0,39

The rows of this triangle are symmetric up to sign. Row sums = 2 after row 0. Unsigned row sums = A116466. Row squared sums = A116467. Central terms of odd rows: T(2*n+1,n+1) = |A064310(n)|.

G.f.: A(x,y) = 1/(1-x*y)+ x*(1+x-2*x^2*y)/(1-x)/(1+x+x*y)/(1-x*y). G.f. of matrix power T^m: 1/(1-x*y)+ m*x*(1+x-2*x^2*y)/(1-x)/(1+x+x*y)/(1-x*y).

Matrix inverse is: T^-1 = 2*I - T.

Matrix log is: LOG(T) = T - I.

Triangle T begins:

1;

1, 1;

1, 0, 1;

1,-1, 1, 1;

1, 0, 0, 0, 1;

1,-1, 0, 0, 1, 1;

1, 0, 1, 0,-1, 0, 1;

1,-1,-1,-1, 1, 1, 1, 1;

1, 0, 2, 2, 0,-2,-2, 0, 1;

1,-1,-2,-4,-2, 2, 4, 2, 1, 1;

1, 0, 3, 6, 6, 0,-6,-6,-3, 0, 1;

1,-1,-3,-9,-12,-6, 6, 12, 9, 3, 1, 1;

1, 0, 4, 12, 21, 18, 0,-18,-21,-12,-4, 0, 1; ...

The g.f. of column k, C_k(x), obeys the recurrence:

C_k = C_{k-1} + (-1)^k*x*(1+2*x)/(1-x)/(1+x)^k with C_0 = 1/(1-x);

so that column g.f.s continue as:

C_1 = C_0 - x*(1+2*x)/(1-x)/(1+x),

C_2 = C_1 + x*(1+2*x)/(1-x)/(1+x)^2,

C_3 = C_2 - x*(1+2*x)/(1-x)/(1+x)^3, ...

(PARI) {T(n, k)=local(x=X+X*O(X^n), y=Y+Y*O(Y^k)); polcoeff(polcoeff( 1/(1-x*y)+ x*(1+x-2*x^2*y)/(1-x)/(1+x+x*y)/(1-x*y), n, X), k, Y)} (PARI) {T(n, k)=local(M=matrix(n+1, n+1)); for(r=1, n+1, for(c=1, r, M[r, c]=if(r==c, 1, if(c==1, 1, if(c>1, (2*M^0-M)[r-1, c-1])+(2*M^0-M)[r-1, c])))); return(M[n+1, k+1])}

Cf. A116466 (unsigned row sums), A116467 (row squared sums), A064310 (central terms); A112555 (variant).

Sequence in context: A074943 A045719 A114906 this_sequence A140666 A130772 A109265

Adjacent sequences: A114697 A114698 A114699 this_sequence A114701 A114702 A114703

sign,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Feb 19 2006