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%I #7 Dec 25 2013 02:22:49
%S 0,1,-3,1,5,-5,1,-7,14,-7,1,-3,9,-6,1,-11,55,-77,44,-11,1,13,-91,182,
%T -156,65,-13,1,1,-8,14,-7,1,17,-204,714,-1122,935,-442,119,-17,1,-19,
%U 285,-1254,2508,-2717,1729,-665,152,-19,1,1,-16,60,-78,44,-11,1,-23,506,-3289,9867,-16445,16744,-10948,4692,-1311,230,-23,1
%N Coefficient table for the minimal polynomials of s(2*l+1)^2 = (2*sin(Pi/(2*l+1)))^2.
%C The length of row l is delta(2*l+1) + 1 = A055034(2*l+1) + 1, l >= 0.
%C See the comments on A232631 (even n case) for s(n) = 2*sin(Pi/n) and the minimal polynomial of s(n)^2. Here n = 2*l+1 and s(2*l+1)^2 = 4 - rho(2*l+1)^2 is an integer in the algebraic number field Q(rho(2*l+1)). The minimal polynomial of s(2*l+1)^2 is then MPs2(2*l+1, x) = product(x - 2*(1 + cos(Pi*rpnodd(2*l+1,j)/(2*l+1))), j=1..delta(2*l+1)), l >= 0, where rpnodd(2*l+1) is the list of positive odd numbers < 2*l+1 and relatively prime to 2*l+1. rpnodd(2*l+1,j) is the j-th member of this increasingly ordered list. Here the identity 4 - (2*cos(Pi*(2*k+1)/(2*l+1)))^2 = 2*(1 - cos(Pi*2*(2*k+1)/(2*l+1))) = 2*(1 - (- cos(Pi*(2*l+1 - 2*(2*k+1))/ (2*l+1)))) has been used, and for 2*k+1 < 2*l+1 and gcd(2*k+1, 2*l+1) = 1 this becomes the product given above because 1 = gcd(-(2*k+1), 2*l+1) = gcd(-2*(2*k+1), 2*l+1) = gcd(2*l+1, -2*(2*k+1) + (2*l+1)).
%C This computation was motivated by a preprint of S. Mustonen, P. Haukkanen and J. K. Merikoski, called ``Polynomials associated with squared diagonals of regular polygons'', Nov 16 2013.
%F a(l,m) = [x^m] MPs2(2*l+1, x), l >= 1, m = 0, 1, ...., delta(l), with the minimal polynomial MPs2(l, x) of (2*sin(Pi/(2*l+1)))^2, explained above in a comment.
%e The table a(l,m) begins (n = 2*l+1):
%e ------------------------------------------------------------------------------------------------------
%e n, l\m 0 1 2 3 4 5 6 7 8 9 10 11 ...
%e 1, 0: 0 1
%e 3, 1: -3 1
%e 5, 2: 5 -5 1
%e 7, 3: -7 14 -7 1
%e 9, 4: -3 9 -6 1
%e 11, 5: -11 55 -77 44 -11 1
%e 13, 6: 13 -91 182 -156 65 -13 1
%e 15, 7: 1 -8 14 -7 1
%e 17, 8: 17 -204 714 -1122 935 -442 119 -17 1
%e 19, 9: -19 285 -1254 2508 -2717 1729 -665 152 -19 1
%e 21, 10: 1 -16 60 -78 44 -11 1
%e 23, 11: -23 506 -3289 9867 -16445 16744 -10948 4692 -1311 230 -23 1
%e 25, 12: 5 -125 875 -2675 4300 -4005 2275 -800 170 -20 1
%e 27, 13: -3 81 -540 1386 -1782 1287 -546 135 -18 1
%e ....
%e n=29, l=14: 29,-1015,10556,-51272,140998,-243542,281010,-224808,127281,-51359,14674, -2900,377,-29,1.
%e n=31, l=15: -31, 1240, -14756, 82212, -260338, 520676, -700910, 660858, -447051, 219604, -78430, 20150, -3627, 434, -31, 1.
%e ...
%e The minimal polynomial of s(5)^2 = (2*sin(Pi/5))^2 = 4 - rho(5)^2
%e = 2*(1 - cos(Pi*2/5)) = 2*(1 + cos(Pi*3/5)), approx. 1.381966, is MPs2(5, x) = product(x - 2*(1 + cos(Pi*rpnodd(5,j)/5)), j=1..2) = (x - 2*(1 + cos(Pi/5))*(x - 2*(1 + cos(Pi*3/5)) = (x - (2 + phi)*(x - (2 + 1 - phi)) = x^2 - 5*x + (6 + phi - phi^2) = x^2 - 5*x +5, where phi = rho(5) is the golden section.
%e The row n=17 checks with WolframAlpha's MinimalPolynomial[(2*sin(Pi/17))^2 ,x] = 17-204 x+714 x^2-1122 x^3+935 x^4-442 x^5+119 x^6-17 x^7+x^8.
%t Flatten[ CoefficientList[ Table[ MinimalPolynomial[ (2*Sin[Pi/(2*l+1)])^2, x], {l, 0, 15}], x]] (adapted from Jean-François Alcover, A187360) - Wolfdieter Lang, Dec 23 2013
%Y Cf. A232631 (even n), A232633 (all n), A055034.
%K sign,tabf,easy
%O 0,3
%A _Wolfdieter Lang_, Dec 18 2013
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