cp's OEIS Frontend

The length ratio (largest diagonal)/side in the regular 7-gon (heptagon) is sigma(7) = S(2, rho(7)) = -1 + rho(7)^2, with rho(7) = 2*cos(Pi/7), which is approx. 1.8019377358 (see A160389 for its decimal expansion, and A049310 for the Chebyshev S-polynomials). sigma(7), approx. 2.2469796, is also the reciprocal of one of the solutions of the minimal polynomial C(7, x) = x^3 - x^2 - 2*x + 1 of rho(7) (see A187360), namely 1/(2*cos(3*Pi/7)).

sigma(7) is the limit of a(n+1)/a(n) for n->infinity for the sequences A006054 and A077998 which can be considered as analogs of the Fibonacci sequence in the pentagon. Thus sigma(7) plays in the heptagon the role of the golden section in the pentagon.

See the Steinbach link.

Let A be the tridiagonal unit-primitive matrix (see [Jeffery]) A = A_{9,1} = [0,1,0,0; 1,0,1,0; 0,1,0,1; 0,0,1,1]. Then a(n)=[A^n](2,3). - _L. Edson Jeffery, Mar 19 2011

From Wolfdieter Lang, Oct 02 2013: (Start)

This sequence a(n) appears in the formula for the nonnegative powers of the algebraic number rho(9) := 2*cos(Pi/9) of degree 3, the ratio of the smallest diagonal/side in the regular 9-gon, in terms of the power basis of the algebraic number field Q(rho(9)) (see A187360, n=9).

rho(9)^n = A(n)*1 + B(n)*rho(9) + C(n)*rho(9)^2, with A(0) = 1, A(1) = 0, A(n) = B(n-2), n >= 2, B(0) = 0, B(n) = a(n-1), n >= 1, C(0) = 0, C(n) = B(n-1), n >= 1. (End)

This constant appears in a historic problem posed by Adriaan van Roomen (Adrianus Romanus) in his Ideae mathematicae from 1593, solved by Viète (see the Vieta link) using trigonometry. See the Havil reference, problem 1 (for a correction see below), pp. 69-74, and the Maor reference for Viète's approach, pp. 58-60.

The problem involves the monic Chebyshev polynomial of the first kind R(45, x) (R coefficients are given in A127672). The present problem was stated as R(45, x) = sqrt(2 + sqrt(2 + sqrt(2 + sqrt(2)))) for x = sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 + sqrt(3))))) (see A302712). This is equivalent to R(45, 2*sin(Pi/96)) = 2*sin(15*Pi/32). It is a special case of the well known identity R(2*k+1, x) = x*(-1)^k*S(2*k, sqrt(4-x^2)), with the Chebyshev S polynomials (see A049310 for the coefficients). Take k = 22, x = 2*sin(Pi/96), and see the Havil reference, p. 71, for the proof of 2*sin(15*Pi/32) = sqrt(2 + sqrt(2 + sqrt(2 + sqrt(2)))). [In the Havil reference on p. 69, the second to last exponent is 43 (not 41), and in the first problem, for the argument x a further +sqrt(2... is missing. In the general identity given on p. 71 a sign factor is missing. It should read, with n = 2*k+1: P_{2*k+1}(2*sin(theta)) = 2*(-1)^k*sin((2*k+1)*theta).]

For the argument x = sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 + sqrt(3))))) = 2*sin(Pi/96) = 0.65438165643552284... see A302712.

R(45, x) factorizes into minimal polynomials of 2*cos(Pi/k), named C(k, x), for short, C[k], with coefficients given in A187360 as follows. R(45, x) = C[90]*C[30]*C[18]*C[10]*C[6]*C[2]. See a comment in A127672.

All 45 zeros of R(45, x), which are real, are 2*cos((2*k+1)*Pi/90), for k = 0..44. See a comment in A127672.

Viète used the iteration, written in terms of R polynomials as R(45, x) = -R(3, -R(3, R(5, x))) (from the semigroup property of Chebyshev T polynomials). See the Maor reference, pp. 58-60. - Wolfdieter Lang, May 05 2018

An algebraic integer of degree 16. - Charles R Greathouse IV, Jan 29 2022

For multiplication Modd n (not to be confused with multiplication mod n) see a comment on A203571.

The trivial solution of x^2 == 1 (Modd n) is x = 1 (Modd n). Note that x = -1 (Modd n) == +1 (Modd n). In the ordinary mod n case the trivial solution is 1 (mod 2) for n=2 (-1 == +1 (mod 2)) and if n>2 the two trivial solutions are 1 (mod n) and the noncongruent -1 (mod n) == n-1 (mod n).

Here multiplication on the reduced residue system Modd p, p an odd prime, with only odd numbers is considered (which is possible, contrary to mod p). In order to have inverses one has to exclude all reduced residue classes Modd p with even numbers. The (p-1)/2 residue classes are then [1],[3],...,[p-2]. For m=1,3,...,p-2, the class [m] Modd p is the union of the ordinary reduced residue classes mod 2p: [m] and [-m]=[2p-m]. Besides the trivial solution x=+1 (Modd p) (note that -1 == +1 (Modd p)) there is a further nontrivial solution if and only if (p-1)/2 is even, i.e. p=A002144(n) (primes of the form 4*k+1), n>=1. The present sequence entry a(n) gives the smallest positive representative for this nontrivial solution Modd A002144(n).

This result uses the fact that every finite group of prime order p is the cyclic group Z_p (Corollary to Lagrange's or also Cauchy's theorem on finite groups or see A000688 for abelian groups). Here for the multiplicative group Modd p (on the odd residue classes) which has order (p-1)/2, for p an odd prime. This turns out to be the Galois group for the minimal polynomial C(p,x), whose coefficients are found in A187360. The sequence {a(n)} arises if one asks for the smallest positive members of the odd restricted residue system Modd p, namely [1],[3],...,[p-2], which are their own inverses besides the trivial element 1 (Modd p).

The row a(n) has in the table for the multiplicative group Modd A002144(n) a 1 on the diagonal, if the table is written for the odd representatives 1,3,...,p-2. The only other diagonal entry 1 appears for the element 1. Note that in the ordinary mod p case, p an odd prime, one always has for the multiplicative group mod p (which has p-1 residue classes) in the multiplication table the diagonal entry 1 only for the representatives 1 and p-1, but p-1 == -1 (mod p) is a trivial solution of x^2 == 1 (mod p).

Middle line of 5-wave sequence A038201.

Let M denotes the 5 X 5 matrix = row by row (1,1,1,1,1)(1,1,1,1,0)(1,1,1,0,0)(1,1,0,0,0)(1,0,0,0,0) and A(n) the vector (x(n),y(n),z(n),t(n),u(n))=M^n*A where A is the vector (1,1,1,1,1) then a(n)=z(n). - Benoit Cloitre, Apr 02 2002

a(n) appears in the formula for 1/rho(11)^n, with rho(11) := 2*cos(Pi/11) (length ratio (smallest diagonal/side) in the regular 11-gom) when written in the power basis of the degree 5 number field Q(rho(11)): 1/rho(11)^n = a(n)*1 + A230080(n)*rho(11) - A230081(n)*rho(11)^2 - A069006(n-1)* rho(11)^3 + a(n-1)*rho(11)^4, n >= 0, with A069006(-1) = 0 = a(-1). See A230080 with the example for n=4. - Wolfdieter Lang, Nov 04 2013

From Wolfdieter Lang, Nov 20 2013: (Start)

The limit a(n+1)/a(n) for n -> infinity is omega(11) := S(4, x) = 1 - 3*x^2 + x^4 with x = rho(11). omega(11) = 1/(2*cos(Pi*5/11)), approx. 3.51333709. For the Chebyshev S-polynomial see A049310. For rho(11) see the preceding comment. The decimal expansion of omega(11) is given in A231186. omega(11) is an integer in Q(rho(11)) with power basis coefficients [1,0,-3,0,1]. It is known to be the length ratio (longest diagonal)/side in the regular 11-gon.

This limit follows from the a(n)-recurrence and the solutions of X^5 - 3*X^4 - 3*X^3 + 4*X^2 + X - 1 = 0, which are given by the inverse of the known solutions of the minimal polynomial C(11, x) of rho(11) (see A187360). The other four X solutions are 1/rho(11), with coefficients [3,3,-4,-1,1] in the power basis of Q(rho(11)), approx. 0.52110856, 1/(2*cos(Pi*3/11)) with coefficients [-1,-1,1,0,0], approx. 0.763521119, 1/(2*cos(Pi*7/11)) with coefficients [0,-3,3,1,-1], approx. -1.20361562, and 1/(2*cos(Pi*9/11)) with coefficients [0,1,3,0,-1], approx. -0.59435114. These solutions for X are therefore irrelevant for this sequence.

The same limit omega(11) is therefore obtained for the sequences A069006, A230080 and A230081. See the Nov 04 2013 comment.

(End)

In the regular (2*l+1)-gon, l >= 1, inscribed in a circle of radius R the length ratio side/R is s(2*l+1) = 2*sin(Pi/(2*l+1)). This can be written as a polynomial in the length ratio (smallest diagonal)/side in the (2*(2*l+1))-gon given by rho(2*(2*l+1)) = 2*cos(Pi/(2*(2*l+1))). This leads, in a first step, to the signed triangle A111125. Because of the minimal polynomial of the algebraic number rho(2*(2*l+1)) of degree delta(2*(2*l+1)) = A055034(2*(2*l+1)), called C(2*(2*l+1),x) (with coefficients given in A187360), one can eliminate all powers rho(2*(2*l+1))^k with k >= delta(2*(2*l+1)) by using C(2*(2*l+1),rho(2*(2*l+1))) = 0. This leads to the present table expressing s(2*(l+1)) in terms of odd powers of rho(2*(2*l+1)) with maximal exponent delta(2*(2*l+1))-1.

This table gives the coefficients of s(2*l+1), related to the (2*l+1)-gon, in the power basis of the algebraic number field Q(rho(2*(2*l+1))) of degree delta(2*(2*l+1)), related to rho from the (2*(2*l+1))-gon, provided one inserts zeros for the even powers, starting each row with a zero and filling zeros at the end in order to obtain the row length delta(2*(2*l+1)). Note that some trailing zeros in the present table (e.g., row l = 10) have been given such that the row length for the s(2*l+1) coefficients in the power basis Q(rho(2*(2*l+1))) becomes just twice the one of this table.

Thanks go to Seppo Mustonen for telling me about his findings regarding the square of the sum of all length in the regular n-gon, which led me to consider this entry (even though for odd n this is not needed because only s(2*l+1)^2 = 4 - rho(2*l+1)^2 enters).

This sequence is fundamental for the coefficient sequences for the nonnegative powers of rho(11) = 2*cos(Pi/n) (length ration (smallest diagonal)/side in the regular 11-gon (Hendecagon)) when written in the power basis of the degree 5 number field Q(rho(11)). See A187360 for the minimal polynomial of rho(11) which is C(11, x) = x^5 - x^4 - 4*x^3 + 3*x^2 + 3*x - 1. See A231182-5 for these coefficient sequences.

Because 2*sin(Pi*4/n) = 2*cos(Pi*abs(n-8)/(2*n)) = 2*cos(Pi*a(n)/b(n)) with gcd(a(n),b(n)) = 1, one has

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

b(n) = A232625(n). This shows that 2*sin(Pi*4/n) is an integer in the algebraic number field Q(rho(b(n))) of degree delta(b(n)), with delta(k) = A055034(k). This degree delta(b(n)) is given in A231193(n), and if gcd(n,2) = 1 it coincides with the one for sin(2*Pi/n) given by A093819(n). See Theorem 3.9 of the I. Niven reference, pp. 37-38, which uses gcd(k, n) = 1. See also the Jan 09 2011 comment on A093819.

a(n) and b(n) = A232625(n) are the k=2 members of a family of pair of sequences p(k,n) and q(k,n), n >= 1, k >= 1, relevant to determine the algebraic degree of 2*sin(Pi*2*k/n) from the trigonometric identity (used in the D. H. Lehmer and I. Niven references) 2*sin(Pi*2*k/n) = 2*cos(Pi*abs(n-4*k)/(2*n)) = 2*cos(Pi*p(k,n)/q(k,n)). This is R(p(k,n), x) (mod C(q(k,n), x)), with x = 2*cos(Pi/q(k,n)) =: rho(q(k,n)). The polynomials R and C have been used above. C(q(k,n), x) is the minimal polynomial of rho(q(k,n)) with degree delta(q(k,n)), which is then the degree, call it deg(k,n), of the integer 2*sin(Pi*2*k/n) in the number field Q(rho(q(k,n))). From Theorem 3.9 of the I. Niven reference deg(k,n) is, for given k, for those n with gcd(k, n) = 1 determined by A093819(n). In general deg(k,n) = A093819(n/gcd(k,n)). For the k=1 instance p(1,n) and q(1,n) see comments on A106609 and A225975.

The length of row n is deg(n) + 1, n >= 1, with the degree deg(1) = deg(2) = 1, and deg(n) = phi(n)/2 = A023022(n) for n >= 3. That is: 2, 2, 2, 2, 3, 2, 4, 3, 4, 3, 6, 3, 7, 4, 5, 5, 9, 4, 10, 5, ...

2*cos(2*Pi/n) = R(2, rho(n)) = -2 + rho(n)^2, with rho(n) = 2*cos(Pi/n) and the monic Chebyshev T-polynomials R(n, x), n>=1, with coefficient table A127672. For even n 2*cos(2*Pi/n) becomes rho(n/2). Therefore, 2*cos(2*Pi/n) is an integer in the algebraic number field Q(rho(n/2)) or Q(rho(n)) if n is even or odd, respectively. The degree deg(n) of the minimal polynomials, call them MPR2(n, x), is delta(n/2) or delta(n) for even or odd n, respectively, with delta(n) = A055034(n). This becomes deg(n) as given above.

These minimal polynomials are C(n/2, x) if n is even, with C(k, x) the minimal polynomials of rho(k) given in A187360.

For odd n the known zeros of C(n, x) are rho(n) and its conjugates, call them rho(n;j), j=1, 2, ..., delta(n), with rho(n;1) = rho(n). These conjugates can be written in the power basis of Q(rho(2*l+1)), l >= 1. See the link to the Q(2cos(Pi/n)) paper in A187360, and there Table 4. Then the (monic) minimal polynomial MPR2(2*l+1, x) = Product_{j=1..delta(2*l+1)} (x - (-2 + rho(2*l+1;j)^2)), l >= 0. After expansion all powers of rho(2*l+1) not smaller than delta(2*l+1) are reduced with the help of C(2*l+1,rho(2*l+1)) = 0, leading automatically to integer coefficients (without using the trigonometric version of rho(2*l+1)).

Compare the present minimal polynomials with the (non-monic) minimal polynomials of cos(2*Pi/n) given in an Artur Jasinski comment from Oct 28 2008 on A023022.

The present monic integer minimal polynomials of 2*cos(2*Pi/n), called MPR2(n, x), are related to the non-monic integer minimal polynomials of 2*cos(2*Pi/n) of A181877, called there psi(n, x) by MPR2(n, x) = psi(n, x/2). See Table 5 of the Wolfdieter Lang link given there. - Wolfdieter Lang, Nov 29 2013

The present minimal polynomials MPR2(n, x) are C(n/2, x) if n is even (see above) and (-1)^degC(n)*C(n, -x) if n is odd, with the C polynomials from A187360 of degree degC(n) = A055034(n). Note that degC(2*k+1) = deg(2*k+1) = A023022(2*k+1), k >= 0. - Wolfdieter Lang, Apr 12 2018

Let {U(n, x)} be defined as: U(0, x) = 0, U(1, x) = 1, U(n, x) = x*U(n-1, x) - U(n-2, x) for n >= 2, then U(n, x) = Product_{k|2n, k>=3} MPR2(k, x) for n > 0, because U(n, x) = Product_{m=1..n-1} (x - 2*cos(Pi*m/n)) for n > 0. - Jianing Song, Jul 08 2019

Conjecture: For odd n > 1, the term of the highest degree of (MPR2(2n, x) - MPR2(n, x))/2 is (-1)^omega(n) * x^(phi(n)/2-n/rad(n)) = A076479(n) * x^(A023022(n)-A003557(n)). For example, for n = 15, (MPR2(30, x) - MPR2(15, x))/2 = x^3 - 4x; for n = 105, (MPR2(210, x) - MPR2(105, x))/2 = -x^23 + ...; for n = 225, (MPR2(450, x) - MPR2(225, x))/2 = x^45 + ... If this is true, then for odd n > 1, a(n,A023022(n)-k) = a(2n,A023022(n)-k) = 0 for k = 1, 3, ..., A003557(n)-2; a(n,A023022(n)-A003557(n)) = -A076479(n) and a(2n,A023022(n)-A003557(n)) = A076479(n). - Jianing Song, Jul 11 2019

Conjecture: Let MPR2(n, x) equal the odd indexed (n) monic polynomial. If the number of roots with negative signs is even, then n is a term in A014659. Example: n = 7 for x^3 + x^2 - 2x - 1, having two negative roots, (-445041..., and -1.801937...). Two is even so the integer 7 is in A014659. n = 9 for the polynomial x^3 - 3x + 1, with one negative root, (-1.87938). The term 9 is in A014657. - Gary W. Adamson, Oct 20 2021

From Gary W. Adamson, Nov 30 2021 (Start)

Given the first (phi(n))/2 terms for odd n, the number of even terms in the set is equal to the number of positive roots in MPR2(n, x). The number of odd terms is equal to the number of negative roots in MPR2(n, x). For n = 11, (phi(11))/2 = 5, and the set is (1, 2, 3, 4, 5); having two even and three odd terms.

Given MPR2(11, x) = x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1, there are two roots with positive signs: 1.682508..., and .830830...; and three roots with negative signs: -1.918985..., -1.309921..., and -.284629....Using the Descartes' rule for signs, MPR2(11, x) has coefficients signed (+ + - - + +); having two sign changes indicating two positive roots. With all real roots there are three (= 5 - 2) roots signed negative. (End)

For Modd n (not to be confused with mod n) see a comment on A203571.

Precisely these numbers n (only the ones <=165 are shown above) have no primitive root Modd n. See the zero entries of A206550, except A206550(1) = 0 which stands for a primitive root 0.

The multiplicative Modd n group is the Galois group Gal(Q(rho(n))/Q), with the algebraic number rho(n) := 2*cos(Pi/n) with minimal polynomial C(n,x), whose coefficients are given in A187360.