cp's OEIS Frontend

A230073 Coefficients of the minimal polynomials of the algebraic numbers sqLhat(2*l) from A230072, l >= 1, related to the square of all length in a regular (2*l)-gon inscribed in a circle of radius 1 length unit.

%I A230073 #15 Jun 22 2024 16:05:35
%S A230073 -1,1,1,-6,1,1,-14,1,1,-28,70,-28,1,1,-44,166,-44,1,1,-60,134,-60,1,1,
%T A230073 -90,911,-2092,911,-90,1,1,-120,1820,-8008,12870,-8008,1820,-120,1,1,
%U A230073 -138,975,-1868,975,-138,1,1,-184,3740,-16136,25414,-16136,3740,-184,1,1,-230,7085,-67528,252242,-394404,252242,-67528,7085,-230,1,1,-248,3612,-16072,25670,-16072,3612,-248,1
%N A230073 Coefficients of the minimal polynomials of the algebraic numbers sqLhat(2*l) from A230072, l >= 1, related to the square of all length in a regular (2*l)-gon inscribed in a circle of radius 1 length unit.
%C A230073 The length of row no. l of this table is delta(2*l) + 1 = A055034(2*l) + 1.
%C A230073 sqLhat(2*l), l >= 1, from A230072, an algebraic number (in fact integer) of degree delta(2*l) over the rationals, gives the square of the sum of the length ratios of all lines/R (also called chords/R) divided by (2*l)^2 in a regular (2*l)-gon inscribed in a circle of radius R. This number lives in the algebraic number field Q(rho(2*l)), with rho(2*l) = 2*cos(Pi/(2*l)) (see A187360 and the W. Lang link below for this number field).
%C A230073 The minimal polynomial for sqLhat(2*l) = Sum_{m=0..delta(2*l)} A230072(l,m)* rho(2*l)^m, called here psqLhat(l, x), is computed from the conjugates rho(2*l)^{(j)}, j = 0, ..., delta(2*l)-1, with rho(2*l)^{(0)} = rho(2*l), by calculating the conjugates sqLhat(2*l)^{(j)}, j = 0, ..., delta(2*l)-1, which are polynomials in rho(2*l) =: z, with the usual rules for conjugation. All results have to be taken modulo the minimal polynomial C(2*l, z) of rho(2*l) (see A187360 table 2 and section 3 for C(n,x)), in order to obtain finally elements written in the power basis of the field Q(rho(2*l)). The conjugates rho(2*l)^{(j)} are just the delta(2*l) roots of C(2*l, z). Therefore, psqLhat(l, x) = (Product_{j=0.. delta(2*l)-1} (x - substitute(rho(2*l) = z, sqLhat(2*l)^{(j)}))) mod C(2*l, z).
%C A230073 Thanks go to _Seppo Mustonen_, who asked me to look into this matter. I thank him for sending the below given link to his work about the square of the sum of all lengths in an n-gon, called there L(n)^2. Here n is even (n=2*l) and sqLhat(2*l) = (L(n)^2)/n^2. The odd n case is obtained from A228780 as L(2*l+1)^2 = (2*l+1)^2*S2(2*l+1) (observing that all distinct line lengths occur precisely n times in the regular n-gon if n is odd). His polynomials given in his eq. (6) (here for the n even case) are in general not monic, and not irreducible. Instead one should consider the minimal (monic, irreducible and integer) polynomials PsqL(l, x) := (2*l)^(2*delta(2*l))* psqLhat(l, x/(2*l)^2), l >= 1 (for n = 2*l).
%C A230073 Mustonen's polynomials from his eq. (6) for the even n case coincide with PsqL(l, x) precisely for l = 2^k, k>=1, and for l = 1 (k=0) one has to take the negative. In all other cases the degrees do not fit (Mustonen's polynomials become reducible over the integers). His conjecture for the coefficients can then be rewritten as a conjecture for the present polynomials psqLhat(2^k, x), k >= 0 (see the formula section).
%C A230073 S. Mustonen also conjectured about the other zeros of his polynomials.
%H A230073 Wolfdieter Lang, <a href="https://arxiv.org/abs/1210.1018">The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon</a>, arXiv:1210.1018 [math.GR], 2012-2017.
%H A230073 Seppo Mustonen, <a href="http://www.survo.fi/papers/Roots2013.pdf"> Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations</a>.
%F A230073 a(l,m) = [x^m](psqLhat(l, x)), l >= 1, m = 0, ..., delta(2*l), with delta(2*l) = A055034(2*l), and the formula for psqLhat(2*l, x) is given in a comment above.
%F A230073 Mustonen's conjecture (adapted, see a comment above) is: a(2^k,m) = ((-1)^m)*binomial(2*2^k,2*m), k >= 1, and for k=0 a(1,0) = -1 and a(1,1) = 1 is trivial.
%e A230073 The table a(l,m) (n = 2*l) starts: (row length A055034(2*l))
%e A230073 l,    n\m 0    1     2      3      4      5     6     7  8
%e A230073 1,   2:  -1    1
%e A230073 2,   4:   1   -6     1
%e A230073 3,   6:   1  -14     1
%e A230073 4,   8:   1  -28    70    -28      1
%e A230073 5,  10:   1  -44   166    -44      1
%e A230073 6,  12:   1  -60   134    -60      1
%e A230073 7,  14:   1  -90   911  -2092    911    -90     1
%e A230073 8,  16:   1 -120  1820  -8008  12870  -8008  1820  -120  1
%e A230073 9,  18:   1 -138   975  -1868    975   -138     1
%e A230073 10, 20:   1 -184  3740 -16136  25414 -16136  3740  -184  1
%e A230073 ...
%e A230073 11, 22:  1 -230 7085 -67528 252242 -394404 252242 -67528 7085  -230   1
%e A230073 12, 24: 1 -248 3612 -16072  25670 -16072 3612 -248 1
%e A230073 13, 26: 1 -324 14626 -215604 1346671 -3965064 5692636 -3965064 1346671 -215604 14626 -324 1
%e A230073 14, 28:  1 -372 18242 -266916 1488367 -3925992 5377436 -3925992 1488367 -266916 18242 -372 1
%e A230073 15, 30:  1 -376  4380  -15944  24134  -15944  4380 -376 1
%e A230073 l = 3, n=6: (hexagon) psqLhat(3, x) = 1 - 14*x + x^2. The two roots are positive: 7 + 4*sqrt(3) = sqLhat(3) and 7 - 4*sqrt(3). For the square of the sum of all length ratios one has PsqL(3, x) = 1296 - 504*x + x^2, with the previous two roots scaled by a factor 36.
%e A230073 l = 5, n=10: (decagon) = psqLhat(5, x) = 1 - 44*x + 166*x^2 - 44*x^3 + x^4 with the four positive roots sqLhat(10) = 7 + 8*phi + 4*sqrt(7+11*phi), 15 - 8*phi + 4*sqrt(18 - 11*phi), 15 - 8*phi - 4*sqrt(18 - 11*phi), 7 + 8*phi - 4*sqrt(7 + 11*phi), approximately 39.86345819, 3.851840015, 0.259616169, 0.02508563, respectively, where phi = rho(5) = (1+sqrt(5))/2 (the golden section). PsqL(5, x) =100000000 - 44000000*x + 1660000*x^2 - 4400*x^3 + x^4, with the previous four roots scaled by a factor 100.
%e A230073 l=6, n = 12: (dodecagon) psqLhat(6, x) = 1 - 60*x + 134*x^2 - 60*x^3 + x^4, with the four positive roots sqLhat(12) = 15 + 6*sqrt(6) + 80*sqrt(3*(49-20*sqrt(6))) + 98*sqrt(2*(49-20*sqrt(6))), 15 + 6*sqrt(6) - 80*sqrt(3*(49-20*sqrt(6))) - 98*sqrt(2*(49-20*sqrt(6))), 15 - 6*sqrt(6) + 2*sqrt(2*(49-20*sqrt(6))), 15 - 6*sqrt(6) - 2*sqrt(2*(49-20*sqrt(6))), approximately 57.69548054, 1.69839638, 0.58879070, 0.01733238, respectively.
%e A230073 Mustonen's conjecture for rows no. l = 2^k, k >= 1 (see a comment above): l = 8 (k=3): ((-1)^m)*binomial(16,2*m), m = 0..8: [1, -120, 1820, -8008, 12870, -8008, 1820, -120, 1], with obvious symmetry.
%Y A230073 Cf. A055034, A187360, A228780, A230072 (sqLhat(2*l)).
%K A230073 sign,tabf
%O A230073 1,4
%A A230073 _Wolfdieter Lang_, Oct 09 2013