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A338321
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Trace of complement matrix for polynomial triangle centers of degree n (on the Nagel line).
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0
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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FORMULA
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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
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EXAMPLE
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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.
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PROG
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(PARI) f(n) = round((n + 3)^2 / 12); \\ A001399
a(n) = 2*f(n) - f(n-1);
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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