

A228781


Irregular triangle read by rows: coefficients of minimal polynomial of a certain algebraic number S2(2*k+1) from Q(2*cos(Pi/n)) related to the regular (2*k+1)gon, k >= 1.


3



3, 1, 5, 10, 1, 7, 35, 21, 1, 3, 27, 33, 1, 11, 165, 462, 330, 55, 1, 13, 286, 1287, 1716, 715, 78, 1, 1, 28, 134, 92, 1, 17, 680, 6188, 19448, 24310, 12376, 2380, 136, 1, 19, 969, 11628, 50388, 92378, 75582, 27132, 3876, 171, 1, 1, 58, 655, 1772, 1423, 186, 1
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OFFSET

1,1


COMMENTS

The row length sequence of this table is delta(2*k+1), with the degree delta(n) = A055034(n) of the algebraic number rho(n):= 2*cos(Pi/n), k >= 1.
The 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 odd n case, n = 2*k + 1 is considered. S2(n) is the square of the sum of the distinct length ratios side/radius or diagonal/radius with the radius of the circle in which a regular ngon 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+1) has degree delta(2*k+1) and is given by
p(2*k+1,x) = Product_{j=1..delta(2*k+1)} (x  S2(2*k+1)^{(j1)} (mod C(2*k+1,delta(n))) = sum(a(k, m)*x^m, m = 0..delta(2*k+1)), where S2(2*k+1)^{(0)} = S2(2*k+1) and S2(2*k+1)^{(j1)} is the (j1)th conjugate of S2(2*k+1). The conjugate of a number alpha(n) = Sum_{j=0..(delta(n)1)} b(n, j)*rho(n)^j in Q(rho(n)) is obtained from the conjugates of rho(n), given in turn by the zeros x(n, j) of the minimal polynomial C(n, x) (see A187360 and the link to the W. Lang Galois paper, tables 2 and 3) as rho(n)^{(j1)} = x(n, j), j = 1..delta(n), with rho(n)^{(0)} = rho(n).
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 ngon. See his paper given as a link below. The author thanks Seppo Mustonen for sending his paper.
If the minimal polynomial of the algebraic number S2(n) in the ngon with n = 2*k+1 is p(n, x) then the minimal polynomial of the square of the sum of the length of all n sides and n*(n3)/2 diagonals is P(n, x) = n^(2*delta(n))*p(n, x/n^2).


LINKS

Table of n, a(n) for n=1..57.
Wolfdieter Lang, The field Q(2cos(pi/n)), its Galois group and length ratios in the regular ngon, arXiv:1210.1018 [math.GR], 20122017.
Seppo Mustonen, Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations.
Seppo Mustonen, Lengths of edges and diagonals and sums of them in regular polygons as roots of algebraic equations [Local copy]


FORMULA

a(k, m) = [x^m] p(2*k+1, x), with the minimal polynomial p(2*k+1, x) of S2(2*k+1) given in the power basis in A228780. p(2*k+1, x) is given in a comment above in terms of the S2(2*k+1) and its conjugates S2(2*k+1)^{(j1)}, j=2, ..., delta(2*k+1), where delta(n) = A055034(n).
Conjecture from Seppo Mustonen, rewritten for the p(n, x) coefficients for odd primes: p(prime(j), x) = Sum_{i=0..imax(j)} (1)^(imax(j  i))* binomial(prime(j), 2*i+1)*x^i, with imax(j) = (prime(j)1)/2. See the adapted eq. (5) of the S. Mustonen paper.


EXAMPLE

The irregular triangle a(k, m) begins:
n k /m 0 1 2 3 4 5 6 7 8
3 1: 3 1
5 2: 5 10 1
7 3: 7 35 21 1
9 4: 3 27 33 1
11 5: 11 165 462 330 55 1
13 6: 13 286 1287 1716 715 78 1
15 7: 1 28 134 92 1
17 8: 17 680 6188 19448 24310 12376 2380 136 1
...
n = 19, L = 9: 19, 969, 11628, 50388, 92378, 75582, 27132, 3876, 171, 1.
n = 21, L = 10: 1, 58, 655, 1772, 1423, 186, 1.
p(5, x) = (x  S2(5))*(x  S2(5)^{(1)}), with S2(5) = 3 + 4*rho(5), where rho(5)=phi, the golden section. C(5, x) = x^2  x  1 = (x  rho(5))*(x  (1rho(5))), hence rho(5)^{(1)} = 1rho(5), and S2(5)^{(1)} = 3 + 4*(1  rho(5)) = 7  4*rho(5). Thus p(5, x) = 16*rho^2 + 21 + 16*rho 10*x + x^2 which becomes modulo C(5,rho(5)), i.e., using rho(5)^2 = rho(5) + 1, finally p(n, 5) = 5  10*x + x^2.
Conjecture (Seppo Mustonen): p(5, x) = binomial(5, 1)  binomial(5, 3)*x + binomial(5, 5)* x^2 = 5  10*x + x^2.


CROSSREFS

Cf. A055034, A187360, A228780, A228782 (even case).
Sequence in context: A146255 A331432 A122366 * A103327 A177463 A065229
Adjacent sequences: A228778 A228779 A228780 * A228782 A228783 A228784


KEYWORD

sign,tabf


AUTHOR

Wolfdieter Lang, Oct 01 2013


STATUS

approved



