%I #11 May 27 2023 08:11:29
%S 1,-1,1,0,-2,1,0,2,-3,1,0,0,4,-4,1,0,0,-4,8,-5,1,0,0,0,-8,12,-6,1,0,0,
%T 0,8,-20,18,-7,1,0,0,0,0,16,-32,24,-8,1,0,0,0,0,-16,48,-56,32,-9,1,0,
%U 0,0,0,0,-32,80,-80,40,-10,1,0,0,0,0,0,32,-112,160,-120,50,-11,1,0,0,0,0,0,0,64,-192,240,-160,60,-12,1
%N Triangular matrix T, read by rows, such that column k is equal (in absolute value) to row (k-1) in the matrix inverse T^-1 (with extrapolated zeros) for k>0, with T(n,n)=1 (n>=0) and T(n,n-1)=-n (n>=1).
%C Row sums are: {1,0,-1,0, 1,0,-1,0, ...}. Absolute row sums form A038754. Let A(x,y) be the g.f. of T and B(x,y) be the g.f. of T^-1; then B(x,y)=1+x*y*A(-1/y,-x*y^2) and A(x,y)=(B(-x^2*y,-1/x)-1)/(x*y).
%F G.f.: A(x, y) = (1 - x + x*y)/(1 + 2*x^2*y - x^2*y^2).
%F Conjectures from _Peter Bala_, May 25 2023: (Start)
%F T(2*n+1,k) = Sum_{i = k-n-1..n} Stirling2(n,i)*Stirling1(i+2,k+1-n) for 0 <= k <= 2*n+1.
%F T(2*n,k) = binomial(n,k-n)*(-2)^(2*n-k) for 0 <= k <= 2*n. Cf. A038207. (End)
%e Rows of T begin:
%e 1;
%e -1, 1;
%e 0, -2, 1;
%e 0, 2, -3, 1;
%e 0, 0, 4, -4, 1;
%e 0, 0, -4, 8, -5, 1;
%e 0, 0, 0, -8, 12, -6, 1;
%e 0, 0, 0, 8, -20, 18, -7, 1; ...
%e The matrix inverse T^-1 equals triangle A104040:
%e 1;
%e 1, 1;
%e 2, 2, 1;
%e 4, 4, 3, 1;
%e 8, 8, 8, 4, 1;
%e 16, 16, 20, 12, 5, 1;
%e 32, 32, 48, 32, 18, 6, 1;
%e 64, 64, 112, 80, 56, 24, 7, 1; ...
%e The rows of T^-1 equal columns of T in absolute value.
%o (PARI) {T(n,k)=local(X=x+x*O(x^n),Y=y+y*O(y^k)); polcoeff(polcoeff((1-X+X*Y)/(1+2*X^2*Y-X^2*Y^2),n,x),k,y)}
%Y Cf. A104040, A038207, A038754.
%K sign,tabl
%O 0,5
%A _Paul D. Hanna_, Mar 02 2005