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For any integer n>=0, 2 * Integral_{t=-1..1} T_n(t/2)*exp(-t)*dt = 4 * Integral_{z=-1/2..1/2} T_n(z)*exp(-2*z)*dz = A300485(n)*exp(1) - a(n)*exp(-1).

For any integer n>=0, 2 * Integral_{t=-1..1} T_n(t/2)*exp(-t)*dt = 4 * Integral_{z=-1/2..1/2} T_n(z)*exp(-2*z)*dz = a(n)*exp(1) - A300483(n)*exp(-1).

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)

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.

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)).

a(n) is the trace of the 2*n-th power of the adjacency matrix M for the simple Lie algebra B_4, given in the Damianou link. M = Matrix[row 1; row 2; row 3; row 4] = Matrix[0,1,0,0; 1,0,1,0; 0,1,0,2; 0,0,1,0]. Equivalently, the trace tr(M^(2*k)) is the sum of the 2*n-th powers of the eigenvalues of M. The eigenvalues are the zeros of the characteristic polynomial of M, which is det(x*I - M) = x^4 - 4*x^2 + 2 = A127672(4,x), and are (+-) sqrt(2 + sqrt(2)) and (+-) sqrt(2 - sqrt(2)), or the four unique values generated by 2*cos((2*n+1)*Pi/8). Compare with A025192 for B_3. The odd power traces vanish.

-log(1 - 4*x^2 + 2*x^4) = 8*x^2/2 + 24*x^4/4 + 80*x^6/6 + ... = Sum_{n>0} tr(M^k) x^k / k = Sum_{n>0} a(n) x^(2k) / 2k gives an aerated version of the sequence a(n), excluding a(0), and exp(-log(1 - 4*x + 2*x^2)) = 1 / (1 - 4*x + 2*x^2) is the e.g.f. for A007070.

As in A025192, the cycle index partition polynomials P_k(x[1],...,x[k]) of A036039 evaluated with the negated power sums, the aerated a(n), are P_2(0,-a(1)) = P_2(0,-8) = -8, P_4(0,-a(1),0,-a(2)) = P_4(0,-8,0,-24) = 48, and all other P_k(0,-a(1),0,-a(2),0,...) = 0 since 1 - 4*x^2 + 2*x^4 = 1 - 8*x^2/2! + 48*x^4/4! = det(I - x M) = exp(-Sum_{k>0} tr(M^k) x^k / k) = exp[P.(-tr(M),-tr(M^2),...)x] = exp[P.(0,-a(1),0,-a(2),...)x].

Because of the inverse relation between the Faber polynomials F_n(b1,b2,...,bn) of A263916 and the cycle index polynomials, F_n(0,-4,0,2,0,0,0,...) = tr(M^n) gives aerated a(n), excluding a(0). E.g., F_2(0,-4) = -2 * -4 = 8, F_4(0,-4,0,2) = -4 * 2 + 2 * (-4)^2 = 24, and F_6(0,-4,0,2,0,0) = -2*(-4)^3 + 6*(-4)*2 = 80.

Signed version of A114525. - Eric W. Weisstein, Apr 07 2017

For n >= 3, also the coefficients of the matching polynomial for the n-cycle graph C_n. - Eric W. Weisstein, Apr 07 2017

This triangle describes the Chebyshev transform of A100047 and following. Chebyshev transform of sequence b is c(n) = Sum_{k=1..n} a(n,k)*b(k). - Christian G. Bower, Jun 12 2005

See A049310 for the coefficient table of Chebyshev's S(n,x)=U(n,x/2) polynomials.

This is a 'repetition triangle' based on a signed version of triangle A059260: a(2*p,2*k) = a(2*p+1,2*k) = A059260(p+k,2*k)*(-1)^(p+k) and a(2*p+1,2*k+1) = a(2*p+2,2*k+1) = A059260(p+k+1,2*k+1)*(-1)^(p+k), k >= 0.

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

This is a signed version of the triangle A061896.

The coefficient table for the monic Chebyshev polynomials of the first kind R(n, x) = 2*T(n, x/2) is given in A127672. For the T-polynomials see A053120. The present table is obtained from the row reversed coefficient table A127672 by deleting all odd numbered columns which have only zeros, and appending in the rows numbered n >= 1 zeros in order to obtain a triangle. This becomes the quasi-Riordan triangle T = ((2-z)/(1-z), -z^2/(1-z)). This means that the o.g.f. of the row polynomials Rrev(n, x) := sqrt(x)^n*R(n, 1/sqrt(x)) = Sum_{k=0..n} T(n, k)*x^k have o.g.f. (2-z)/(1 - z + x*z^2) like for ordinary Riordan triangles. However this is not a Riordan triangle (or lower triangular infinite dimensional matrix) in the usual sense because it is not invertible. Therefore, this lower triangular matrix is not a member of the Riordan group.

The row sums give repeat(2,1,-1,-2,-1) which is A057079(n+1), n >= 0. The alternating row sums give the Lucas numbers A000032.