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

0,5

Comments

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).

Links

Table of n, a(n) for n=0..90.

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)

Example

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.

Prog

(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)}

Crossrefs

Cf. A104040, A038207, A038754.

Sequence in context: A031135 A037181 A051070 * A104402 A261897 A131084

Adjacent sequences: A104038 A104039 A104040 * A104042 A104043 A104044

Keyword

sign,tabl

Author

Paul D. Hanna, Mar 02 2005

Status

approved