A338321 Trace of complement matrix for polynomial triangle centers of degree n (on the Nagel line).

1, 3, 4, 5, 6, 9, 9, 12, 14, 16, 18, 22, 23, 27, 30, 33, 36, 41, 43, 48, 52, 56, 60, 66, 69, 75, 80, 85, 90, 97, 101, 108, 114, 120, 126, 134, 139, 147, 154, 161, 168, 177, 183, 192, 200, 208, 216, 226, 233, 243, 252, 261, 270, 281, 289, 300, 310, 320, 330

Offset: 1

Author

Suren Suren, Oct 22 2020

Keywords

• nonn
• easy

Comments

Let ABC be a triangle with sidelengths a,b,c. Treating a,b,c as variables, a polynomial triangle center is defined to be a point with barycentric coordinates of the form f(a,b,c) : f(b,c,a) : f(c,a,b), where p(a,b,c) is a polynomial satisfying these two conditions (homogeneity and bisymmetry):

f(t*a, t*b, t*c) = t^n * f(a,b,c), where n is the degree of the polynomial;

f(a,b,c) = f(a,c,b).

Examples include the incenter, I = a : b : c, the centroid, G = 1 : 1 : 1, and the circumcenter, O = f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = a^2 (b^2 + c^2 - a^2).

The Nagel line of ABC is the line IG, which consists of all points x:y:z: such that (b-c)*x + (c-a)*y + (a-b)*z = 0.

The complement of a point P is the point U such that PU : UG = 3 : -1. If P has barycentric coordinates p : q : r, then its complement is U = q+r : r+p : p+q.

For n >= 1, Kimberling found a basis for polynomials triangle centers of degree n on the Nagel line; the basis consists of symmetric polynomials of degree n and a*(symmetric polynomials of degree n-1.

Seeing that complement is a linear transformation on the vector space of polynomial triangle centers of degree n, we can represent it as a matrix, which we call the "complement matrix for polynomial triangle centers of degree n".

The eigenvalues are 2 and -1 with multiplicity A001399(n) and A001399(n-1) respectively.

In homogenous barycentric coordinates, the points p:q:r and kp:kq:kr are the same, hence the eigenvectors are the fixed points of the complement transformation applied to polynomial triangle centers on the Nagel line. Although there are more than 2 eigenvectors, there are only 2 fixed points, the centroid and the intersection of the Nagel line and the line at infinity (the points m:m:m, where m is a symmetric function of a, b, c is the same as the centroid (1:1:1), this is why the same point appears as different eigenvectors). The eigenvalues associated with the centroid as a fixed point are equal to 2, and the eigenvalues associated with the point at infinity as the fixed point are equal to -1.

Examples

			S represents the sum taken over all distinct permutations of a, b, c, for example, s(a^2*b) means a^2*b+a^2*c+b^2*c+b^2*a+c^2*a+c^2*b, s(a) means a+b+c.
For n=1:
The basis is {a+b+c, a}.
Let T represent the complement transformation, then
  T(a) = b+c = (a+b+c) - a
  T(a+b+c) = 2(a+b+c)
So the corresponding matrix is
  (2  1)
  (0 -1)
Its trace is 1, giving a(1) = 1.
For n=2:
The basis is {s(a^2), s(a*b), a*s(a) }.
  T(s(a^2)) =2s(a^2)
  T(s(ab)) =2s(a*b)
  T(a*s(a)) =s(a^2) +2s(a*b) -a*s(a)
The corresponding matrix is
  (-1, 0, 0)
  ( 1, 2, 0)
  ( 2, 0, 2)
Its trace is 3, giving a(2) = 3.
For n=3:
The basis is {s(a^3), s(a^2*b), a*b*c, a*s(a^2), a*s(a*b) }.
  T(s(a^3)) =2s(a^3)
  T(s(a^2*b)) =2s(a^2*b)
  T(a*b*c) =2a*b*c
  T(a*s(a^2)) =s(a^2*b) +s(a^3) -a*s(a^2)
  T(a*s(a*b)) =s(a^2*b) +3a*b*c -a*s(ab)
Thus the corresponding matrix is
  (2, 0, 0, 1, 0)
  (0, 2, 0, 1, 1)
  (0, 0, 2, 0, 3)
  (0, 0, 0,-1, 0)
  (0, 0, 0, 0,-1)
Its trace is 4, thus a(3) =4.
		

Links

• Clark Kimberling, A Combinatorial Classification of Triangle Centers on the Line at Infinity, J. Int. Seq., Vol. 22 (2019), Article 19.5.4.
• Index entries for linear recurrences with constant coefficients, signature (1,1,0,-1,-1,1).

Crossrefs

Cf. A001399.

Programs

PARI

f(n) = round((n + 3)^2 / 12); \\ A001399
a(n) = 2*f(n) - f(n-1);

Formula

a(n) = 2*A001399(n) - A001399(n-1).

G.f.: -x*(2*x^5-2*x^4-2*x^3+2*x+1)/((x+1)*(x^2+x+1)*(x-1)^3). - Alois P. Heinz, Oct 22 2020