OFFSET
0,1
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)
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.
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
EXAMPLE
x_1^5+x_2^5 = -p^5 + 5p^3q - 5pq^2, so row 5 reads -1, 5, -5.
CROSSREFS
KEYWORD
easy,sign,tabf
AUTHOR
Floor van Lamoen, Oct 22 2005
STATUS
approved