cp's OEIS Frontend

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.

Showing 1-3 of 3 results.

A228786 Table of coefficients of the minimal polynomials of 2*sin(Pi/n), n >= 1.

Original entry on oeis.org

0, 1, -2, 1, -3, 0, 1, -2, 0, 1, 5, 0, -5, 0, 1, -1, 1, -7, 0, 14, 0, -7, 0, 1, 2, 0, -4, 0, 1, -3, 0, 9, 0, -6, 0, 1, -1, 1, 1, -11, 0, 55, 0, -77, 0, 44, 0, -11, 0, 1, 1, 0, -4, 0, 1, 13, 0, -91, 0, 182, 0, -156, 0, 65, 0, -13, 0, 1, 1, -2, -1, 1, 1, 0, -8, 0, 14, 0, -7, 0, 1, 2, 0, -16, 0, 20, 0, -8, 0, 1, 17, 0, -204, 0, 714, 0, -1122, 0, 935, 0, -442, 0, 119, 0, -17, 0, 1, 1, -3, 0, 1
Offset: 1

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Author

Wolfdieter Lang, Oct 07 2013

Keywords

Comments

s(n) := 2*sin(Pi/n) is, for n >= 2, the length ratio side/R of the regular n-gon inscribed in a circle of radius R. This algebraic number s(n), n >= 1, has the degree gamma(n) := A055035(n), and the row length of this table is gamma(n) + 1.
s(n) has been given in the power basis of the relevant algebraic number field in A228783 for even n (bisected into n == 0 (mod 4) and n == 2 (mod 4)), and in A228785 for odd n.
For the computation of the minimal polynomials ps(n,x), using the coefficients of s(n) in the relevant number field, and the conjugates of the corresponding algebraic numbers rho (giving the length ratios (smallest diagonal)/side in the relevant regular polygons see a comment on A228781. Note that the product of the gamma(n) linear factors (x - conjugates) has to be computed modulo the minimal polynomial of the relevant rho(k) = 2*cos(Pi/k) (called C(k,x=rho(k)) in A187360).
Thanks go to Seppo Mustonen, who asked a question about the square of the sum of all lengths in the regular n-gon, which led to this computation of s(n) and its minimal polynomial.
It would be interesting to find out which length ratios in the regular n-gon give the other positive zeros of the minimal polynomial ps(n,x). See some examples below.
The zeros of the row polynomials ps(n,x) are 2*cos(2*Pi*k/c(2*n))) for gcd(k, c(2*n)) = 1, where c(n) = A178182(n), and k from {0, ..., floor(c(2*n)/2)}, for n >= 1. The number of these solutions is gamma(n) = A055035(n). See the formula section. This results from the zeros of the minimal polynomials of sin(2*Pi/n), with coefficients given in A181872/A181873. - Wolfdieter Lang, Oct 30 2019

Examples

			The table a(n, m) starts:
n\m   0  1    2 3   4 5     6 7   8 9   10 12  13 14  15 16 17
1:    0  1
2:   -2  1
3:   -3  0    1
4:   -2  0    1
5:    5  0   -5 0   1
6:   -1  1
7:   -7  0   14 0  -7 0     1
8:    2  0   -4 0   1
9:   -3  0    9 0  -6 0     1
10:  -1  1    1
11: -11  0   55 0 -77 0    44 0 -11 0    1
12:   1  0   -4 0   1
13:  13  0  -91 0 182 0  -156 0  65 0  -13 0   1
14:   1 -2   -1 1
15:   1  0   -8 0  14 0    -7 0   1
16:   2  0  -16 0  20 0    -8 0   1
17:  17  0 -204 0 714 0 -1122 0 935 0 -442  0 119  0 -17  0  1
18:   1 -3    0 1
...
n = 19: [-19, 0, 285, 0, -1254, 0, 2508, 0, -2717, 0, 1729, 0, -665, 0, 152, 0, -19, 0, 1],
n = 20: [1, 0, -12, 0, 19, 0, -8, 0, 1]
n = 5: ps(5,x) = 5 -5*x^2 +1*x^4, with the zeros s(5) = sqrt(3 - tau), sqrt(2 + tau) = tau*s(5) and their negative values, where tau =rho(5) is the golden section. tau*s(5) is the length ratio diagonal/radius in the pentagon.
n = 7: ps(7,x) = -7 + 14*x^2 -7*x^4 + 1*x^6, with the positive zeros s(7) (side/R) about 0.868, s(7)*rho(7) (smallest diagonal/R) about 1.564, and s(7)*(rho(7)^2-1) (longer diagonal/R) about 1.950 in the heptagon inscribed in a circle with radius R.
n = 8: ps(8,x) = 2 -4*x^2 + x^4, with the positive zeros s(8) = sqrt(2-sqrt(2)) and rho(8) = sqrt(2+sqrt(2)) (smallest diagonal/side).
n = 10: ps(10,x) = -1 + x + x^2 with the positive zero s(10) = tau - 1 (the negative solution is -tau).
		

Crossrefs

Formula

a(n, m) = [x^m](minimal polynomial ps(n, x) of 2*sin(Pi/n) over the rationals), n >= 1, m = 0, ..., gamma(n), with gamma(n) = A055035(n).
ps(n,x) = Product_{k=0..floor(c(2*n)/n) and gcd(k, c(2*n)) = 1} (x - 2*cos(2*Pi*k/c(2*n)), with c(2*n) = A178182(2*n), for n >= 1. There are gamma(n) = A055035(n) zeros. - Wolfdieter Lang, Oct 30 2019

A228780 Power basis components of the algebraic numbers S2(n) in Q(2*cos(Pi/n)), where S2(n) is the square of the sum of the lengths of the distinct line segments (side and diagonals) in the regular n-gon.

Original entry on oeis.org

4, 3, 6, 4, 3, 4, 12, 6, -1, 4, 4, -2, -4, 6, 4, 3, 8, 4, -16, -8, 12, 6, 3, -4, -8, 4, 4, 0, 4, 10, 4, 3, 8, -8, -12, 4, 4, 28, 14, -40, -20, 12, 6, -1, 8, 12, 4, -2, -8, 28, 28, -26, -20, 6, 4, -1, -8, 16, 28, -16, -20, 4, 4, 4, 2, -12, -6, 8, 4, 3, -8, -24, 28, 44, -20, -24, 4, 4, 0, 8, 28, -4, -40, -12, 10, 4, -1, -16, -24, 0, 12, 4
Offset: 2

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Author

Wolfdieter Lang, Oct 01 2013

Keywords

Comments

The length of row n of this irregular array is the degree of the algebraic number rho(n):= 2*cos(Pi/n), given in A055034(n). See a Jul 19 2011 comment there.
The regular n-gon, inscribed in a circle of radius defining the length unit 1, has distinct line segments (chords) (V_0, V_j), j=1, ... , floor(n/2), with the n-gon vertices V_j, j=0, ... , n-1 distributed on the circle in the counterclockwise sense. The corresponding length ratios are denoted by L(n,j)/radius. The side length is s(n) = (V_0, V_1) = 2*sin(Pi/n), and for n >= 4 the first (the smallest) diagonal has length s(n)*rho(n), with rho(n) of degree delta(n) given above. s(2) = 2 is the ratio of the diameter of the circle. rho(2) = 0, but we use here rho(2)^0 = 1.
For n = 3: rho(3) = 1, s(3)^2 = 3. The algebraic number field Q(rho(n)) is the subject of the W. Lang link given below.
S2(n) := (sum(L(n,j)/radius, j=1, ... ,floor(n/2))^2 is seen below to be a number in the field Q(rho(n)) of degree delta(n), namely S2(n) = sum(a(n,k)*rho(n)^k, k=0..(delta(n)-1)). From the definition one has S2(n) = (s(n)*sum(S(j-1,rho(n)), j=1..floor(n/2)))^2, with the Chebyshev S-polynomials (see A049310). Due to s(n) = s(2*n)*rho(2*n), rho(2*n) = sqrt(2 + rho(n)) and an S-identity this becomes S2(n) = (s(2*n)*S(floor(n/2)-1, rho(2*n))*S(floor(n/2), rho(2*n)))^2. This can also be written as S2(n) = 4*(1 - T(2*floor(n/2), rho(2*n)/2))*(1 - T(2*(floor(n/2)+1), rho(2*n)/2))/(4-rho(2*n)^2), with Chebyshev's T-polynomials (see A053120). S2(n), written as a function of rho(n), has to be computed modulo the minimal polynomial C(n,rho(n)) of degree delta(n). These minimal polynomials are treated in A187360 (see the link to a Galois paper there, with its Table 2 and Section 3). The result is then the above given representation of S2(n) in the power basis of Q(rho(n)).
This computation was inspired by an email exchange with Seppo Mustonen. The author thanks him for sending the paper given as a link below. In this connection one should consider the even and odd n cases separately in order to find the square of the total length segments/radius in the regular n-gon, noticing that in the odd n case each distinct chord (side or diagonal) appears 2*(n/2) = n times, whereas in the even n case the longest diagonal of length 2 (in units of the radius) appears only n/2 times and the other chords appear n times.

Examples

			The table a(n,k) begins:
n\k     0    1    2    3    4    5 ...
2:      4
3:      3
4:      6    4
5:      3    4
6:     12    6
7:     -1    4    4
8:     -2   -4    6    4
9:      3    8    4
10:   -16   -8   12    6
11:     3   -4   -8    4    4
12:     0    4   10    4
13:     3    8   -8  -12    4    4
14:    28   14  -40  -20   12    6
15:    -1    8    2    4
...
n=5: S2(5) = (4-rho(5)^2)*(Sum_{j=1..2} S(j-1,rho(5)))^2 = 4 + 8*rho(5) + 3*rho(5)^2 - 2*rho(5)^3 - rho(5)^4, reduced with C(5,x) = x^2 - x - 1, with x = rho(5), using C(5,rho(5)) = 0, to eliminate all powers of rho(5) starting with power 2.
This leads to S2(5) = 3*1 + 4*rho(5). rho(5) = phi, the golden section.
The exact or approximate real values for S2(n) are, for n = 2, ..., 15: 4, 3, 11.65685426, 9.472135960, 22.39230484, 19.19566936, 36.32882142, 32.16343753, 53.49096128, 48.37415020, 73.88698896, 67.82742928, 97.52047276, 90.52313112.
		

Crossrefs

Cf. A228781, A228782 (minimal polynomials for odd and even n).

Formula

a(n,k) = [rho^k] (S2(n) modulo C(n,rho(n)), with S2(n) the square of the sum of the distinct length/radius ratios in the regular n-gon, with rho(n) = 2*cos(Pi/n) given above in a comment, and C(n,x) the minimal polynomial of rho(n) given in A187360 (see Table 2 and section 3 of the paper given in the W. Lang link below).

A228782 Irregular triangle read by rows: coefficients of minimal polynomial of a certain algebraic number S2(2*k) from Q(2*cos(Pi/(2*k))) related to the regular (2*k)-gon.

Original entry on oeis.org

-4, 1, 4, -12, 1, 36, -24, 1, 16, -96, 136, -40, 1, 16, -96, 136, -56, 1, 16, -320, 456, -80, 1, 3136, -12544, 14896, -7168, 1484, -112, 1, 256, -7168, 41216, -73472, 53344, -17472, 2576, -144, 1, 64, -1152, 5424, -6080, 2124, -168, 1, 256, -13312, 62720, -104192, 76384, -26048, 3920, -208, 1
Offset: 1

Views

Author

Wolfdieter Lang, Oct 01 2013

Keywords

Comments

The row length sequence of this table is delta(2*k), k >= 1, with the degree delta(n) = A055034(n) of the algebraic number rho(n):=2*cos(Pi/n).
The algebraic 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 even case, n = 2*k, is considered. S2(n) is the square of the sum of the distinct length ratios side/radius and diagonal/radius with the radius of the circle in which a regular n-gon 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) has degree delta(2*k) and is given by p(2*k,x) = Product_{j=1..delta(2*k)} (x - S2(2*k)^{(j-1)}) (mod C(2*k, rho(2*k))) = Sum_{m=0..delta(2*k)} a(k, m)*x^m, where S2(2*k)^{(0)} = S2(2*k) and S2(2*k)^{(j-1)} is the (j-1)-th conjugate of S2(2*L). For the conjugate of an algebraic number in Q(rho(n)) see a comment on A228781.
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 n-gon. See his paper given as a link below. The author thanks Seppo Mustonen for sending his paper.

Examples

			The irregular triangle a(k, m) begins:
  n   k / m   0      1      2      3      4       5     6    7   8
  2   1:     -4      1
  4   2:      4    -12      1
  6   3:     36    -24      1
  8   4:     16    -96    136    -40      1
  10  5:     16    -96    136    -56      1
  12  6:     16   -320    456    -80      1
  14  7:   3136 -12544  14896  -7168   1484   -112    1
  16  8:    256  -7168  41216 -73472  53344 -17472 2576 -144  1
  ...
n = 18, k = 9: 64, -1152, 5424, -6080, 2124, -168, 1;
n = 20, k = 10: 256, -13312, 62720, -104192, 76384, -26048, 3920, -208, 1.
n = 6, k = 3: p(6, x) = (x - S2(6))*(x - S2(6)^{(1)}),
with S2(6) = 12 + 6*rho(6), where rho(6) = sqrt(3). C(6, x) = x^2 - 3 = (x - rho(6))*(x - (-rho(6))), hence rho(6)^{(1)} - -rho(6) and S2(6)^{(1)} = 12 - 6*rho(6). Thus p(6, x) = 144 - 36*rho(6)^2 - 24*x + x^2, reduced with C(6, rho(6)) = 0, i.e., rho(6)^2 = 3; this becomes finally 36 - 24*x + x^2.
		

Crossrefs

Cf. A055034, A187360, A228780, A228781 (odd case).

Formula

a(k,m) = [x^m] p(2*k, x), with the minimal polynomial p(2*k, x) of S2(2*k) given in the power basis in A228780.
p(2*k, x) is given in a comment above in terms of the S2(2*k) and its conjugates S2(2*k)^{(j-1)}, j = 2, ..., delta(2*k), where delta(2*k) = A055034(2*k).
Showing 1-3 of 3 results.