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A228782
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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.
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2
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-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
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OFFSET
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1,1
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COMMENTS
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The row length sequence of this table is delta(2*k), k >= 1, with the degree delta(n) = A055034(n) of the algebraic number rho(n):=2*cos(Pi/n).
The algebraic numbers S2(n) have been given in A228780 in the power basis of the degree delta(n) number field Q(rho(n)), with rho(n):=2*cos(Pi/n), n >= 2. Here the even case, n = 2*k, is considered. S2(n) is the square of the sum of the distinct length ratios side/radius and diagonal/radius with the radius of the circle in which a regular n-gon is inscribed. For two formulas for S2(n) in terms of powers of rho(n) see the comment section of A228780.
The minimal (monic) polynomial of S2(2*k) has degree delta(2*k) and is given by p(2*k,x) = Product_{j=1..delta(2*k)} (x - S2(2*k)^{(j-1)}) (mod C(2*k, rho(2*k))) = Sum_{m=0..delta(2*k)} a(k, m)*x^m, where S2(2*k)^{(0)} = S2(2*k) and S2(2*k)^{(j-1)} is the (j-1)-th conjugate of S2(2*L). For the conjugate of an algebraic number in Q(rho(n)) see a comment on A228781.
The motivation to look into this problem originated from emails by Seppo Mustonen who found experimentally polynomials which had as one zero the square of the total length/radius of all chords (sides and diagonals) in the regular n-gon. See his paper given as a link below. The author thanks Seppo Mustonen for sending his paper.
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LINKS
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FORMULA
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a(k,m) = [x^m] p(2*k, x), with the minimal polynomial p(2*k, x) of S2(2*k) given in the power basis in A228780.
p(2*k, x) is given in a comment above in terms of the S2(2*k) and its conjugates S2(2*k)^{(j-1)}, j = 2, ..., delta(2*k), where delta(2*k) = A055034(2*k).
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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sign,tabf
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AUTHOR
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STATUS
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approved
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