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Repeated terms of A016825 are in the positions 1,2,3,6,5,10,... (A043547).

From Wolfdieter Lang, Dec 04 2013: (Start)

This sequence a(n), n>=1, appears in the formula 2*sin(2*Pi/n) = R(p(n), x) modulo C(a(n), x), with x = rho(a(n)) = 2*cos(Pi/a(n)), the R-polynomials given in A127672 and the minimal C-polynomials of rho given in A187360. This follows from the identity 2*sin(2*Pi/n) = 2*cos(Pi*p(n)/a(n)) with gcd(p(n), a(n)) = 1. For p(n) see a comment on A106609,

Because R is an integer polynomial it shows that 2*sin(2*Pi/n) is an integer in the algebraic number field Q(rho(a(n))) of degree delta(a(n)) (the degree of C(a(n), x)), with delta(k) = A055034(k). This degree is given in A093819. For the coefficients of 2*sin(2*Pi/n) in the power basis of Q(rho(a(n))) see A231189 . (End)

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

Because A007521(n) are the primes congruent 5 (mod 8) it is clear that a(n) is congruent 1 (mod 2), that is odd.

2*a(n) = A055034(A007521(n)), the degree of the minimal polynomial C(A007521(n), x) of 2*rho(Pi/A007521(n)) (see A187360).

The numerators are given in A231190. See the comments there on 2*sin(Pi*4/n).

2*sin(Pi*4/n) = R(b(n), x) (mod C(b(n), x)), with x = 2*cos(Pi/a(n)) =: rho(a(n)). The integer Chebyshev R and C polynomials are found in A127672 and A187360, respectively. b(n) = A231190(n).

delta(a(n)) = deg(2,n), with delta(k) = A055034(k), is the degree of the algebraic number 2*sin(Pi*4/n) given in A232626.

The length of row n of this table is 1 + A023022(n), n >= 0, that is 2, 2, 2, 2, 3, 2, 4, 3, 4, 3, 6, 3, 7, 4, 5, 5, 9, 4,...

s(n):= 2*sin(Pi/n) is for n >= 2 the length ratio side/R of a regular n-gon inscribed in a circle of radius R (in some units). s(1) = 0. In general s(n)^2 = 4 - rho(n)^2 with rho(n):= 2*cos(Pi/n), for n>=2 this is the length ratio (smallest diagonal)/s(n) in the regular n-gon. If n is even, say 2*l, l>=1, then s(2*l)^2 = 2 - rho(l) (because rho(2*l)^2 = rho(l) +2). Therefore, if n is even s(n)^2 is an integer in the algebraic number field Q(rho(n/2)), and if n is odd then it is an integer in Q(rho(n)). The coefficient tables for the minimal polynomials of s(n)^2, called MPs2(n, x), for even and odd n have been given in A232631 and A232632, respectively. See these entries for details, and the link to the Q(2 cos(pi/n)) paper, Table 4, in A187360 for the power basis representation of the zeros of the minimal polynomial C(n, x) of rho(n).

The degree deg(n) of MPs2(n, x) is therefore delta(n/2) or delta(n) for n even or odd, respectively, where delta(n) = A055034(n). This means that deg(1) = deg(2) =1 and deg(n) = phi(n)/2 = A023022(n), n >= 3. deg(n) = A023022(n).

Especially MPs2(p, x) = Product_{j=0..(p-3)/2} (x - 2*(1 + cos(Pi*(2*j+1)/p))), for p an odd prime (A065091).

This computation was motivated by a preprint of S. Mustonen, P. Haukkanen and J. K. Merikoski, called ``Polynomials associated with squared diagonals of regular polygons'', Nov 16 2013.

This is a member of the six sequences which appear for the instance N=7 of the general formula 2*exp(2*Pi*n*I/N) = R(n, x^2-2) + x*S(n-1, x^2-2)*s(N)*I, for n >= 0, with I = sqrt(-1), s(N) = sqrt(2-x)*sqrt(2+x), x = rho(N) := 2*cos(Pi/N) and R and S are the monic Chebyshev polynomials whose coefficient tables are given in A127672 and A049310. If powers x^k with k >= delta(N) = A055034(N) enter in R or x*S then C(N, x), the minimal polynomial of x = rho(N) (see A187360) is used for a reduction. If delta(N) = 2 it may happen that sqrt(2+x) or sqrt(2-x) is an integer in the number field Q(rho(N)). See the N=5 case comment on A164116.

For N=7 with delta(7) = 3, and C(7, x) = x^3 - x^2 - 2*x + 1 the final result becomes 2*exp(2*Pi*n*I/7) = (a(n) + b(n)*x + c(n)*x^2) + (A(n) + B(n)*x + C(n)*x^2)*s(7)*I, with x = rho(7) = 2*cos(Pi/7), a(n) the present sequence, b(n) = A234045(n), c(n) = A234046(n), A(n) = A238468(n), B(n) = A238469(n) and C(n) = A238470(n). The a, b, c and A, B, C brackets are integers in Q(rho(7)).

From Wolfdieter Lang, Oct 24 2013: (Start)

Each a(n) is of course congruent 3 (mod 4).

a(n) = A055034(p7m8(n)), with p7m8(n) := A007522(n). This is the degree of the minimal polynomial of rho(p7m8(n)):= 2*cos(Pi/p7m8(n)), called C(p7m8(n), x) in A187360. (End)

a(n) is also the number of linearly dependent diagonal/side length ratios R(n,k), in the regular n-gon. The following will explain this. In the regular n-gon inscribed in a circle the number of distinct diagonals including the side is floor(n/2). Not all of the corresponding length ratios R(n,k) = d(n,k)/d(n,1), k = 1..floor(n/2), with d(n,1) = s(n) (the length of the side), d(n,2) the length of the smallest diagonal, etc., are linearly independent because C(n,R(n,2)) = 0, where C is the minimal polynomial of R(n,2) = 2*cos(Pi/n) (see A187360) with degree delta(n) = A055034(n). Thus every ratio R(n,j), with j = delta(n)+1, ..., floor(n/2) can be expressed as a linear combination of the independent R(n,k), k=1, ..., delta(n). See the comment from Sep 21 2013 on A053121 for powers of R(n,2) (called there rho(N)). Therefore, a(n) = floor(n/2) - delta(n) is, for n>=2, the number of linearly dependent ratios R(n,k) in the regular n-gon. - Wolfdieter Lang, Sep 23 2013

From Wolfdieter Lang, Nov 23 2020: (Start)

This sequence gives the difference between the number of odd numbers in the smallest nonnegative residue system modulo n (called here RS(n)) and the smallest nonnegative restricted residue system (called here RRS(n), but RRS(1) = {1}, not {0}).

This sequence can be used to find sequence A111774 by recording the positions of the entries >= 1. See a W. Lang comment there, and also A337940, for the proof. Hence the complement of A111774, given in A174090, is given by the numbers m with a(m) = 0. (End)

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.

In the regular (2*l)-gon inscribed in a circle of radius R the length ratio side/R is s(2*l) = 2*sin(Pi/(2*l)). This can be written as a polynomial in the length ratio (smallest diagonal)/side given by rho(2*l) = 2*cos(Pi/(2*l)). (For the 2-gon there is no such diagonal and rho(2) = 0). This leads, in a first step, to the triangle A127672 (see the Oct 05 2013 comment there referring also to the bisections signed A111125 and A127677). Because the minimal polynomial of the algebraic number rho(2*l) of degree delta(2*l) = A055034(2*l), called C(2*l,x) (with coefficients given in A187360) one can eliminate all powers rho(2*l)^k with k >= delta(2*l) by using C(2*l,rho(2*l)) = 0. Furthermore, because for odd l only even powers of rho(2*l) appear, but rho(2*l)^2 = 2 + rho(l), one will obtain a reduced table for s(2*l) with powers rho(2*l)^(2*k+1), k= 0, ..., (delta(2*l)-2)/2 if l is even, and with powers rho(l)^m, m=0, ... , delta(l)-1 if l is odd.

This leads to a reduction of the triangle A127672, when applied for the s(2*l) computation, giving the present table with row length delta(4*L) = A055034(4*L) = phi(8*L)/2 if l =2*L, if L >= 1, and phi(2*L+1)/2 = A055035(4*L+2), if l = 2*L + 1, L >= 1, where phi(n) = A000010(n) (Euler totient).

This table gives the coefficients of s(2*l) in the power basis of the algebraic number field Q(rho(2*l)) of degree delta(2*l) = A055034(2*l) if l is even, and in Q(rho(l)) of degree delta(2*l)/2 if l is odd. s(2) and s(6) are rational integers of degree 1.

Thanks go to Seppo Mustonen whose question about the square of the sum of all length in a regular n-gon, led me to this computation.

If l = 2*L+1, L >= 0, that is n == 2 (mod 4), one can obtain s(2*l) more directly in powers of rho(l) by s(2*l) = R(l-1, rho(l)) (mod C(l,rho(l))), with the monic (except for l=1) Chebyshev T-polynomials, called R, in A127672, and the C polynomials from A187360. - Wolfdieter Lang, Oct 10 2013