

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.


1



1, 1, 1, 6, 1, 1, 14, 1, 1, 28, 70, 28, 1, 1, 44, 166, 44, 1, 1, 60, 134, 60, 1, 1, 90, 911, 2092, 911, 90, 1, 1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1, 1, 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
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OFFSET

1,4


COMMENTS

The length of row No. l of this table is delta(2*l) + 1 = A055034(2*l) + 1.
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 lengths 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).
The minimal polynomial for sqLhat(2*l) = sum(A230072(l,m)* rho(2*l)^m, m=0..delta(2*l)), 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(x  substitute(rho(2*l) = z, sqLhat(2*l)^{(j)}), j = 0.. delta(2*l)1) (mod C(2*l, z)).
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 ngon, 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 length come precisely n times in the regular ngon 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).
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).
S. Mustonen also conjectured about the other zeros of his polynomials.


LINKS

Table of n, a(n) for n=1..75.
Wolfdieter Lang, The field Q(2cos(pi/n)), its Galois group and length ratios in the regular ngon.
Seppo Mustonen, Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations.


FORMULA

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.
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.


EXAMPLE

The table a(l,m) (n = 2*l) starts: (row length A055034(2*l))
l, n\m 0 1 2 3 4 5 6 7 8
1, 2: 1 1
2, 4: 1 6 1
3, 6: 1 14 1
4, 8: 1 28 70 28 1
5, 10: 1 44 166 44 1
6, 12: 1 60 134 60 1
7, 14: 1 90 911 2092 911 90 1
8, 16: 1 120 1820 8008 12870 8008 1820 120 1
9, 18: 1 138 975 1868 975 138 1
10, 20: 1 184 3740 16136 25414 16136 3740 184 1
...
11, 22: 1 230 7085 67528 252242 394404 252242 67528 7085 230 1
12, 24: 1 248 3612 16072 25670 16072 3612 248 1
13, 26: 1 324 14626 215604 1346671 3965064 5692636 3965064 1346671 215604 14626 324 1
14, 28: 1 372 18242 266916 1488367 3925992 5377436 3925992 1488367 266916 18242 372 1
15, 30: 1 376 4380 15944 24134 15944 4380 376 1
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.
l = 5, n=10: (decagon) = psqLhat(5, x) = 1  44*x + 166*x^2  44*x^3 + x^4 with thefour 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.
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*(4920*sqrt(6))) +
98*sqrt(2*(4920*sqrt(6))), 15 + 6*sqrt(6)  80*sqrt(3*(4920*sqrt(6)))  98*sqrt(2*(4920*sqrt(6))), 15  6*sqrt(6) + 2*sqrt(2*(4920*sqrt(6))), 15  6*sqrt(6)  2*sqrt(2*(4920*sqrt(6))), approximately 57.69548054, 1.69839638, 0.58879070, 0.01733238, respectively.
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.


CROSSREFS

Cf. A055034, A187360, A228780, A230072 (sqLhat(2*l)).
Sequence in context: A202875 A203956 A082105 * A143210 A205133 A152238
Adjacent sequences: A230070 A230071 A230072 * A230074 A230075 A230076


KEYWORD

sign,tabf


AUTHOR

Wolfdieter Lang, Oct 09 2013


STATUS

approved



