A117906 Inverse of number triangle A117904.
1, -1, 1, 0, 0, 1, 0, -1, 0, 1, 0, 0, 0, -1, 1, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, -1, 0, 1, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 1
Offset: 0
Examples
Triangle begins 1; -1, 1; 0, 0, 1; 0, -1, 0, 1; 0, 0, 0, -1, 1; 0, 0, -1, 0, 0, 1; 0, 0, 0, 0, -1, 0, 1; 0, 0, 0, 0, 0, 0, -1, 1; 0, 0, 0, 0, 0, -1, 0, 0, 1; 0, 0, 0, 0, 0, 0, 0, -1, 0, 1; 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1; 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 1; 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 1; 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1; 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 1;
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Programs
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Mathematica
M[n_, k_]:= M[n, k]= If[k>n, 0, If[Abs[JacobiSymbol[Binomial[n, 2], 3] - JacobiSymbol[Binomial[k, 2], 3]]==0, 1, 0]]; m:= m= With[{q=20}, Table[M[n, k], {n,0,q}, {k,0,q}]]; T[n_, k_]:= Inverse[m][[n+1, k+1]]; Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 20 2021 *)
Formula
G.f.: (1 -x*(1-y) +x^2*y^2 -x^3*y -x^5*y^2)/(1-x^3*y^3).
Comments