A104041 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).
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, 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, 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
Offset: 0
Examples
Rows of T begin: 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, 0, 8, -20, 18, -7, 1; ... The matrix inverse T^-1 equals triangle A104040: 1; 1, 1; 2, 2, 1; 4, 4, 3, 1; 8, 8, 8, 4, 1; 16, 16, 20, 12, 5, 1; 32, 32, 48, 32, 18, 6, 1; 64, 64, 112, 80, 56, 24, 7, 1; ... The rows of T^-1 equal columns of T in absolute value.
Programs
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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)}
Formula
G.f.: A(x, y) = (1 - x + x*y)/(1 + 2*x^2*y - x^2*y^2).
Conjectures from Peter Bala, May 25 2023: (Start)
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.
T(2*n,k) = binomial(n,k-n)*(-2)^(2*n-k) for 0 <= k <= 2*n. Cf. A038207. (End)
Comments