A228782 Irregular triangle read by rows: coefficients of minimal polynomial of a certain algebraic number S2(2*k) from Q(2*cos(Pi/(2*k))) related to the regular (2*k)-gon.
-4, 1, 4, -12, 1, 36, -24, 1, 16, -96, 136, -40, 1, 16, -96, 136, -56, 1, 16, -320, 456, -80, 1, 3136, -12544, 14896, -7168, 1484, -112, 1, 256, -7168, 41216, -73472, 53344, -17472, 2576, -144, 1, 64, -1152, 5424, -6080, 2124, -168, 1, 256, -13312, 62720, -104192, 76384, -26048, 3920, -208, 1
Offset: 1
Examples
The irregular triangle a(k, m) begins: n k / m 0 1 2 3 4 5 6 7 8 2 1: -4 1 4 2: 4 -12 1 6 3: 36 -24 1 8 4: 16 -96 136 -40 1 10 5: 16 -96 136 -56 1 12 6: 16 -320 456 -80 1 14 7: 3136 -12544 14896 -7168 1484 -112 1 16 8: 256 -7168 41216 -73472 53344 -17472 2576 -144 1 ... n = 18, k = 9: 64, -1152, 5424, -6080, 2124, -168, 1; n = 20, k = 10: 256, -13312, 62720, -104192, 76384, -26048, 3920, -208, 1. n = 6, k = 3: p(6, x) = (x - S2(6))*(x - S2(6)^{(1)}), with S2(6) = 12 + 6*rho(6), where rho(6) = sqrt(3). C(6, x) = x^2 - 3 = (x - rho(6))*(x - (-rho(6))), hence rho(6)^{(1)} - -rho(6) and S2(6)^{(1)} = 12 - 6*rho(6). Thus p(6, x) = 144 - 36*rho(6)^2 - 24*x + x^2, reduced with C(6, rho(6)) = 0, i.e., rho(6)^2 = 3; this becomes finally 36 - 24*x + x^2.
Links
- Wolfdieter Lang, The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon, arXiv:1210.1018 [math.GR], 2012-2017.
- Seppo Mustonen, Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations.
- Seppo Mustonen, Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations. [Local copy]
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