This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A327921 #30 Dec 11 2021 02:34:12
%S A327921 1,0,1,5,1,3,1,3,7,9,1,1,3,5,9,11,13,1,3,5,7,1,5,7,11,13,17,1,2,1,3,5,
%T A327921 7,9,13,15,17,19,21,1,5,7,11,1,3,5,7,9,11,15,17,19,21,23,25,1,3,5,1,7,
%U A327921 11,13,17,19,23,29,1,3,5,7,9,11,13,15
%N A327921 Irregular triangle T read by rows: row n gives the values determining the zeros of the minimal polynomial ps(n, x) of 2*sin(Pi/n) (coefficients in A228786), for n >= 1.
%C A327921 The minimal polynomials of the algebraic number s(n) = 2*sin(Pi/n) of degree gamma(n) = A055035(n) = A093819(2*n) has the zeros 2*cos(2*Pi*T(n,m)/c(2*n)), with c(2*n) = A178182(2*n), for m = 1, 2, ..., gamma(n) and n >= 1.
%C A327921 The number s(n) is the length ratio side(n)/R of the regular n-gon inscribed in a circle of radius R.
%C A327921 The motivation to look at these zeros came from the book of Carl Schick, and the paper by Brändli and Beyne. There, only length ratios diagonals/R in 2*(2*m + 1)-gons, for m >= 1, are considered.
%C A327921 If one is interested in length ratios diagonals/side then the minimal polynomials of rho(n) := 2*cos(Pi/n) (smallest diagonal/side) are important. These are given in A187360, called there C(n, x).
%D A327921 Carl Schick, Trigonometrie und unterhaltsame Zahlentheorie, ISBN 3-9522917-0-6, Bobos Druck, Zürich, 2003.
%H A327921 Gerold Brändli and Tim Beyne, <a href="https://arxiv.org/abs/1504.02757">Modified Congruence Modulo n with Half the Amount of Residues</a>, arXiv:1504.02757v2 [math.NT], 2015 and 2016.
%F A327921 Row n gives the first gamma(n) = A055035(n) members of RRS(c(2*n)), for n >= 1, where RRS(k) is the smallest nonnegative restricted residue system modulo k.
%F A327921 The numbers with odd c(2*n) are n = 2 + 8*k, k >= 0.
%F A327921 The zeros x0^{(n)}_m := 2*cos(2*Pi*T(n,m)/c(2*n)) can be written as polynomials of rho(n) := 2*cos(Pi/n) for even n, and as polynomials of rho(2*n) for odd n as follows. x0^{(n)}_m = R(t*T(n,m), rho(b*n)), with b = 1 or 2 for n even or odd, respectively, and t = 1 for n == 1 (mod 2) and 0 (mod 4), t = 2 and 4 for n == 6 and 2 (mod 8), respectively. Here the monic Chebyshev T polynomials R(n, x) enter, with coefficients given in A127672. This results from 2*n/c(2*n) = 4, 2, 1, 1/2 for n == 2, 6 (mod 8), 0 (mod 4), 1 (mod 2), respectively. Note that rho(n)^2 = 4 - s(n)^2.
%F A327921 In terms of s(n) = 2*sin(Pi/n) the zeros x0^{(n)}_m are written with Chebyshev S (A049310) and R polynomials (A127672) as follows.
%F A327921 x0^{(n)}_m = sqrt(4 - s(b*n)^2) * {S((T(n,m)-1)/2, -R(2, s(bn))) - S((T(n,m)-3)/2, -R(2, s(b*n)))}, for n == 1 (mod 2) with b(n) = 2, and for n == 0 (mod 4) with b = 1,
%F A327921 x0^{(n)}_m = (2 - s(n)^2) * {S((T(n,m)-1)/2, R(4, s(n))) - S((T(n,m)-3)/2, R(4, s(n)))}, for n == 6 (mod 8), and
%F A327921 x0^{(n)}_m = R(T(n,m), R(4, sqrt(4 - s(n)^2))), for n == 2 (mod 8).
%e A327921 The irregular triangle T(n,m) begins:
%e A327921 n\m 1 2 3 4 5 6 7 8 9 10 11 12 ... A178182(2*n) A055035(n)
%e A327921 -------------------------------------------------------------------------
%e A327921 1: 1 4 1
%e A327921 2: 0 1 1
%e A327921 3: 1 5 12 2
%e A327921 4: 1 3 8 2
%e A327921 5: 1 3 7 9 20 4
%e A327921 6: 1 6 1
%e A327921 7: 1 3 5 9 11 13 28 6
%e A327921 8: 1 3 5 7 16 4
%e A327921 9: 1 5 7 11 13 17 36 6
%e A327921 10: 1 2 5 2
%e A327921 11: 1 3 5 7 9 13 15 17 19 21 44 10
%e A327921 12: 1 5 7 11 24 4
%e A327921 13: 1 3 5 7 9 11 15 17 19 21 23 25 52 12
%e A327921 14: 1 3 5 14 3
%e A327921 15: 1 7 11 13 17 19 23 29 60 8
%e A327921 16: 1 3 5 7 9 11 13 15 32 8
%e A327921 ...
%e A327921 --------------------------------------------------------------------------
%e A327921 Some zeros are:
%e A327921 n = 1: 2*cos(2*Pi*1/4) = 0 = s(1),
%e A327921 n = 2: 2*cos(2*Pi*1/4) = 2 = s(2) (diameter/R),
%e A327921 n = 3: 2*cos(2*Pi*1/12) = -2*cos(2*Pi*5/12) = sqrt(3) = s(3),
%e A327921 n = 5: 2*cos(2*Pi*1/20) = -2*cos(2*Pi*9/20) = sqrt(2 + tau),
%e A327921 2*cos(2*Pi*3/20) = -2*cos(2*Pi*7/20) = sqrt(tau - 3) = s(5),
%e A327921 with the golden ratio tau = A001622,
%e A327921 n = 10: 2*cos(2*Pi*1/5) = tau - 1 = s(10), -2*cos(2*Pi*2/5) = -tau.
%e A327921 --------------------------------------------------------------------------
%Y A327921 Cf. A001622, A049310, A055035, A093819, A127672, A178182, A187360, A228786.
%K A327921 nonn,easy,tabf
%O A327921 1,4
%A A327921 _Wolfdieter Lang_, Nov 02 2019