A348729 Decimal expansion of the positive root of Shanks's simplest cubic associated with the prime p = 163.

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

Offset: 2

Author

Peter Bala, Nov 06 2021

Keywords

• nonn
• cons
• easy

Comments

Let a be a natural number 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. The polynomial has three real roots, one positive and two negative. In the case a = 11, corresponding to the prime p = 163, the three real roots of Shanks' cubic x^3 - 11*x^2 - 14*x - 1 in descending order are r_0 = 12.1582466687..., r_1 = - -0.0759979672... and r_2 = -1.0822487014.... Here we consider the positive root r_1.

The linear fractional transformation z -> - 1/(1 + z) cyclically permutes the three roots r_0, r_1 and r_2: the quadratic mapping z -> z^2 - 12*z - 2 also cyclically permutes the roots.

The algebraic number field Q(r_0) is a totally real cubic field with class number 4 and discriminant equal to 163^2. The Galois group of Q(r_0) over Q is a cyclic group of order 3. The real numbers r_0 and 1 + r_0 are two independent fundamental units of the field Q(r_0). See Shanks.

Examples

			12.15824666871213538260037124700042984524658480470748 ...
		

Links

• D. Shanks, The simplest cubic fields, Math. Comp., 28 (1974), 1137-1152.
• Index entries for algebraic numbers, degree 3.

Crossrefs

Cf. A005471, A160389, A255240, A255241, A255249, A348720 - A348728.

Programs

Maple

R := convert([seq(mod(n^3, 163), n = 1..162)], set):
P := k -> sqrt( mul(sin((1/163)*k*r*Pi), r in R) ):
evalf(sqrt(P(3)/P(1)), 105);

Mathematica

rs = Union@Mod[Range[1, 162]^3, 163]; f[k_] := Sqrt[Product[Sin[k*r*Pi/163], {r, rs}]]; RealDigits[Sqrt[f[3]/f[1]], 10, 100][[1]] (* Amiram Eldar, Nov 08 2021 *)

PARI

polrootsreal(x^3 - 11*x^2 - 14*x - 1)[3] \\ Charles R Greathouse IV, Feb 04 2025

Formula

Let R = {1, 5, 6, 8, ..., 155, 157, 158, 162} denote the multiplicative subgroup of nonzero cubic residues in the finite field Z_163, with cosets 2*R = {2, 7, 9, 10, ..., 153, 154, 156, 161} and 3*R = {3, 4, 11, 14, ..., 149, 152, 159, 160}.

Define P(k) = Product_{r in R, r <= (163-1)/2} sin(k*r*Pi/163). The three roots of the cubic x^3 - 11*x^2 - 14*x - 1 are

r_0 = sqrt(P(3)/P(1)) = 12.1582466687....

r_1 = -sqrt(P(1)/P(2)) = -0.0759979672....

r_2 = -sqrt(P(2)/P(3)) = -1.0822487014....