%I #9 Apr 08 2024 09:21:26
%S 1,1,1,-3,2,1,-11,15,3,1,25,-44,-18,4,1,41,-115,-110,50,5,1,-43,246,
%T 375,-220,-45,6,1,29,315,861,-805,-385,105,7,1,-335,232,-1204,2296,
%U 1750,-616,-84,8,1,-1199,3033,1044,3780,5166,-2898,-924,180,9,1
%N 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.
%C The matrix inverse of H*[C^-1]*H is H*C*H = A118438, 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 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.
%e Triangle begins:
%e 1;
%e 1, 1;
%e -3, 2, 1;
%e -11, 15, 3, 1;
%e 25,-44,-18, 4, 1;
%e 41,-115,-110, 50, 5, 1;
%e -43, 246, 375,-220,-45, 6, 1;
%e 29, 315, 861,-805,-385, 105, 7, 1;
%e -335, 232,-1204, 2296, 1750,-616,-84, 8, 1;
%e -1199, 3033, 1044, 3780, 5166,-2898,-924, 180, 9, 1;
%e ...
%e The matrix log, log(T) = A118441, starts:
%e 0;
%e 1, 0;
%e -4, 2, 0;
%e -12, 12, 3, 0;
%e 32,-48,-24, 4, 0;
%e 80,-160,-120, 40, 5, 0;
%e ...
%e where matrix square, log(T)^2, is a single diagonal:
%e 0;
%e 0,0;
%e 2,0,0;
%e 0,6,0,0;
%e 0,0,12,0,0;
%e 0,0,0,20,0,0;
%e ...
%t nmax = 12;
%t h[n_, k_] := Binomial[n, k]*(-1)^(Quotient[n+1, 2] - Quotient[k, 2]+n-k);
%t H = Table[h[n, k], {n, 0, nmax}, {k, 0, nmax}];
%t Cn = Table[Binomial[n, k], {n, 0, nmax}, {k, 0, nmax}];
%t Tn = H.Inverse[Cn].H;
%t T[n_, k_] := Tn[[n+1, k+1]];
%t Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Apr 08 2024 *)
%o (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]}
%Y Cf. A118436 (column 0), A118437 (row sums), A118438 (matrix inverse), A118441 (matrix log), A118433 (self-inverse H).
%K sign,tabl
%O 0,4
%A _Paul D. Hanna_, Apr 28 2006