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.

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A227622 Primes p of the form m^2 + 27.

Original entry on oeis.org

31, 43, 127, 223, 283, 811, 1051, 1471, 1627, 2143, 2731, 3163, 3391, 4651, 5503, 6427, 8863, 9631, 16411, 16927, 18523, 23131, 23743, 27583, 28927, 29611, 33151, 37663, 42463, 43291, 44971, 45823, 56671, 65563, 70783, 78427, 80683, 84127, 87643, 106303, 110251, 122527, 123931, 131071
Offset: 1

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Author

William P. Orrick, Jul 17 2013

Keywords

Comments

Orders for which residues mod p of the form x^i, i congruent to 0, 1, or 3 (mod 6), form a difference set with parameters (v,k,lambda)=(p,(p-1)/2,(p-3)/4), where x is a primitive root such that 3=x^j, with j congruent to 1 (mod 6). This construction is due to Marshall Hall. Such a difference set has the same parameters as the difference set formed by quadratic residues, that is, the Paley difference set, but is not equivalent to it. Both difference sets give rise to Hadamard matrices of size p+1.
From Peter Bala, Nov 19 2021: (Start)
2 is a cube mod p (a particular case of a more general result of Gauss). See A014752.
Primes of the form a^2 + 6*a + 36, where a is an integer.
Let p == 1 (mod 6) be a prime. There are integers c and d, unique up to sign, such that 4*p = c^2 + 27*d^2 [see, for example, Ireland and Rosen, Proposition 8.3.2]. This sequence lists those primes with d = 2. Cf. A005471 (case d = 1) and A349461 (case d = 3).
Primes p of the form m^2 + 27 are related to cyclic cubic fields in several ways:
(1) The cubic polynomial X^3 - p*X + 2*p, with discriminant 4*m^2*p^2, a square, is irreducible over Q by Eisenstein's criteria. It follows that the Galois group of the polynomial over Q is the cyclic group C_3 (apply Conrad, Corollary 2.5).
Note that the roots of the cubic X^3 - p*X + 2*p, are the differences n_0 - n_1, n_1 - n_2 and n_2 - n_0 of the cubic Gaussian periods n_i for the modulus p.
(2) The cubic 2*X^3 + p*(X + 2)^2, with discriminant 64*m^2*p^2, a square, is irreducible over Q by Eisenstein's criteria, and so the Galois group of the polynomial over Q is the cyclic group C_3.
(3) The cubic X^3 - (m-3)*X^2 - 2*(m+3)*X - 8, has discriminant (2*p)^2, a square. (This is the polynomial g_3(m-3, 0, -2; X) in the notation of Hashimoto and Hoshi.) The cubic is irreducible over Q for nonzero m by the Rational Root Theorem and hence the Galois group of the polynomial over Q is the cyclic group C_3. (End)

Examples

			For p=31, using x=3 as primitive root, the set of residues {1,2,3,4,6,8,12,15,16,17,23,24,27,29,30} is a difference set.
2 a cube mod p: 4^3 == 2 (mod 31); 20^3 == 2 (mod 43); 8^3 == 2 (mod 127); 68^3 == 2 (mod 223). - _Peter Bala_, Nov 19 2021
		

References

  • K. Ireland and M. Rosen, A classical introduction to modern number theory, vol. 84, Graduate Texts in Mathematics. Springer-Verlag. [Prop. 8.3.2, p. 96]
  • Thomas Storer, Cyclotomy and difference sets. Markham, Chicago, 1967, pages 73-76.

Crossrefs

Programs

  • Mathematica
    Select[Table[m^2+27,{m,0,100}],PrimeQ]
  • PARI
    for(n=0,10^3,my(p=n^2+27);if(isprime(p),print1(p,", "))); \\ Joerg Arndt, Jul 18 2013

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) ).

A349461 Primes of the form m^2 + 9*m + 81.

Original entry on oeis.org

61, 67, 73, 103, 151, 193, 271, 367, 523, 613, 661, 991, 1117, 1321, 1543, 1621, 1783, 1867, 2131, 2713, 3253, 3967, 4093, 4483, 6067, 6703, 7717, 8803, 9181, 10567, 11617, 11833, 13171, 13633, 14341, 15313
Offset: 1

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Author

Peter Bala, Nov 18 2021

Keywords

Comments

3 is a cube mod p for all primes in this list; this is a particular case of a result of Gauss. See Ireland and Rosen, Chapter 9, Exercise 23, p. 135. Some examples are given below.
Primes p such that 4*p - 243 is a square. Let p == 1 (mod 6) be a prime. There are integers c and d such that 4*p = c^2 + 27*d^2 (see, for example, Ireland and Rosen, Proposition 8.3.2). This sequence lists the primes with d = 3. Cf. A005471 (case d = 1) and A227622 (case d = 2).
Primes p of the form m^2 + 9*m + 81 are related to cyclic cubic fields in several ways:
(1) The cubic x^3 - p*x + 3*p, with discriminant ((2*m + 9)*p)^2, is irreducible over Q by Eisenstein's criteria. It follows that the Galois group of the polynomial over Q is the cyclic group C_3 (apply Conrad, Corollary 2.5).
Note that the roots of x^3 - p*x + 3*p are the differences n_0 - n_1, n_1 - n_2 and n_2 - n_0, where n_0, n_1 and n_2 are the three cubic Gaussian periods for the modulus p.
(2) The cubic x^3 - m*x^2 - 3*(m + 9)*x - 27 has discriminant (3*p)^2, a square. This is the polynomial g_3(a, 0, -3; X) in the notation of Hashimoto and Hoshi. The cubic is irreducible over Q by the Rational Root Theorem and hence the Galois group of the polynomial over Q is the cyclic group C_3.
(3) The cubic 3*x^3 + p*(x + 3)^2, with discriminant 81*p^2*(4*p - 243), a square, is irreducible over Q by Eisenstein's criteria. It follows that the Galois group of the polynomial over Q is the cyclic group C_3.

Examples

			61 = (-4)^2 + 9*(-4) + 81; 67 = (-2)^2 + 9*(-2) + 81; 73 = (-1)^2 + 9*(-1) + 81; 103 = (2)^2 + 9*(2) + 81.
3 is a cube mod p:
4^3 == 3 (mod 61); 18^3 == 3 (mod 67); 25^3 == 3 (mod 73); 67^3 == 3 (mod 103).
		

References

  • K. Ireland and M. Rosen, A classical introduction to modern number theory, Vol. 84, Graduate Texts in Mathematics, Springer-Verlag.

Crossrefs

Programs

  • Mathematica
    Select[Table[m^2+9*m+81, {m, -4, 120}], PrimeQ]
  • PARI
    for (m = -4, 120, my(p = m^2 + 9*m + 81); if (isprime(p), print1(p,", ")));

A084711 -21*zeta_K(-1), where K runs through the simplest cubic fields.

Original entry on oeis.org

1, 7, 21, 147, 1393, 2569, 7525, 13748, 30171, 43491, 100639, 123456, 224175, 296991, 495367, 658684, 1013008, 1438440, 1954357, 2310462, 3813903, 4092029, 5916929, 6974592, 10276475, 10989267, 14919105, 21957309, 26812232, 36865003
Offset: 1

Views

Author

N. J. A. Sloane, Jul 03 2003

Keywords

Crossrefs

See A005471 for discriminants. Cf. A084719, A084734.

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Jul 03 2003

A084719 8190*zeta_K(-3), where K runs through the simplest cubic fields.

Original entry on oeis.org

3033, 213796, 3237611, 346757031, 70148893905, 295137877145, 3661984928043, 11211220470548, 82812514410205, 204891119186297, 1080348973121981, 2303987041770259, 9457781291826393, 18239333904161593, 62303386892919535, 111016989991835620, 328979435981818027
Offset: 1

Views

Author

N. J. A. Sloane, Jul 03 2003

Keywords

Crossrefs

See A005471 for discriminants. Cf. A084711, A084734.

Extensions

Terms a(5) and beyond from Kim and Kim added by Andrey Zabolotskiy, Mar 25 2021

A084734 -3591*zeta_K(-5), where K runs through the simplest cubic fields.

Original entry on oeis.org

421401, 372237052, 24797946347, 37916777337111, 159408405876911409, 1524411665780115233, 79758314837247277491, 459985258296743678924, 10707168309538015545757, 44491781846495075733473, 602111794355444803103621, 1993855069179097491346675, 18336798043602484502981169
Offset: 1

Views

Author

N. J. A. Sloane, Jul 03 2003

Keywords

Crossrefs

See A005471 for discriminants. Cf. A084719, A084711.

Extensions

Terms a(4) and beyond from Kim and Kim added by Andrey Zabolotskiy, Mar 25 2021

A128878 Primes of the form 47*n^2 - 1701*n + 10181.

Original entry on oeis.org

10181, 8527, 6967, 5501, 4129, 2851, 1667, 577, 379, 1451, 2617, 3877, 5231, 6679, 8221, 9857, 11587, 13411, 15329, 17341, 19447, 21647, 31387, 34057, 36821, 39679, 45677, 48817, 52051, 65927, 81307, 89561, 102647, 107197, 116579, 126337, 131357
Offset: 1

Views

Author

Douglas Winston (douglas.winston(AT)srupc.com), Apr 17 2007

Keywords

Comments

Primes are given in the order in which they arise for increasing n.
Polynomial generates 22 primes for 0 <= n <= 42, i.e., for n = 0, 1, 2, 3, 4, 5, 6, 7, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42.
If the definition is replaced by "Numbers n of the form 47*k^2 - 1701*k + 10181 such that either n or -n is a prime" we get (essentially) A050267.

Examples

			47k^2 - 1701k + 10181 = 21647 for k = 42.
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, 3rd edition, Springer, 2004, ISBN 0-387-20860-7, Section A17, page 59.

Crossrefs

Programs

  • Mathematica
    Select[Table[47*n^2 - 1701*n + 10181, {n, 0, 100}], # > 0 && PrimeQ[#] &] (* T. D. Noe, Aug 02 2011 *)

Extensions

Edited by Klaus Brockhaus, Apr 22 2007 and by N. J. A. Sloane, May 05 2007 and May 06 2007

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}.
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