cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-9 of 9 results.

A348720 Decimal expansion of 4*cos(2*Pi/13)*cos(3*Pi/13).

Original entry on oeis.org

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

Views

Author

Peter Bala, Oct 31 2021

Keywords

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. The cubic polynomial has discriminant equal to p^2 and has three real roots, one positive and two negative. Here we consider the positive root in the case a = 1 corresponding to the prime p = 13. See A348721 and A348722 for the negative roots.
Let r_0 = 4*cos(2*Pi/13)*cos(3*Pi/13): r_0 is the positive root of the cubic equation x^3 - x^2 - 4*x - 1 = 0. The negative roots are 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....
The roots of the cubic are permuted by the linear fractional transformation x -> - 1/(1 + x) of order 3:
r_1 = - 1/(1 + r_0); r_2 = - 1/(1 + r_1); r_0 = - 1/(1 + r_2).
The quadratic mapping z -> z^2 - 2*z - 2 also cyclically permutes the roots. The mapping z -> - z^2 + z + 3 gives the inverse cyclic permutation of the roots.
The algebraic number field Q(r_0) is a totally real cubic field with class number 1 and discriminant equal to 13^2. The Galois group of Q(r_0) over Q is a cyclic group of order 3. See Shanks, Table 1, entry corresponding to a = 1.
The real numbers r_0 and 1 + r_0 are two independent fundamental units of the field Q(r_0). See Shanks. In Cusick and Schoenfeld, r_0 and r_1 (denoted there by E_1 and E_2) are taken as a fundamental pair of units (see case 4 in the table).

Examples

			2.651093408937175306253240337787615403132441075705596684018767...
		

Crossrefs

Programs

  • Maple
    evalf(4*cos(2*Pi/13)*cos(3*Pi/13), 100);
  • Mathematica
    RealDigits[4*Cos[2*Pi/13]*Cos[3*Pi/13], 10, 100][[1]] (* Amiram Eldar, Nov 08 2021 *)
  • PARI
    polrootsreal(x^3 - x^2 - 4*x - 1)[3] \\ Charles R Greathouse IV, Oct 30 2023

Formula

r_0 = 2*(cos(Pi/13) + cos(5*Pi/13)).
r_0 = sin(4*Pi/13)*sin(6*Pi/13) / (sin(2*Pi/13)*sin(3*Pi/13)).
r_0 = Product_{n >= 0} (13*n+4)*(13*n+6)*(13*n+7)*(13*n+9)/( (13*n+2)*(13*n+3)*(13*n+10)*(13*n+11) ).
r_1 = 2*(cos(3*Pi/13) - cos(2*Pi/13)).
r_1 = - sin(Pi/13)*sin(5*Pi/13)/(sin(4*Pi/13)*sin(6*Pi/13)).
r_1 = - Product_{n >= 0} (13*n+1)*(13*n+5)*(13*n+8)*(13*n+12)/( (13*n+4)*(13*n+6)*(13*n+7)*(13*n+9) ).
r_2 = 2*(cos(7*Pi/13) - cos(4*Pi/13)).
r_2 = - sin(2*Pi/13)*sin(3*Pi/13)/(sin(Pi/13)*sin(5*Pi/13)).
r_2 = - 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
r_0 = abs( (1 - z^4)*(1 - z^6)/((1 - z^2)*(1 - z^3)) );
r_1 = - abs( (1 - z)*(1 - z^5)/((1 - z^4)*(1 - z^6)) );
r_2 = - 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)^(1/13) + (-1)^(5/13) - (-1)^(8/13) - (-1)^(12/13). - Peter Luschny, Nov 08 2021

A348728 Decimal expansion of the absolute value of one of the negative roots of Shanks' simplest cubic associated with the prime p = 37.

Original entry on oeis.org

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

Views

Author

Peter Bala, Oct 31 2021

Keywords

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. The polynomial has three real roots, one positive and two negative.
In the case a = 4, corresponding to the prime p = 37, the three real roots of the cubic x^3 - 4*x^2 - 7*x - 1 in descending order are r_0 = 5.3447123654..., r_1 = - 0.1576115578... and r_2 = - 1.1871008076....
Here we consider the absolute value of the root r_2. See A348726 (r_0) and A348727 (|r_1|) for the other two roots.

Examples

			1.18710080760640920168337009872276109935284715168366...
		

Crossrefs

Programs

  • Maple
    R := k -> sin(k*Pi/37)*sin(6*k*Pi/37)*sin(8*k*Pi/37)*sin(10*k*Pi/37)* sin(11*k*Pi/37)*sin(14*k*Pi/37): evalf(-R(3)/R(1), 100);
  • Mathematica
    f[ks_,m_] := Product[Sin[k*Pi/m], {k,ks}]; ks = {1, 6, 8, 10, 11, 14}; RealDigits[f[3*ks,37]/f[ks,37], 10, 100][[1]] (* Amiram Eldar, Nov 08 2021 *)

Formula

|r_2| = 2*(-cos(Pi/37) + cos(6*Pi/37) + cos(8*Pi/37) + cos(10*Pi/37) - cos(11*Pi/37) + cos(14*Pi/37)) - 1.
|r_2| = |R(3)/R(1)|, where R(k) = sin(k*Pi/37)*sin(6*k*Pi/37)*sin(8*k*Pi/37)* sin(10*k*Pi/37)*sin(11*k*Pi/37)*sin(14*k*Pi/37).
Let R = <1, 6, 8, 10, 11, 14, 23, 26, 27, 29, 31, 36> denote the multiplicative subgroup of nonzero cubic residues in the finite field Z_37, with cosets 2*R = {2, 9, 12, 15, 16, 17, 20, 21, 22, 25, 28, 35} and 3*R = {3, 4, 5, 7, 13, 18, 19, 24, 30, 32, 33, 34}. Then the constant equals Product_{n >= 0} ( Product_{k in the coset 3*R} (37*n+k) )/( Product_{k in the group R} (37*n + k) ).

A348721 Decimal expansion of 4*cos(4*Pi/13)*cos(6*Pi/13).

Original entry on oeis.org

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

Views

Author

Peter Bala, Oct 31 2021

Keywords

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_1.
See A348720 and A348722 for the other two roots.

Examples

			0.27389055496421759453148984462749498951809365234341 ...
		

Crossrefs

Programs

  • Maple
    evalf(4*cos(4*Pi/13)*cos(6*Pi/13), 100);
  • Mathematica
    RealDigits[4*Cos[4*Pi/13]*Cos[6*Pi/13], 10, 100][[1]] (* Amiram Eldar, Nov 08 2021 *)

Formula

Equals 2*(cos(2*Pi/13) - cos(3*Pi/13)).
Equals sin(Pi/13)*sin(5*Pi/13)/(sin(4*Pi/13)*sin(6*Pi/13)).
Equals Product_{n >= 0} (13*n+1)*(13*n+5)*(13*n+8)*(13*n+12)/( (13*n+4)*(13*n+6)*(13*n+7)*(13*n+9) ).
Equivalently, let z = exp(2*Pi*i/13). Then the constant = abs( (1 - z)*(1 - z^5)/ ((1 - z^4)*(1 - z^6)) ).
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)^(2/13) - (-1)^(3/13) + (-1)^(10/13) - (-1)^(11/13). - Peter Luschny, Nov 08 2021

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

Original entry on oeis.org

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

Views

Author

Peter Bala, Oct 31 2021

Keywords

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 ...
		

Crossrefs

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

A348723 Decimal expansion of the positive root of Shanks' simplest cubic associated with the prime p = 19.

Original entry on oeis.org

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

Views

Author

Peter Bala, Oct 31 2021

Keywords

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 = 2, corresponding to the prime p = 19, Shanks' cyclic cubic is x^3 - 2*x^2 - 5*x - 1 of discriminant 19^2. The polynomial has three real roots, one positive and two negative. Let r_0 = 3.507018644... denote the positive root. The other roots are r_1 = - 1/(1 + r_0) = - 0.2218761622... and r_2 = - 1/(1 + r_1) = - 1.2851424818.... See A348724 and A348725.
The quadratic mapping z -> z^2 - 3*z - 2 cyclically permutes the roots of the cubic: the mapping z -> - z^2 + 2*z + 4 gives the inverse cyclic permutation of the three roots.
The algebraic number field Q(r_0) is a totally real cubic field of class number 1 and discriminant equal to 19^2. 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. In Cusick and Schoenfeld, r_0 and r_1 (denoted there by E_1 and E_2) are taken as a fundamental pair of units (see case 9 in the table).

Examples

			3.50701864409297629866079992371567802902597642013036...
		

Crossrefs

Programs

  • Maple
    evalf(sin(4*Pi/19)*sin(6*Pi/19)*sin(9*Pi/19)/(sin(Pi/19)*sin(7*Pi/19)*sin(8*Pi/19)), 100);
  • Mathematica
    RealDigits[Sin[4*Pi/19]*Sin[6*Pi/19]*Sin[9*Pi/19]/(Sin[Pi/19]*Sin[7*Pi/19]*Sin[8*Pi/19]), 10, 100][[1]] (* Amiram Eldar, Nov 08 2021 *)
  • PARI
    sin(4*Pi/19)*sin(6*Pi/19)*sin(9*Pi/19)/(sin(Pi/19)*sin(7*Pi/19)*sin(8*Pi/19)) \\ Michel Marcus, Nov 08 2021

Formula

r_0 = 2*(cos(4*Pi/19) + cos(6*Pi/19) - cos(9*Pi/19)) + 1.
r_0 = sin(4*Pi/19)*sin(6*Pi/19)*sin(9*Pi/19)/(sin(Pi/19)*sin(7*Pi/19)*sin(8*Pi/19)).
r_0 = 1/(8*cos(4*Pi/19)*cos(6*Pi/19)*cos(9*Pi/19)).
r_0 = Product_{n >= 0} (19*n+4)*(19*n+6)*(19*n+9)*(19*n+10)*(19*n+13)*(19*n+15)/( (19*n+1)*(19*n+7)*(19*n+8)*(19*n+11)*(19*n+12)*(19*n+18) ).
r_1 = - sin(2*Pi/19)*sin(3*Pi/19)*sin(5*Pi/19)/(sin(4*Pi/19)*sin(6*Pi/19)*sin(9*Pi/19)) = - 1/(8*cos(2*Pi/19)*cos(3*Pi/19)*cos(5*Pi/19)).
r_1 = - Product_{n >= 0} (19*n+2)*(19*n+3)*(19*n+5)*(19*n+14)*(19*n+16)*(19*n+17)/( (19*n+4)*(19*n+6)*(19*n+9)*(19*n+10)*(19*n+13)*(19*n+15) ).
r_2 = - sin(Pi/19)*sin(7*Pi/19)* sin(8*Pi/19)/(sin(2*Pi/19)*sin(3*Pi/19)*sin(5*Pi/19)) = - 1/(8*cos(Pi/19)*cos(7*Pi/19)*cos(8*Pi/19)).
r_2 = - Product_{n >= 0} (19*n+1)*(19*n+7)*(19*n+8)*(19*n+11)*(19*n+12)*(19*n+18)/( (19*n+2)*(19*n+3)*(19*n+5)*(19*n+14)*(19*n+16)*(19*n+17) ).
Let z = exp(2*Pi*i/19). Then
r_0 = abs( (1 - z^4)*(1 - z^6)*(1 - z^9)/((1 - z)*(1 - z^7)*(1 - z^8)) ).
Note: C = {1, 7, 8, 11, 12, 18} is the subgroup of nonzero cubic residues in the finite field Z_19 with cosets 2*C = {2, 3, 5, 14, 16, 17} and 4*C = {4, 6, 9, 10, 13, 15}.

A348724 Decimal expansion of the absolute value of one of the negative roots of Shanks' simplest cubic associated with the prime p = 19.

Original entry on oeis.org

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

Views

Author

Peter Bala, Oct 31 2021

Keywords

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 = 2, corresponding to the prime p = 19, Shanks' cyclic cubic is x^3 - 2*x^2 - 5*x - 1 of discriminant 19^2. The polynomial has three real roots, one positive and two negative. Let r_0 = 3.507018644... denote the positive root. The other roots are r_1 = - 1/(1 + r_0) = - 0.2218761622... and r_2 = - 1/(1 + r_1) = - 1.2851424818.... See A348723 and A348725.
Here we consider the absolute value of the root r_1. In Cusick and Schoenfeld r_1 is denoted by E_2. See case 9 in the table.

Examples

			0.22187616226319093426668005018505061559919549440775...
		

Crossrefs

Programs

  • Maple
    evalf(sin(2*Pi/19)*sin(3*Pi/19)*sin(5*Pi/19)/(sin(4*Pi/19)*sin(6*Pi/19)*sin(9*Pi/19)), 100);
  • Mathematica
    RealDigits[Sin[2*Pi/19]*Sin[3*Pi/19]*Sin[5*Pi/19]/(Sin[4*Pi/19]*Sin[6*Pi/19]*Sin[9*Pi/19]), 10, 100][[1]] (* Amiram Eldar, Nov 08 2021 *)
  • PARI
    sin(2*Pi/19)*sin(3*Pi/19)*sin(5*Pi/19)/(sin(4*Pi/19)*sin(6*Pi/19)*sin(9*Pi/19)) \\ Michel Marcus, Nov 08 2021

Formula

|r_1| = 2*(cos(3*Pi/19) + cos(5*Pi/19) - cos(2*Pi/19)) - 1.
|r_1| = sin(2*Pi/19)*sin(3*Pi/19)*sin(5*Pi/19)/(sin(4*Pi/19)*sin(6*Pi/19) *sin(9*Pi/19)) = 1/(8*cos(2*Pi/19)*cos(3*Pi/19)*cos(5*Pi/19)).
|r_1| = Product_{n >= 0} (19*n+2)*(19*n+3)*(19*n+5)*(19*n+14)*(19*n+16)*(19*n+17)/( (19*n+4)*(19*n+6)*(19*n+9)*(19*n+10)*(19*n+13)*(19*n+15) ).
Let z = exp(2*Pi*i/19). Then
|r_1| = abs( (1 - z^2)*(1 - z^3)*(1 - z^5)/((1 - z^4)*(1 - z^6)*(1 - z^9)) ).
Note: C = {1, 7, 8, 11, 12, 18} is the subgroup of nonzero cubic residues in the finite field Z_19 with cosets 2*C = {2, 3, 5, 14, 16, 17} and 4*C = {4, 6, 9, 10, 13, 15}.
Equals -1 - (-1)^(2/19) + (-1)^(3/19) + (-1)^(5/19) - (-1)^(14/19) - (-1)^(16/19) + (-1)^(17/19). - Peter Luschny, Nov 08 2021

A348725 Decimal expansion of the absolute value of one of the negative roots of Shanks' simplest cubic associated with the prime p = 19.

Original entry on oeis.org

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

Views

Author

Peter Bala, Oct 31 2021

Keywords

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 = 2, corresponding to the prime p = 19, Shanks' cyclic cubic is x^3 - 2*x^2 - 5*x - 1 of discriminant 19^2. The polynomial has three real roots, one positive and two negative. Let r_0 = 3.507018644... denote the positive root. The other roots are r_1 = - 1/(1 + r_0) = - 0.2218761622... and r_2 = - 1/(1 + r_1) = - 1.2851424818.... See A348723 (r_0) and A348724 (|r_1|).
Here we consider the absolute value of the root r_2.

Examples

			1.28514248182978536439411987353062741342678092572261 ...
		

Crossrefs

Programs

  • Maple
    evalf(sin(Pi/19)*sin(7*Pi/19)*sin(8*Pi/19)/(sin(2*Pi/19)*sin(3*Pi/19)*sin(5*Pi/19)), 100);
  • Mathematica
    RealDigits[Sin[Pi/19]*Sin[7*Pi/19]*Sin[8*Pi/19]/(Sin[2*Pi/19]*Sin[3*Pi/19]*Sin[5*Pi/19]), 10, 100][[1]] (* Amiram Eldar, Nov 08 2021 *)

Formula

|r_2| = sin(Pi/19)*sin(7*Pi/19)*sin(8*Pi/19)/(sin(2*Pi/19)*sin(3*Pi/19)* sin(5*Pi/19)) = 1/(8*cos(Pi/19)*cos(7*Pi/19)*cos(8*Pi/19)).
|r_2| = Product_{n >= 0} (19*n+1)*(19*n+7)*(19*n+8)*(19*n+11)*(19*n+12)*(19*n+18)/ ( (19*n+2)*(19*n+3)*(19*n+5)*(19*n+14)*(19*n+16)*(19*n+17) ).
|r_2| = 2*(cos(Pi/19) + cos(7*Pi/19) - cos(8*Pi/19)) - 1.
Let z = exp(2*Pi*i/19). Then
|r_2| = abs( (1 - z)*(1 - z^7)*(1 - z^8)/((1 - z^2)*(1 - z^3)*(1 - z^5)) ).
Note: C = {1, 7, 8, 11, 12, 18} is the subgroup of nonzero cubic residues in the finite field Z_19 with cosets 2*C = {2, 3, 5, 14, 16, 17} and 4*C = {4, 6, 9, 10, 13, 15}.
Equals -1 + (-1)^(1/19) + (-1)^(7/19) - (-1)^(8/19) + (-1)^(11/19) - (-1)^(12/19) - (-1)^(18/19). - Peter Luschny, Nov 08 2021

A348726 Decimal expansion of the positive root of Shanks' simplest cubic associated with the prime p = 37.

Original entry on oeis.org

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

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Author

Peter Bala, Oct 31 2021

Keywords

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. The polynomial has three real roots, one positive and two negative.
In the case a = 4, corresponding to the prime p = 37, the three real roots of the cubic x^3 - 4*x^2 - 7*x - 1 in descending order are r_0 = 5.3447123654..., r_1 = - 0.1576115578... and r_2 = - 1.1871008076.... Here we consider the positive root r_0. See A348727 (|r_1|) and A348728 (|r_2|) for the other two roots.
The algebraic number field Q(r_0) is a totally real cubic field with class number 1 and discriminant equal to 37^2. 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. In Cusick and Schoenfeld, r_0 and r_1 (denoted there by E_1 and E_2) are taken as a fundamental pair of units (see case 37 in the table).
Let R = <1, 6, 8, 10, 11, 14, 23, 26, 27, 29, 31, 36> denote the multiplicative subgroup of nonzero cubic residues in the finite field Z_37, with cosets 2*R = {2, 9, 12, 15, 16, 17, 20, 21, 22, 25, 28, 35} and 3*R = {3, 4, 5, 7, 13, 18, 19, 24, 30, 32, 33, 34}.
Define R(k) = sin(k*Pi/37)*sin(6*k*Pi/37)*sin(8*k*Pi/37)*sin(10*k*Pi/37)* sin(11*k*Pi/37)*sin(14*k*Pi/37). Then the three roots of the cubic x^3 - 4*x^2 - 7*x - 1 are
r_0 = - R(2)/R(3) = 5.3447123654..., r_1 = - R(1)/R(2) = - 0.1576115578... and r_2 = R(3)/R(1) = - 1.1871008076....
The linear fractional transformation z -> - 1/(1 + z) cyclically permutes the three roots of the cubic polynomial.
The quadratic mapping z -> z^2 - 5*z - 2 also cyclically permutes the roots of the cubic: the inverse cyclic permutation of the roots is given by z -> - z^2 + 4*z + 6.

Examples

			5.34471236545183496316569141884698646995869587081422 ...
		

Crossrefs

Programs

  • Maple
    R := k -> sin(k*Pi/37)*sin(6*k*Pi/37)*sin(8*k*Pi/37)*
    sin(10*k*Pi/37)*sin(11*k*Pi/37)*sin(14*k*Pi/37): evalf(-R(2)/R(3), 100);
  • Mathematica
    f[ks_,m_] := Product[Sin[k*Pi/m], {k,ks}]; ks = {1, 6, 8, 10, 11, 14}; RealDigits[f[2*ks,37]/f[3*ks,37], 10, 100][[1]] (* Amiram Eldar, Nov 08 2021 *)

Formula

r_0 = 1 + 2*(cos(3*Pi/37) - cos(4*Pi/37) + cos(5*Pi/37) + cos(7*Pi/37) + cos(13*Pi/37) - cos(18*Pi/37)).
r_0 = |R(2)/R(3)| = Product_{n >= 0} ( Product_{k in the coset 2*R} (37*n+k) )/( Product_{k in the coset 3*R} (37*n + k) );
|r_1| = |R(1)/R(2)| = Product_{n >= 0} ( Product_{k in the group R} (37*n+k) )/( Product_{k in the coset 2*R} (37*n + k) );
|r_2| = |R(3)/R(1)| = Product_{n >= 0} ( Product_{k in the coset 3*R} (37*n+k) )/( Product_{k in the group R} (37*n + k) ).
R(2)/R(1) + R(2)/R(3) = 1 = R(3)/R(2) - R(3)/R(1) = R(1)/R(2) - R(1)/R(3).

A348727 Decimal expansion of the absolute value of one of the negative roots of Shanks' simplest cubic associated with the prime p = 37.

Original entry on oeis.org

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

Views

Author

Peter Bala, Oct 31 2021

Keywords

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. The polynomial has three real roots, one positive and two negative.
In the case a = 4, corresponding to the prime p = 37, the three real roots of the cubic x^3 - 4*x^2 - 7*x - 1 in descending order are r_0 = 5.344712365..., r_1 = - 0.1576115578... and r_2 = - 1.187100807.... Here we consider the absolute value of the root r_1 (|E_2| in the notation of Cusick and Schoenfeld). See A348726 (r_0) and A348728 (|r_2|) for the other two roots.

Examples

			0.15761155784542576148232132012422537060584871913055 ...
		

Crossrefs

Programs

  • Maple
    R := k -> sin(k*Pi/37)*sin(6*k*Pi/37)*sin(8*k*Pi/37)*sin(10*k*Pi/37)* sin(11*k*Pi/37)*sin(14*k*Pi/37): evalf(R(1)/R(2), 100);
  • Mathematica
    f[ks_,m_] := Product[Sin[k*Pi/m], {k,ks}]; ks = {1, 6, 8, 10, 11, 14}; RealDigits[f[ks,37]/f[2*ks,37], 10, 100][[1]] (* Amiram Eldar, Nov 08 2021 *)

Formula

|r_1| = 1/((2^6)*cos(Pi/37)*cos(6*Pi/37)*cos(8*Pi/37)*cos(10*Pi/37)*cos(11*Pi/37)* cos(14*Pi/37)).
|r_1| = 2*(cos(2*Pi/37) - cos(9*Pi/37) + cos(12*Pi/37) - cos(15*Pi/37) + cos(16*Pi/37) - cos(17*Pi/37)) - 1.
|r_1| = R(1)/R(2), where R(k) = sin(k*Pi/37)*sin(6*k*Pi/37)* sin(8*k*Pi/37)*sin(10*k*Pi/37)*sin(11*k*Pi/37)*sin(14*k*Pi/37).
Let R = <1, 6, 8, 10, 11, 14, 23, 26, 27, 29, 31, 36> denote the multiplicative subgroup of nonzero cubic residues in the finite field Z_37, with cosets 2*R = {2, 9, 12, 15, 16, 17, 20, 21, 22, 25, 28, 35} and 3*R = {3, 4, 5, 7, 13, 18, 19, 24, 30, 32, 33, 34}. Then constant equals
Product_{n >= 0} ( Product_{k in the coset 2*R} (37*n+k) )/( Product_{k in the group R} (37*n + k) ).
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