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A118438 Triangle T, read by rows, equal to the matrix product T = H*C*H, where H is the self-inverse triangle A118433 and C is Pascal's triangle.

%I A118438 #7 Apr 08 2024 12:16:25
%S A118438 1,-1,1,5,-2,1,11,-9,-3,1,-23,44,30,-4,1,-41,125,110,-30,-5,1,45,-246,
%T A118438 -345,220,75,-6,1,-29,-301,-861,875,385,-63,-7,1,337,-232,1260,-2296,
%U A118438 -1610,616,140,-8,1,1199,-3015,-1044,-3612,-5166,3150,924,-108,-9,1
%N A118438 Triangle T, read by rows, equal to the matrix product T = H*C*H, where H is the self-inverse triangle A118433 and C is Pascal's triangle.
%C A118438 The matrix inverse of H*C*H is H*[C^-1]*H = A118435, where H^2 = I (identity). The matrix log, log(T) = -A118441, is a matrix square root of a triangular matrix with a single diagonal (two rows down from the main diagonal).
%F A118438 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 = A118435.
%e A118438 Triangle begins:
%e A118438 1;
%e A118438 -1, 1;
%e A118438 5,-2, 1;
%e A118438 11,-9,-3, 1;
%e A118438 -23, 44, 30,-4, 1;
%e A118438 -41, 125, 110,-30,-5, 1;
%e A118438 45,-246,-345, 220, 75,-6, 1;
%e A118438 -29,-301,-861, 875, 385,-63,-7, 1;
%e A118438 337,-232, 1260,-2296,-1610, 616, 140,-8, 1;
%e A118438 1199,-3015,-1044,-3612,-5166, 3150, 924,-108,-9, 1; ...
%t A118438 nmax = 12;
%t A118438 h[n_, k_] := Binomial[n, k]*(-1)^(Quotient[n+1, 2] - Quotient[k, 2]+n-k);
%t A118438 H = Table[h[n, k], {n, 0, nmax}, {k, 0, nmax}];
%t A118438 Cn = Table[Binomial[n, k], {n, 0, nmax}, {k, 0, nmax}];
%t A118438 Tn = H.Cn.H;
%t A118438 T[n_, k_] := Tn[[n+1, k+1]];
%t A118438 Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Apr 08 2024 *)
%o A118438 (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*M)[n+1,k+1]}
%Y A118438 Cf. A118439 (column 0), A118440 (row sums), A118435 (matrix inverse), A118441 (-matrix log); A118433 (self-inverse H).
%K A118438 sign,tabl
%O A118438 0,4
%A A118438 _Paul D. Hanna_, Apr 28 2006