cp's OEIS Frontend

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.

%I A113279 #20 Jan 14 2016 05:08:58
%S A113279 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,
%T A113279 9,-27,30,-9,1,-10,35,-50,25,-2,-1,11,-44,77,-55,11,1,-12,54,-112,105,
%U A113279 -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
%N 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.
%C A113279 The triangle begins: 2 .-1 1..-2 .-1..3 1..-4.2
%C A113279 From _Tom Copeland_, Nov 07 2015: (Start)
%C A113279 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.
%C A113279 The Girard-Newton-Waring identities (Wikipedia) express the power sums in terms of the elementary symmetric polynomials, giving
%C A113279 p_0 = x1^0 + x0^0 = 2
%C A113279 p_1 = x1 + x2 = e1 = -p = F(1,p,q,0,..)
%C A113279 p_2 = x1^2 + x2^2 =  e1^2 - 2 e2 = p^2 - 2 q = F(2,p,q,0,..)
%C A113279 p_3 = e1^3 - 3 e2 e1 = -p^3 + 3 pq = F(3,p,q,0,..)
%C A113279 p_4 = p^4 - 4 p^2 q + 2 q^2 = F(4,p,q,0,..)
%C A113279 ... .
%C A113279 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.
%C A113279 Let p = q = t, then
%C A113279 F(1,t,t,0,..)/t = -1
%C A113279 F(2,t,t,0,..)/t = -2 + t
%C A113279 F(3,t,t,0,..)/t^2 = 3 - t
%C A113279 F(4,t,t,0,..)/t^2 = 2 - 4 t + t^2
%C A113279 ... .
%C A113279 Or,
%C A113279 t * F(1,1/t,1/t,0,..) = -1
%C A113279 t^2 * F(2,1/t,1/t,0,..) = 1 -2 t
%C A113279 t^3 * F(3,1/t,1/t,0,..) = -1 + 3 t
%C A113279 t^4 * F(4,1/t,1/t,0,..) = 1 - 4 t + 2 t^2
%C A113279 ... .
%C A113279 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
%C A113279   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.
%C A113279 (End)
%D A113279 M. Herkenhoff Konersmann, Sprokkel XXXI: x_1^n+x_2^n, Nieuw Tijdschr. Wisk, 42 (1954-55) 180.
%H A113279 T. Copeland, <a href="http://tcjpn.wordpress.com/2015/10/12/the-elliptic-lie-triad-kdv-and-ricattt-equations-infinigens-and-elliptic-genera/">Addendum to Elliptic Lie Triad</a>
%H A113279 Wikipedia, <a href="https://en.wikipedia.org/wiki/Newton%27s_identities">Newton's identities</a>.
%F A113279 T(n, k) = (-1)^(n+k)*A034807(n, k).
%F A113279 O.g.f.: 2-(2xt+1)xt/(t+xt+(xt)^2) = (2+x)/(1+x+x^2/t). - _Tom Copeland_, Nov 07 2015
%e A113279 x_1^5+x_2^5 = -p^5 + 5p^3q - 5pq^2, so row 5 reads -1, 5, -5.
%Y A113279 Cf. A034807, A263916.
%K A113279 easy,sign,tabf
%O A113279 0,1
%A A113279 _Floor van Lamoen_, Oct 22 2005