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%I #47 Nov 05 2020 15:36:58
%S 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,
%T 75,80,85,90,97,101,108,114,120,126,134,139,147,154,161,168,177,183,
%U 192,200,208,216,226,233,243,252,261,270,281,289,300,310,320,330
%N Trace of complement matrix for polynomial triangle centers of degree n (on the Nagel line).
%C 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):
%C f(t*a, t*b, t*c) = t^n * f(a,b,c), where n is the degree of the polynomial;
%C f(a,b,c) = f(a,c,b).
%C 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).
%C 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.
%C 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.
%C 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.
%C 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".
%C The eigenvalues are 2 and -1 with multiplicity A001399(n) and A001399(n-1) respectively.
%C 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.
%H Clark Kimberling, <a href="https://www.emis.de/journals/JIS/VOL22/Kimberling/kimb9.html">A Combinatorial Classification of Triangle Centers on the Line at Infinity</a>, J. Int. Seq., Vol. 22 (2019), Article 19.5.4.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,0,-1,-1,1).
%F a(n) = 2*A001399(n) - A001399(n-1).
%F 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
%e 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.
%e For n=1:
%e The basis is {a+b+c, a}.
%e Let T represent the complement transformation, then
%e T(a) = b+c = (a+b+c) - a
%e T(a+b+c) = 2(a+b+c)
%e So the corresponding matrix is
%e (2 1)
%e (0 -1)
%e Its trace is 1, giving a(1) = 1.
%e For n=2:
%e The basis is {s(a^2), s(a*b), a*s(a) }.
%e T(s(a^2)) =2s(a^2)
%e T(s(ab)) =2s(a*b)
%e T(a*s(a)) =s(a^2) +2s(a*b) -a*s(a)
%e The corresponding matrix is
%e (-1, 0, 0)
%e ( 1, 2, 0)
%e ( 2, 0, 2)
%e Its trace is 3, giving a(2) = 3.
%e For n=3:
%e The basis is {s(a^3), s(a^2*b), a*b*c, a*s(a^2), a*s(a*b) }.
%e T(s(a^3)) =2s(a^3)
%e T(s(a^2*b)) =2s(a^2*b)
%e T(a*b*c) =2a*b*c
%e T(a*s(a^2)) =s(a^2*b) +s(a^3) -a*s(a^2)
%e T(a*s(a*b)) =s(a^2*b) +3a*b*c -a*s(ab)
%e Thus the corresponding matrix is
%e (2, 0, 0, 1, 0)
%e (0, 2, 0, 1, 1)
%e (0, 0, 2, 0, 3)
%e (0, 0, 0,-1, 0)
%e (0, 0, 0, 0,-1)
%e Its trace is 4, thus a(3) =4.
%o (PARI) f(n) = round((n + 3)^2 / 12); \\ A001399
%o a(n) = 2*f(n) - f(n-1);
%Y Cf. A001399.
%K nonn,easy
%O 1,2
%A _Suren Suren_, Oct 22 2020
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