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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. 1
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, 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, 11, 13, 17, 19, 23, 29, 1, 3, 5, 7, 9, 11, 13, 15 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

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.

The number s(n) is the length ratio side(n)/R of the regular n-gon inscribed in a circle of radius R.

The motivation to loook at these zeros came from the book of Carl Schick, and the paper by Brändli and Beyne. There length ratios diagonals/R in 2*(2*m + 1)-gons, for m >= 1, are only considered.

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).

REFERENCES

Carl Schick, Trigonometrie und unterhaltsame Zahlentheorie, ISBN 3-9522917-0-6, Bobos Druck, Zürich, 2003.

LINKS

Table of n, a(n) for n=1..74.

Gerold Brändli and Tim Beyne, Modified Congruence Modulo n with Half the Amount of Residues, arXiv:1504.02757v2 [math.NT], 2015 and 2016.

FORMULA

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.

The numbers with odd c(2*n) are n = 2 + 8*k, k >= 0.

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.

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.

  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,

  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

  x0^{(n)}_m = R(T(n,m), R(4, sqrt(4 - s(n)^2))), for n == 2 (mod 8).

EXAMPLE

The irregular triangle T(n,m) begins:

n\m   1 2  3  4  5  6  7  8  9 10 11 12 ...      A178182(2*n)  A055035(n)

-------------------------------------------------------------------------

1:    1                                                4          1

2:    0                                                1          1

3:    1 5                                             12          2

4:    1 3                                              8          2

5:    1 3  7  9                                       20          4

6:    1                                                6          1

7:    1 3  5  9 11 13                                 28          6

8:    1 3  5  7                                       16          4

9:    1 5  7 11 13 17                                 36          6

10:   1 2                                              5          2

11:   1 3  5  7  9 13 15 17 19 21                     44         10

12:   1 5  7 11                                       24          4

13:   1 3  5  7  9 11 15 17 19 21 23 25               52         12

14:   1 3  5                                          14          3

15:   1 7 11 13 17 19 23 29                           60          8

16:   1 3  5  7  9 11 13 15                           32          8

...

--------------------------------------------------------------------------

Some zeros are:

n = 1:  2*cos(2*Pi*1/4) = 0 = s(1),

n = 2:  2*cos(2*Pi*1/4) = 2 = s(2) (diameter/R),

n = 3:  2*cos(2*Pi*1/12) = -2*cos(2*Pi*5/12) = sqrt(3) = s(3),

n = 5:  2*cos(2*Pi*1/20) = -2*cos(2*Pi*9/20) = sqrt(2 + tau),

        2*cos(2*Pi*3/20) = -2*cos(2*Pi*7/20) = sqrt(tau - 3) = s(5),

with the golden ratio tau = A001622,

n = 10: 2*cos(2*Pi*1/5) = tau - 1 = s(10),  -2*cos(2*Pi*2/5) = -tau.

--------------------------------------------------------------------------

CROSSREFS

Cf. A001622, A049310, A055035, A093819, A127672, A178182, A187360, A228786.

Sequence in context: A229181 A121267 A111740 * A155059 A206076 A329374

Adjacent sequences:  A327918 A327919 A327920 * A327922 A327923 A327924

KEYWORD

nonn,easy,tabf

AUTHOR

Wolfdieter Lang, Nov 02 2019

STATUS

approved

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Last modified July 30 06:21 EDT 2021. Contains 346348 sequences. (Running on oeis4.)