cp's OEIS Frontend

For any integer n>=0, 2 * Integral_{t=-2..2} T_n(t/2)*exp(-t)*dt = 4 * Integral_{z=-1..1} T_n(z)*exp(-2*z)*dz = A102761(n)*exp(2) - a(n)*exp(-2).

If X is an n-set and Y a fixed 2-subset of X then a(n-6) is equal to the number of (n-6)-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007

a(n-1) = risefac(n+1,6)/6! - risefac(n+1,4)/4! is for n >=1 also the number of independent components of a symmetric traceless tensor of rank 6 and dimension n. Here risefac is the rising factorial. Put a(0) = 0. - Wolfdieter Lang, Dec 10 2015

15-gon is the first m-gon with odd composite m which is constructible (see A003401) in virtue of the fact that 15 is the product of two distinct Fermat primes (A019434). The next such case is 51-gon (m=3*17), followed by 85-gon (m=5*17), 771-gon (m=3*257), etc.

From Wolfdieter Lang, Apr 29 2018: (Start)

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 Havil reference, problem 4, pp. 69-74. See also the comments in A302711 with a link to Romanus' book, Exemplum quaesitum.

This problem is equivalent to R(45, 2*sin(Pi/675)) = 2*sin(Pi/15), with a special case of monic Chebyshev polynomials of the first kind, named R, given in A127672. For the constant 2*sin(Pi/675) see A302716. (End)

a(m,n) is a polynomial in m of degree n.

For any integers m>=0, n>=0, 2 * Integral_{t=-m..m} T_n(t/2)*exp(-t)*dt = 4 * Integral_{z=-m/2..m/2} T_n(z)*exp(-2*z)*dz = A300481(m,n)*exp(m) - a(m,n)*exp(-m).

Although negative values of m are not present here or in A300480, the two arrays are connected with the formula: a(m,n) = A300480(-m,n). Thus, they essentially represent two "halves" of the same array indexed by integers m.

a(m,n) is a polynomial in m of degree n.

For any integers m>=0, n>=0, 2 * Integral_{t=-m..m} T_n(t/2)*exp(-t)*dt = 4 * Integral_{z=-m/2..m/2} T_n(z)*exp(-2*z)*dz = a(m,n)*exp(m) - A300480(m,n)*exp(-m).

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)