%I #19 May 18 2024 06:11:51
%S 1,1,-1,-1,2,1,-1,3,3,-1,1,-4,-6,4,1,1,-5,-10,10,5,-1,-1,6,15,-20,-15,
%T 6,1,-1,7,21,-35,-35,21,7,-1,1,-8,-28,56,70,-56,-28,8,1,1,-9,-36,84,
%U 126,-126,-84,36,9,-1,-1,10,45,-120,-210,252,210,-120,-45,10,1
%N Self-inverse triangle H, read by rows; a nontrivial matrix square-root of identity: H^2 = I, where H(n,k) = C(n,k)*(-1)^(floor((n+1)/2) - floor(k/2) + n - k) for n >= k >= 0.
%C There are an infinite number of integer square-roots of the identity matrix.
%F E.g.f.: A(x,y) = cos(x)*exp(-x*y) + sin(x)*exp(x*y).
%F O.g.f.: A(x,y) = (1 + x*(1-y) + x^2*(1+2*y-y^2) + x^3*(1+y+y^2+y^3)) / (1 + 2*x^2*(1-y^2) + x^4*(1+y^2)^2).
%e Triangle H begins:
%e 1;
%e 1, -1;
%e -1, 2, 1;
%e -1, 3, 3, -1;
%e 1, -4, -6, 4, 1;
%e 1, -5, -10, 10, 5, -1;
%e -1, 6, 15, -20, -15, 6, 1;
%e -1, 7, 21, -35, -35, 21, 7, -1;
%e 1, -8, -28, 56, 70, -56, -28, 8, 1;
%e 1, -9, -36, 84, 126, -126, -84, 36, 9, -1;
%e -1, 10, 45, -120, -210, 252, 210, -120, -45, 10, 1; ...
%e G.f.s for columns:
%e k=0: (x + 1)/(1+x^2);
%e k=1: (x^2 + 2*x - 1)/(1+x^2)^2;
%e k=2: (-x^3 - 3*x^2 + 3*x + 1)/(1+x^2)^3;
%e k=3: (-x^4 - 4*x^3 + 6*x^2 + 4*x - 1)/(1+x^2)^4;
%e k=4: (x^5 + 5*x^4 - 10*x^3 - 10*x^2 + 5*x + 1)/(1+x^2)^5;
%e k=5: (x^6 + 6*x^5 - 15*x^4 - 20*x^3 + 15*x^2 + 6*x - 1)/(1+x^2)^6.
%e The g.f. of column k is thus:
%e G_k(x) = (Sum_{j=0..k+1} -H(k+1,j)*(-x)^(k+1-j))/(1+x^2)^(k+1).
%e The triangle formed from above polynomial numerators of column g.f.s is described by the e.g.f.: cos(x*y)*exp(-x) - sin(x*y)*exp(x).
%t H[n_, k_] := Binomial[n, k]*(-1)^(Quotient[n+1, 2]-Quotient[k, 2]+n-k);
%t Table[H[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Apr 08 2024 *)
%o (PARI) {H(n,k)=binomial(n,k)*(-1)^((n+1)\2-k\2+n-k)}
%o for(n=0,12,for(k=0,n,print1(H(n,k),", "));print(""))
%o (PARI) /* Using E.G.F.: */
%o {H(n,k)=local(x=X+X*O(X^n),y=Y+Y*O(Y^k));n!*polcoeff(polcoeff( cos(x)*exp(-x*y)+sin(x)*exp(x*y),n,X),k,Y)}
%o for(n=0,12,for(k=0,n,print1(H(n,k),", "));print(""))
%o (PARI) /* Using O.G.F.: */
%o {H(n,k)=polcoeff(polcoeff((1+x*(1-y)+x^2*(1+2*y-y^2)+x^3*(1+y+y^2+y^3))/(1+2*x^2*(1-y^2)+x^4*(1+y^2)^2+x*O(x^n)+y*O(y^k)),n,x),k,y)}
%o for(n=0,12,for(k=0,n,print1(H(n,k),", "));print(""))
%Y Cf. A118434 (row sums), A118435 (H*[C^-1]*H).
%K sign,tabl
%O 0,5
%A _Paul D. Hanna_, Apr 28 2006