A174127 Triangle T(n, k) = (n-k)^3 * binomial(n-1, k-1)^3 with T(n, 0) = T(n, n) = 1, read by rows.
1, 1, 1, 1, 1, 1, 1, 8, 8, 1, 1, 27, 216, 27, 1, 1, 64, 1728, 1728, 64, 1, 1, 125, 8000, 27000, 8000, 125, 1, 1, 216, 27000, 216000, 216000, 27000, 216, 1, 1, 343, 74088, 1157625, 2744000, 1157625, 74088, 343, 1, 1, 512, 175616, 4741632, 21952000, 21952000, 4741632, 175616, 512, 1
Offset: 0
Examples
Triangle begins as: 1; 1, 1; 1, 1, 1; 1, 8, 8, 1; 1, 27, 216, 27, 1; 1, 64, 1728, 1728, 64, 1; 1, 125, 8000, 27000, 8000, 125, 1; 1, 216, 27000, 216000, 216000, 27000, 216, 1; 1, 343, 74088, 1157625, 2744000, 1157625, 74088, 343, 1; 1, 512, 175616, 4741632, 21952000, 21952000, 4741632, 175616, 512, 1;
Links
- G. C. Greubel, Rows n = 0..100 of the triangle, flattened
Programs
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Magma
T:= func< n,k,q | k eq 0 or k eq n select 1 else (n-k)^q*Binomial(n-1,k-1)^q >; [T(n,k,3): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 11 2021
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Mathematica
(* First program *) c[n_]:= If[n<2, 1, Product[(i-1)^3, {i,2,n}]]; T[n_, k_]:= c[n]/(c[k]*c[n-k]); Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* Second program *) T[n_, k_, q_]:= If[k==0 || k==n, 1, (n-k)^q*Binomial[n-1, k-1]^q]; Table[T[n,k,3], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 10 2021 *) -
Sage
def T(n,k,q): return 1 if (k==0 or k==n) else (n-k)^q*binomial(n-1,k-1)^q flatten([[T(n,k,3) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 11 2021
Formula
Let c(n) = Product_{i=2..n} (i-1)^3 for n > 2 otherwise 1. The number triangle is given by T(n, k) = c(n)/(c(k)*c(n-k)).
From G. C. Greubel, Feb 11 2021: (Start)
T(n, k) = (n-k)^3 * binomial(n-1, k-1)^3 with T(n, 0) = T(n, n) = 1.
Sum_{k=0..n} T(n, k) = 2 + (n-1)^3*A000172(n-2) - [n=0]. (End)
Extensions
Edited by G. C. Greubel, Feb 11 2021
Comments