120, 540, 120, 1, 1, 330, 4950, 4950, 330, 1, 1, 792, 32670, 108900, 32670, 792, 1, 1, 1716, 169884, 1557270, 1557270, 169884, 1716, 1, 1, 3432, 736164, 16195608, 44537922, 16195608, 736164, 3432, 1
Offset: 0
Examples
Triangle begins as: 1; 1, 1; 1, 8, 1; 1, 36, 36, 1; 1, 120, 540, 120, 1; 1, 330, 4950, 4950, 330, 1; 1, 792, 32670, 108900, 32670, 792, 1; 1, 1716, 169884, 1557270, 1557270, 169884, 1716, 1; 1, 3432, 736164, 16195608, 44537922, 16195608, 736164, 3432, 1; 1, 6435, 2760615, 131589315, 868489479, 868489479, 131589315, 2760615, 6435, 1;
Links
- Seiichi Manyama, Rows n = 0..139, flattened
- Johann Cigler, Pascal triangle, Hoggatt matrices, and analogous constructions, arXiv:2103.01652 [math.CO], 2021.
- Johann Cigler, Some observations about Hoggatt triangles, Universität Wien (Austria, 2021).
Crossrefs
Programs
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Magma
[(&*[Binomial(n+j,k)/Binomial(k+j,k): j in [0..6]]): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 13 2022
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Mathematica
T[n_, k_]:= Product[Binomial[n+j, k]/Binomial[k+j, k], {j,0,6}]; Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Nov 13 2022 *)
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PARI
T(n, k) = prod(j=0, 6, binomial(n+j, k)/binomial(k+j, k)); \\ Seiichi Manyama, Apr 01 2021
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SageMath
def A142467(n,k): return product(binomial(n+j,k)/binomial(k+j,k) for j in (0..6)) flatten([[A142467(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Nov 13 2022
Extensions
Edited by the Associate Editors of the OEIS, May 17 2009
Comments