A172090 Triangle T(n, k) = f(n-k) + f(k) - f(n), where f(n) = -3*n with f(0) = 1, f(1) = -2, read by rows.
1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1
Offset: 0
Examples
Triangle begins as: 1; 1, 1; 1, 2, 1; 1, 1, 1, 1; 1, 1, 0, 1, 1; 1, 1, 0, 0, 1, 1; 1, 1, 0, 0, 0, 1, 1; 1, 1, 0, 0, 0, 0, 1, 1; 1, 1, 0, 0, 0, 0, 0, 1, 1; 1, 1, 0, 0, 0, 0, 0, 0, 1, 1; 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1;
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Crossrefs
Row sums are A151798.
Programs
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Mathematica
(* First program *) f[n_]:= f[n]= If[n < 2, (-1)^n*(n+1), -3*n]; T[n_, k_]:= f[n-k] +f[k] -f[n]; Table[T[n, k], {n, 0, 15}, {k, 0, n}]//Flatten (* modified by G. C. Greubel, Apr 29 2021 *) (* Second program *) T[n_, k_]:= If[n<3, Binomial[n, k], If[n==3 || k<2 || k>n-2, 1, 0]]; Table[T[n, k], {n, 0, 15}, {k, 0, n}]//Flatten (* G. C. Greubel, Apr 29 2021 *)
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Sage
def f(n): return (-1)^n*(n+1) if (n<2) else -3*n def T(n,k): return f(n-k) + f(k) - f(n) flatten([[T(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Apr 29 2021
Formula
T(n, k) = f(n-k) + f(k) - f(n), where f(n) = -3*n with f(0) = 1, f(1) = -2.
From G. C. Greubel, Apr 29 2021: (Start)
T(n, k) is defined by T(n, 0) = T(n, 1) = T(n, n-1) = T(n, n) = T(3, k) = 1, T(2, 1) = 2 and 0 otherwise.
Sum_{k=0..n} T(n,k) = A151798(n). (End)
Extensions
Edited by G. C. Greubel, Apr 29 2021