A118435 Triangle T, read by rows, equal to the matrix product T = H*[C^-1]*H, where H is the self-inverse triangle A118433 and C is Pascal's triangle.
1, 1, 1, -3, 2, 1, -11, 15, 3, 1, 25, -44, -18, 4, 1, 41, -115, -110, 50, 5, 1, -43, 246, 375, -220, -45, 6, 1, 29, 315, 861, -805, -385, 105, 7, 1, -335, 232, -1204, 2296, 1750, -616, -84, 8, 1, -1199, 3033, 1044, 3780, 5166, -2898, -924, 180, 9, 1
Offset: 0
Examples
Triangle begins: 1; 1, 1; -3, 2, 1; -11, 15, 3, 1; 25,-44,-18, 4, 1; 41,-115,-110, 50, 5, 1; -43, 246, 375,-220,-45, 6, 1; 29, 315, 861,-805,-385, 105, 7, 1; -335, 232,-1204, 2296, 1750,-616,-84, 8, 1; -1199, 3033, 1044, 3780, 5166,-2898,-924, 180, 9, 1; ... The matrix log, log(T) = A118441, starts: 0; 1, 0; -4, 2, 0; -12, 12, 3, 0; 32,-48,-24, 4, 0; 80,-160,-120, 40, 5, 0; ... where matrix square, log(T)^2, is a single diagonal: 0; 0,0; 2,0,0; 0,6,0,0; 0,0,12,0,0; 0,0,0,20,0,0; ...
Crossrefs
Programs
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Mathematica
nmax = 12; h[n_, k_] := Binomial[n, k]*(-1)^(Quotient[n+1, 2] - Quotient[k, 2]+n-k); H = Table[h[n, k], {n, 0, nmax}, {k, 0, nmax}]; Cn = Table[Binomial[n, k], {n, 0, nmax}, {k, 0, nmax}]; Tn = H.Inverse[Cn].H; T[n_, k_] := Tn[[n+1, k+1]]; Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 08 2024 *) -
PARI
{T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1)*(-1)^(r\2- (c-1)\2+r-c))),C=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1))));(M*C^-1*M)[n+1,k+1]}
Formula
Since T + T^-1 = C + C^-1, then [T^-1](n,k) = (1+(-1)^(n-k))*C(n,k) - T(n,k) is a formula for the matrix inverse T^-1 = A118438.
Comments