A348722 Decimal expansion of 4*cos(8*Pi/13)*cos(12*Pi/13).

1, 3, 7, 7, 2, 0, 2, 8, 5, 3, 9, 7, 2, 9, 5, 7, 7, 1, 1, 7, 2, 1, 7, 5, 0, 4, 9, 3, 1, 6, 0, 1, 2, 0, 4, 1, 3, 6, 1, 4, 3, 4, 7, 4, 2, 3, 3, 6, 2, 1, 7, 9, 1, 4, 8, 5, 5, 3, 2, 2, 2, 6, 5, 1, 1, 6, 8, 7, 5, 2, 5, 1, 8, 1, 1, 6, 5, 0, 2, 1, 7, 7, 6, 8, 2, 2, 3, 3, 1, 9, 6, 0, 9, 2, 5, 6, 8, 5, 5, 7

Offset: 1

Author

Peter Bala, Oct 31 2021

Keywords

• nonn
• easy
• cons

Comments

Let a be an integer and let p be a prime of the form a^2 + 3*a + 9 (see A005471). Shanks introduced a family of cyclic cubic fields generated by the roots of the polynomial x^3 - a*x^2 - (a + 3)*x - 1.

In the case a = 1, corresponding to the prime p = 13, Shanks' cyclic cubic is x^3 - x^2 - 4*x - 1 of discriminant 13^2. The three real roots of the cubic are r_0 = 4*cos(2*Pi/13)*cos(3*Pi/13) = 2.6510934089..., r_1 = - 4*cos(4*Pi/13)*cos(6*Pi/13) = - 0.2738905549... and r_2 = - 4*cos(8*Pi/13)*cos(12*Pi/13) = - 1.3772028539.... Here we consider the absolute value of the root r_2.

See A348720 and A348721 for the other two roots.

Examples

			1.3772028539729577117217504931601204136143474233621 ...
		

Links

• T. W. Cusick and Lowell Schoenfeld, A table of fundamental pairs of units in totally real cubic fields, Math. Comp. 48 (1987), 147-158 (see case 4 in the Table)
• D. Shanks, The simplest cubic fields, Math. Comp., 28 (1974), 1137-1152

Crossrefs

Cf. A005471, A160389, A255240, A255241, A255249, A348720 - A348729.

Programs

Maple

evalf(4*cos(8*Pi/13)*cos(12*Pi/13), 100);

Mathematica

RealDigits[4*Cos[8*Pi/13]*Cos[12*Pi/13], 10, 100][[1]] (* Amiram Eldar, Nov 08 2021 *)

Formula

Equals 4*cos(Pi/13)*cos(5*Pi/13).

Equals 2*(cos(4*Pi/13) + cos(6*Pi/13)).

Equals 2*(cos(Pi/13) + cos(5*Pi/13) - cos(2*Pi/13) - cos(10*Pi/13)) - 1.

Equals sin(2*Pi/13)*sin(3*Pi/13)/(sin(Pi/13)*sin(5*Pi/13)).

Equals Product_{n >= 0} (13*n+2)*(13*n+3)*(13*n+10)*(13*n+11)/( (13*n+1)*(13*n+5)*(13*n+8)*(13*n+12) ).

Equivalently, let z = exp(2*Pi*i/13). Then the constant equals abs( (1 - z^2)*(1 - z^3)/((1 - z)*(1 - z^5)) ).

Note: C = {1, 5, 8, 12} is the subgroup of nonzero cubic residues in the finite field Z_13 with cosets 2*C = {2, 3, 10, 11} and 4*C = {4, 6, 7, 9}.

Equals (-1)^(4/13) + (-1)^(6/13) - (-1)^(7/13) - (-1)^(9/13). - Peter Luschny, Nov 08 2021