A113279 Triangle T(n,k) of coefficients of r_1^n+r_2^n in terms of p and q, where r_1,r_2 are the roots of x^2+px+q=0.

2, -1, 1, -2, -1, 3, 1, -4, 2, -1, 5, -5, 1, -6, 9, -2, -1, 7, -14, 7, 1, -8, 20, -16, 2, -1, 9, -27, 30, -9, 1, -10, 35, -50, 25, -2, -1, 11, -44, 77, -55, 11, 1, -12, 54, -112, 105, -36, 2, -1, 13, -65, 156, -182, 91, -13, 1, -14, 77, -210, 294, -196, 49, -2, -1, 15, -90, 275, -450, 378, -140, 15

Offset: 0

Author

Floor van Lamoen, Oct 22 2005

Keywords

• easy
• sign
• tabf

Comments

The triangle begins: 2 .-1 1..-2 .-1..3 1..-4.2

From Tom Copeland, Nov 07 2015: (Start)

The row polynomials are the power sums p_n = x1^n + x2^n of the solutions of 0 = (x-x1)(x-x2) = x^2 - (x1+x2) x + x1x2 = x^2 + px + q, so -p = x1+x2 = e1(x1,x2), the first order elementary symmetric polynomial for two variables, or indeterminates, and q = x1*x2 = e2(x1,x2), the second order elementary symmetric polynomial.

The Girard-Newton-Waring identities (Wikipedia) express the power sums in terms of the elementary symmetric polynomials, giving

p_0 = x1^0 + x0^0 = 2

p_1 = x1 + x2 = e1 = -p = F(1,p,q,0,..)

p_2 = x1^2 + x2^2 = e1^2 - 2 e2 = p^2 - 2 q = F(2,p,q,0,..)

p_3 = e1^3 - 3 e2 e1 = -p^3 + 3 pq = F(3,p,q,0,..)

p_4 = p^4 - 4 p^2 q + 2 q^2 = F(4,p,q,0,..)

... .

These bivariate partition polynomials are the Faber polynomials F(n,b1,b2,..,bn) of A263916 with b1 = -e1 = p, b2 = e2 = q, and all other indeterminates set to zero.

Let p = q = t, then

F(1,t,t,0,..)/t = -1

F(2,t,t,0,..)/t = -2 + t

F(3,t,t,0,..)/t^2 = 3 - t

F(4,t,t,0,..)/t^2 = 2 - 4 t + t^2

... .

Or,

t * F(1,1/t,1/t,0,..) = -1

t^2 * F(2,1/t,1/t,0,..) = 1 -2 t

t^3 * F(3,1/t,1/t,0,..) = -1 + 3 t

t^4 * F(4,1/t,1/t,0,..) = 1 - 4 t + 2 t^2

... .

The sequence of Faber polynomials F(n,1/t,1/t,0,..) is obtained from the logarithmic generator -log[1+(x+x^2)/t] = sum{n>=1, F(n,1/t,1/t,0,..) x^n/n}, so

2-(2xt+1)xt/(t+xt+(xt)^2) = 2 + sum{n>=1, t^n F(n,1//t,1/t,0,..) x^n} is an o.g.f. for the row polynomials of this entry.

(End)

Examples

			x_1^5+x_2^5 = -p^5 + 5p^3q - 5pq^2, so row 5 reads -1, 5, -5.
		

References

• M. Herkenhoff Konersmann, Sprokkel XXXI: x_1^n+x_2^n, Nieuw Tijdschr. Wisk, 42 (1954-55) 180.

Links

• T. Copeland, Addendum to Elliptic Lie Triad
• Wikipedia, Newton's identities.

Crossrefs

Cf. A034807, A263916.

Formula

T(n, k) = (-1)^(n+k)*A034807(n, k).

O.g.f.: 2-(2xt+1)xt/(t+xt+(xt)^2) = (2+x)/(1+x+x^2/t). - Tom Copeland, Nov 07 2015