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This algebraic number rho(9)^2 of degree 3 is a root of its minimal polynomial x^3 - 6*x^2 + 9*x - 1.

The other two roots are x2 = (2*cos(5*Pi/9))^2 = (2*cos(4*Pi/9))^2 = (R(4,rho(9)))^2 = 2 - rho(9) = 0.120614758..., and x3 = (2*cos(7*Pi/9))^2 = (2*cos(7*Pi/9))^2 = (R(7,rho(9)))^2 = 4 + rho(9) - rho(9)^2 = 2.347296355... = A130880 + 2, with rho(9) = 2*cos(Pi/9) = A332437, the monic Chebyshev polynomials R (see A127672), and the computation is done modulo the minimal polynomial of rho(9) which is x^3 - 3*x - 1 (see A187360).

This gives the representation of these roots in the power basis of the simple field extension Q(rho(9)). See the linked W. Lang paper in A187360, sect. 4.

This number rho(9)^2 appears as limit of the quotient of consecutive numbers af various sequences, e.g., A094256 and A094829.

The algebraic number rho(9)^2 - 2 = 1.532088898... of Q(rho(9)) has minimal polynomial x^3 - 3*x + 1 over Q. The other roots are -rho(9) = -A332437 and 2 + rho(9) - rho(9)^2 = A130880. - Wolfdieter Lang, Sep 20 2022

Or, numbers k such that k^2 == 1 (mod 9).

Or, numbers k such that the iterative cycle j -> sum of digits of j^2 when started at k contains a 1. E.g., 8 -> 6+4 = 10 -> 1+0+0 = 1 and 17 -> 2+8+9 = 19 -> 3+6+1 = 10 -> 1+0+0 = 1. - Asher Auel, May 17 2001

It is well known that the length sin 70° (cos 20°) is not constructible with ruler and compass, since it is a root of the irreducible polynomial 8x^3 - 6x - 1 and 3 fails to divide any power of 2. - Jean-François Alcover, Aug 10 2014 [cf. the Maxfield ref.]

A cubic number with denominator 2. - Charles R Greathouse IV, Aug 27 2017

From Peter Bala, Oct 21 2021: (Start)

The minimal polynomial of cos(Pi/9) is 8*x^3 - 6*x - 1 with discriminant (2^6)*(3^4), a square: hence the Galois group of the algebraic number field Q(sin(70°) over Q is the cyclic group of order 3.

The two other zeros of the minimal polynomial are cos(5*Pi/9) = - A019819 and cos(7*Pi/9) = - A019859. The mapping z -> 1 - 2*z^2 cyclically permutes the three zeros. The inverse permutation is given by the mapping z -> 2*z^2 - z - 1. (End)

Also: a bond percolation threshold probability on the triangular lattice.

Also: the edge length of a regular 18-gon with unit circumradius. Such an m-gon is not constructible using a compass and a straightedge (see A004169). With an even m, in fact, it would be constructible only if the (m/2)-gon were constructible, which is not true in this case (see A272488). - Stanislav Sykora, May 01 2016

In general, a(n) = (2/m)*Sum_{r=1..m-1} sin(r*j*Pi/m)*sin(r*k*Pi/m)*(2*cos(r*Pi/m))^(2n+1) counts (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < m and |s(i)-s(i-1)| = 1 for i = 1,2,...,2n+1, s(0) = j, s(2n+1) = k.

a(n)/a(n-1) tends to 3.53208888...; = 2 + 2*cos(2*Pi/9) = A332438. - Gary W. Adamson, May 29 2008

From Wolfdieter Lang, Mar 27 2020: (Start)

The explicit form is written in terms of r = rho(9) = 2*cos(Pi/9) = A332437 as a Binet - de Moivre type formula a(n+2) = r^(2*(n+1))*(A(r) + Bp(r)*Cp(r)^(n+1)) + Bm(r)*Cm(r)^(n+1)), with A(r) = (1/9)*(2 + 5*r -r^2), approx. 0.87387081, Bp(r) = (1/18)*((14*r^2 - 5*r - 42)*sqrt(3*(3*r + 1)*(r - 1)) + r^2 - 5*r - 2) = (1/9)*(8 - r - 4*r^2), approx. -0.88974898, Cp(r) = (1/2)*(9*r^2 - 3*r - 26)*(3*r - 1 + sqrt(3*(3*r+1)*(r-1))) = 32 + 4*r - 11*r^2, approx. 0.66456322, Bm(r) = (1/18)*(-(14*r^2 - 5*r - 42)*sqrt(3*(3*r + 1)*(r - 1)) + r^2 - 5*r - 2) = (1/9)*(-10 -4*r + 5*r^2), approx. 0.01587816, and Cm(r) = (1/2)*(9*r^2 - 3*r - 26)*(3*r - 1 - sqrt(3*(3*r + 1)*(r - 1))) = 21 + 2*r - 7*r^2, approx. 0.03414828, for n >= 0.

Proof by partial fraction decomposition of the g.f. using the roots of 1 - 6*x + 9*x^2 - x^3 written in terms of r, which are X1(r) = 1/r^2 = 9 + r - 3*r^2, approx. 0.28311858, Xp(r) = (r/2)*(3*r - 1 + sqrt((3*(3*r+1))*(r-1))) = 1 + 2*r + r^2, approx. 8.29085937, Xm(r) = (r/2)*(3*r - 1 - sqrt((3*(3*r + 1))*(r - 1))) = -1 - 3*r + 2*r^2, approx. 0.42602205. Xp(r)*Xm(r) = r^2. The reduction with the minimal polynomial of r = rho(9), i.e., C(9, x) = x^3 - 3*x - 1 (see A187360) has been used to avoid all powers of r larger than 2. The reciprocal roots turn out to be the roots of the minimal polynomial of r^2, see A332438. 1/X1(r) = r^2, 1/Xp(r) = 2 - r, and 1/Xm(r) = 4 + r - r^2.

This proves the above stated limit of a(n+3)/a(n+2) for n to infinity, namely r^2 = A332438, approx. 3.53208889.

(End)

Previous name was: Let M = the 4 X 4 matrix [0 1 0 0 / 0 0 1 0 / 0 0 0 1 / -1 10 -15 7]. Perform M^n * [1 0 0 0] = [p q r s]. Then a(n-3), a(n-2), a(n-1), a(n) = -p, -q, -r, -s respectively.

a(n)/a(n-1) tends to 3.53208888624... = 4*cos^2(Pi/9), which is an eigenvalue of the matrix and a root of the polynomial x^4 - 6x^3 + 15x^2 -10x + 1 = 0 (having roots 4*cos^2(r*Pi/9), with r = 1,2,3,4).

Number of (s(0), s(1), ..., s(2n+4)) such that 0 < s(i) < 9 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+4, s(0) = 1, s(2n+4) = 7. - Herbert Kociemba, Jun 13 2004

From Wolfdieter Lang, Mar 27 2020: (Start)

This sequence, with offset -5, starting with -85, -10, -1, 0, 0, 0, 1, 7, ... appears in the formula for the n-th power of the 4 X 4 tridiagonal matrix given in A332602 as M_4 = matrix([1,1,0,0], [1,2,1,0], [0,1,2,1], [0,0,1,2]): (M_4)^n = a(n-2)*(M_4)^3 + b(n)*(M_4)^2 + c(n)*M_4 - a(n-3)*1_4, for n >= 0, with the 4 X 4 unit Matrix 1_4, b(n) = -15*a(n-3) + 10*a(n-4) - a(n-5), and c(n) = 10*a(n-3) - a(n-4). Proof from the characteristc polynomial of M_4 (see a comment in A332602) and the Cayley-Hamilton theorem.

From the proof that A094829(n+3)/A094829(n+2) -> rho(9)^2 = A332438 for n-> infinitiy, with rho(9) = 2*cos(Pi/9) = A332437 (see a comment in A094829), and a formula given below the same limit is obtained for a(n+1)/a(n) for n -> infinity, as stated in a comment above. (End)

From Wolfdieter Lang, Mar 27 2020: (Start)

a(n) also gives the upper left entry of the n-th power of the 4 X 4 tridiagonal matrix M_4, given in A332602: M_4 = Matrix([1,1,0,0], [1,2,1,0], [0,1,2,1], [0,0,1,2]): a(n) = (M_4)^n[1,1]. Proof from the formula for (M_4)^n, given in a comment in A094256, derived from the Cayley-Hamilton theorem, which leads to the recurrence. The formula for a(n) in terms of A094256 is given below.

For A094256(n+1)/A094256(n), like for A094829(n+1)/A094829(n), the limit for n -> infinity is rho(9)^2 = A332438 = 3.53208888..., with rho(9) = 2*cos(Pi/9) = A332437. Therefore the formula of a(n) in terms of A094256 shows that the same limit is reached for a(n+1)/a(n). See this conjecture by Gary W. Adamson in A332602.

(End)

From Artur Jasinski, Apr 30 2010: (Start)

Numbers congruent to 4 or 5 mod 9.

Numbers which are not the sum of 3 cubes.

Complement to A060464. (End)

Numbers k such that A010888(k^2) = 7. - V.J. Pohjola, Aug 18 2012

Consider an infinite graph where vertices are selected with probability p. The site percolation threshold is a unique value p_c such that if p > p_c an infinite connected component of selected vertices will almost surely exist, and if p < p_c an infinite connected component will almost surely not exist. This sequence gives p_c for the (3,6,3,6) Kagome Archimedean lattice.

This is one of the three real roots of x^3 - 3x^2 + 1. The other roots are 1 + A332437 = 2.879385241... and -(A332438 - 3) = - 0.5320888862... . - Wolfdieter Lang, Dec 13 2022