A104509 Matrix inverse of triangle A104505, which is the right-hand side of triangle A084610 of coefficients in (1 + x - x^2)^n.

1, 1, -1, 3, -2, 1, 4, -6, 3, -1, 7, -12, 10, -4, 1, 11, -25, 25, -15, 5, -1, 18, -48, 60, -44, 21, -6, 1, 29, -91, 133, -119, 70, -28, 7, -1, 47, -168, 284, -296, 210, -104, 36, -8, 1, 76, -306, 585, -699, 576, -342, 147, -45, 9, -1, 123, -550, 1175, -1580, 1485, -1022, 525, -200, 55, -10, 1, 199, -979, 2310, -3454, 3641

Offset: 0

Author

Paul D. Hanna, Mar 11 2005

Keywords

• sign
• tabl

Comments

Riordan array ( (1 + x^2)/(1 - x - x^2), -x/(1 - x - x^2) ) belonging to the hitting time subgroup of the Riordan group (see Peart and Woan). - Peter Bala, Jun 29 2015

The sums of absolute values along steep diagonals in this triangle are: 1, 1, 3, 4 + |-1|, 7 + |-2|, 11 + |-6|, 18 + |-12| + 1, ... and these are the tribonacci numbers A000213 that begin with 1, 1, 1, 3. To see this, replace the y in the g.f. A(x,y) = (1 + x^2)/(1-x-x^2 + x*y) with y=-x^2, multiply by x, and add 1, to obtain the g.f. (1 - x^2)/(1-x-x^2-x^3) for A000213. - Noah Carey and Greg Dresden, Nov 02 2021

Examples

			Rows begin:
   1;
   1,   -1;
   3,   -2,   1;
   4,   -6,   3,   -1;
   7,  -12,  10,   -4,   1;
  11,  -25,  25,  -15,   5,   -1;
  18,  -48,  60,  -44,  21,   -6,   1;
  29,  -91, 133, -119,  70,  -28,   7,  -1;
  47, -168, 284, -296, 210, -104,  36,  -8, 1;
  76, -306, 585, -699, 576, -342, 147, -45, 9, -1; ...
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10152 (rows 0 to 141, flattened).
• P. Peart and W.-J. Woan, A divisibility property for a subgroup of Riordan matrices, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263.
• Wikipedia, Lucas polynomials.

Crossrefs

Leftmost column is A000204 (Lucas numbers). Other columns include: A045925, A067988. Row sums are: {1,0,2,0,2,0,2,...}. Absolute row sums form: A099425. Antidiagonal sums are: {1,1,2,2,2,2,2,...}. Absolute antidiagonal sums are: A084214.

Cf. A104505, A084610, A000213.

Programs

Maple

S:= series((1 + x^2)/(1-x-x^2 + x*y),x, 20):
for n from 0 to 19 do R[n]:= coeff(S,x,n) od:
seq(seq(coeff(R[n],y,j),j=0..n), n=0..19); # Robert Israel, Jun 30 2015

Mathematica

nmax = 11;
T[n_, k_] := Coefficient[(1 + x - x^2)^n, x, n + k];
M = Table[T[n, k], {n, 0, nmax}, {k, 0, nmax}] // Inverse;
Table[M[[n+1, k+1]], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 27 2019 *)

PARI

{ T(n,k) = my(X=x+x*O(x^n),Y=y+y*O(y^k)); polcoeff(polcoeff((1 + X^2)/(1-X-X^2 + X*Y),n,x),k,y); }

PARI

{ tabl(nn) = my(m = matrix(nn, nn, n, k, n--; k--; if((nMichel Marcus, Jun 30 2015

PARI

{ A104509(n,k) = if(n==0, k==0, (-1)^k * sum(i=0, (n-k)\2, n/(n-i) * binomial(n-k-i,i) * binomial(n-i,k) )); } \\ Max Alekseyev, Oct 11 2021

Formula

For n>=1, a(n,k) = (-1)^k * Sum_{i=0..[(n-k)/2]} n/(n-i) * binomial(n-i,i) * binomial(n-2*i,k) = (-1)^k * Sum_{i=0..[(n-k)/2]} n/(n-i) * binomial(n-k-i,i) * binomial(n-i,k). - Max Alekseyev, Oct 11 2021

G.f.: A(x, y) = (1 + x^2)/(1-x-x^2 + x*y).

G.f. for column k: g_k(x) = -(x^2+1)*x^k/(x^2+x-1)^(k+1). - Robert Israel, Jun 30 2015

G.f. for row n>=1 is the Lucas polynomial L_n(1-x). - Max Alekseyev, Oct 11 2021