0,5

There are an infinite number of integer square-roots of the identity matrix.

E.g.f.: A(x,y) = cos(x)*exp(-x*y) + sin(x)*exp(x*y). 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).

Triangle H begins:

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, 6, 1;

-1, 7, 21,-35,-35, 21, 7,-1;

1, -8,-28, 56, 70,-56,-28, 8, 1;

1, -9,-36, 84, 126,-126,-84, 36, 9,-1;

-1, 10, 45,-120,-210, 252, 210,-120,-45, 10, 1; ...

G.f.s for columns are:

k=0: (x + 1)/(1+x^2);

k=1: (x^2 + 2*x - 1)/(1+x^2)^2;

k=2: (-x^3 - 3*x^2 + 3*x + 1)/(1+x^2)^3;

k=3: (-x^4 - 4*x^3 + 6*x^2 + 4*x - 1)/(1+x^2)^4;

k=4: (x^5 + 5*x^4 - 10*x^3 - 10*x^2 + 5*x + 1)/(1+x^2)^5;

k=5: (x^6 + 6*x^5 - 15*x^4 - 20*x^3 + 15*x^2 + 6*x - 1)/(1+x^2)^6.

The g.f. of column k is thus:

G_k(x) = [ Sum_{j=0..k+1} -H(k+1,j)*(-x)^(k+1-j) ]/(1+x^2)^(k+1).

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).

(PARI) {H(n, k)=binomial(n, k)*(-1)^((n+1)\2-k\2+n-k)} (PARI) /* Using E.G.F.: */ {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)} (PARI) /* Using O.G.F.: */ {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)}

Cf. A118434 (row sums), A118435 (H*[C^-1]*H).

Adjacent sequences: A118430 A118431 A118432 this_sequence A118434 A118435 A118436

Sequence in context: A034932 A094495 A117440 this_sequence A007318 A108086 A130595

sign,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Apr 28 2006