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 1-10 of 24 results. Next

A362922 Decimal expansion of cos(2*Pi/7) = sin(3*Pi/14) = A255249/2.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Jun 25 2023

Keywords

Comments

This number, negated, is a zero of the polynomial 8*x^3 - 4*x^2 - 4*x + 1 that arises in the dissection of a regular heptagon. The other two zeros are cos(Pi/7) (A073052) and sin(Pi/14) (A232736).
The old definition was: Decimal expansion of 1/(8*cos(Pi/7)*sin(Pi/14)).

Examples

			0.6234898018587335305250048840042398106322747308964021053655...
		

Crossrefs

Programs

  • Maple
    Digits := 110: evalf(((-1)^(2/7) - (-1)^(5/7))/2, Digits)*10^96:
    ListTools:-Reverse(convert(floor(%), base, 10)); # Peter Luschny, Jun 25 2023
  • Mathematica
    First@ RealDigits[1/(8*Cos[Pi/7]*Sin[Pi/14]), 10, 96] (* Michael De Vlieger, Jun 25 2023 *)

Formula

Equals 1/(4*cos(Pi/7)-2) = 1/(2*A160389-2). - Alois P. Heinz, Jun 25 2023
Equals (i^(4/7) - i^(10/7))/2. - Peter Luschny, Jun 26 2023

Extensions

Simpler definition from Alois P. Heinz, Jun 25 2023.

A160389 Decimal expansion of 2*cos(Pi/7).

Original entry on oeis.org

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

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Author

Harry J. Smith, May 31 2009

Keywords

Comments

Arises in the approximation of 14-fold quasipatterns by 14 Fourier modes.
Let DTS(n^c) denote the set of languages accepted by a deterministic Turing machine with space n^(o(1)) and time n^(c+o(1)), and let SAT denote the Boolean satisfiability problem. Then (1) SAT is not in DTS(n^c) for any c < 2*cos(Pi/7), and (2) the Williams inference rules cannot prove that SAT is not in DTS(n^c) for any c >= 2*cos(Pi/7). These results also apply to the Boolean satisfiability problem mod m where m is in A085971 except possibly for one prime. - Charles R Greathouse IV, Jul 19 2012
rho(7):= 2*cos(Pi/7) is the length ratio (smallest diagonal)/side in the regular 7-gon (heptagon). The algebraic number field Q(rho(7)) of degree 3 is fundamental for the 7-gon. See A187360 for the minimal polynomial C(7, x) of rho(7). The other (larger) diagonal/side ratio in the heptagon is sigma(7) = -1 + rho(7)^2, approx. 2.2469796. (see the decimal expansion in A231187). sigma(7) is the limit of a(n+1)/a(n) for n->infinity for the sequences like A006054 and A077998 which can be considered as analogs of the Fibonacci sequence in the pentagon. Thus sigma(7) plays in the heptagon the role of the golden section in the pentagon. See the P. Steinbach reference. - Wolfdieter Lang, Nov 21 2013
An algebraic integer of degree 3 with minimal polynomial x^3 - x^2 - 2x + 1. - Charles R Greathouse IV, Nov 12 2014
The other two solutions of the minimal polynomial of rho(7) = 2*cos(Pi/7) are 2*cos(3*Pi/7) and 2*cos(5*Pi/7). See eq. (20) of the W. Lang link. - Wolfdieter Lang, Feb 11 2015
The constant is the square root of 3.24697... (cf. A116425). It is the fifth-longest diagonal in the regular 14-gon with unit radius, which equals 2*sin(5*Pi/14). - Gary W. Adamson, Feb 14 2022

Examples

			1.801937735804838252472204639014890102331838324263714300107124846398864...
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 207.

Crossrefs

Cf. A039921 (continued fraction).
Cf. A003558 (the constant is cyclic with period 3, for N = 7).

Programs

  • Magma
    R:= RealField(200); Reverse(Intseq(Floor(10^110*2*Cos(Pi(R)/7)))); // Marius A. Burtea, Nov 13 2019
  • Maple
    evalf(2*cos(Pi/7), 100); # Wesley Ivan Hurt, Feb 01 2017
  • Mathematica
    RealDigits[2 Cos[Pi/7], 10, 111][[1]] (* Robert G. Wilson v, Jun 11 2013 *)
  • PARI
    default(realprecision, 20080); x=2*cos(Pi/7); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b160389.txt", n, " ", d));
    

Formula

Equals 2*A073052. - Michel Marcus, Nov 21 2013
Equals (Re((-(4*7)*(1 + 3*sqrt(3)*i))^(1/3)) + 1)/3, with the real part Re, and i = sqrt(-1). - Wolfdieter Lang, Feb 24 2015
Equals i^(2/7) - i^(12/7). - Peter Luschny, Apr 04 2020
From Peter Bala, Oct 20 2021: (Start)
Equals 2 - (1 - z)*(1 - z^6)/((1 - z^3)*(1 - z^4)), where z = exp(2*Pi*i/7).
The other two zeros of the minimal polynomial x^3 - x^2 - 2*x + 1 of 2*cos(Pi/7) are given by 2 - (1 - z^3)*(1 - z^4)/((1 - z^2)*(1 - z^5)) = 2*cos(3*Pi/7) = A255241 and 2 - (1 - z^2)*(1 - z^5)/((1 - z)*(1 - z^6)) = cos(5*Pi/7) = -A362922.
Equals Product_{n >= 0} (7*n+2)*(7*n+5)/((7*n+1)*(7*n+6)) = 1 + Product_{n >= 0} (7*n+2)*(7*n+5)/((7*n+3)*(7*n+4)) = 1/A255240.
The linear fractional mapping r -> 1/(1 - r) cyclically permutes the three zeros of the minimal polynomial x^3 - x^2 - 2*x + 1. The inverse mapping is r -> (r - 1)/r.
The quadratic mapping r -> 2 - r^2 also cyclically permutes the three zeros. The inverse mapping is r -> r^2 - r - 1. (End)
Equals i^(2/7) + i^(-2/7). - Gary W. Adamson, Feb 11 2022
From Amiram Eldar, Nov 22 2024: (Start)
Equals Product_{k>=1} (1 - (-1)^k/A047336(k)).
Equals 1 + cosec(3*Pi/14)/2 = 1 + Product_{k>=1} (1 + (-1)^k/A047341(k)). (End)
Equals sqrt(A116425). - Hugo Pfoertner, Nov 22 2024

A255241 Decimal expansion of 2*cos(3*Pi/7).

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, Mar 13 2015

Keywords

Comments

This is also the decimal expansion of 2*sin(Pi/14).
rho_2 := 2*cos(3*Pi/7) and rho(7) := 2*cos(Pi/7) (see A160389) are the two positive zeros of the minimal polynomial C(7, x) = x^3 - x^2 - 2*x + 1 of the algebraic number rho(7), the length ratio of the smaller diagonal and the side in the regular 7-gon (heptagon). See A187360 and a link to the arXiv paper given there, eq. (20) for the zeros of C(n, x). The other zero is negative, rho_3 := 2*cos(5*Pi/n). See -A255249.
Also the edge length of a regular 14-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 A272487). - Stanislav Sykora, May 01 2016

Examples

			0.445041867912628808577805128993589518932711137529089910623974031...
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 207.

Crossrefs

Edge lengths of other nonconstructible n-gons: A272487 (n=7), A272488 (n=9), A272489 (n=11), A130880 (n=18), A272491 (n=19). - Stanislav Sykora, May 01 2016

Programs

  • Magma
    R:= RealField(120); 2*Cos(3*Pi(R)/7); // G. C. Greubel, Sep 04 2022
    
  • Mathematica
    RealDigits[N[2Cos[3Pi/7], 100]][[1]] (* Robert Price, May 01 2016 *)
  • PARI
    2*sin(Pi/14)
    
  • PARI
    polrootsreal(x^3 - x^2 - 2*x + 1)[2] \\ Charles R Greathouse IV, Oct 30 2023
    
  • SageMath
    numerical_approx(2*cos(3*pi/7), digits=120) # G. C. Greubel, Sep 04 2022

Formula

2*cos(3*Pi/7) = 2*sin(Pi/14) = 2*A232736 = 1/A231187 = 0.4450418679...
See A232736 for the decimal expansion of cos(3*Pi/7) = sin(Pi/14).
Equals i^(6/7) - i^(8/7). - Peter Luschny, Apr 04 2020
From Peter Bala, Oct 11 2021: (Start)
Equals 2 - (1 - z^3)*(1 - z^4)/((1 - z^2)*(1 - z^5)), where z = exp(2*Pi*i/7).
Equals 1 - A255240. (End)

Extensions

Offset corrected by Stanislav Sykora, May 01 2016

A047385 Numbers that are congruent to {2, 5} mod 7.

Original entry on oeis.org

2, 5, 9, 12, 16, 19, 23, 26, 30, 33, 37, 40, 44, 47, 51, 54, 58, 61, 65, 68, 72, 75, 79, 82, 86, 89, 93, 96, 100, 103, 107, 110, 114, 117, 121, 124, 128, 131, 135, 138, 142, 145, 149, 152, 156, 159, 163, 166, 170
Offset: 1

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Author

Keywords

Comments

Also, numbers n such that (n^2+3)/7 is a nonnegative integer. - Bruno Berselli, Jan 16 2016

Crossrefs

Programs

Formula

G.f.: x*(2 + 3*x + 2*x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Dec 05 2011
a(n) = (14*n - (-1)^n - 7)/4. - Bruno Berselli, Dec 05 2011
a(n) = 4*n - 2 - floor(n/2). - Wesley Ivan Hurt, Oct 16 2013
E.g.f.: 2 + ((14*x - 7)*exp(x) - exp(-x))/4. - David Lovler, Sep 01 2022
From Amiram Eldar, Sep 26 2022: (Start)
a(n) = A113804(n)/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(3*Pi/14)*Pi/7. (End)
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2*sin(3*Pi/14) (A255249).
Product_{n>=1} (1 + (-1)^n/a(n)) = 1/(2*cos(Pi/7)) (A255240). (End)

A116425 Decimal expansion of 2 + 2*cos(2*Pi/7).

Original entry on oeis.org

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

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Author

Eric W. Weisstein, Feb 15 2006

Keywords

Comments

A root of the equation x^3 - 5*x^2 + 6*x - 1 = 0. - Arkadiusz Wesolowski, Jan 13 2016
The other two roots of this minimal polynomial of the present algebraic number (rho(7))^2, with rho(7) = 2*cos(Pi/7) = A160389 are (2*cos(3*Pi/7))^2 = (A255241)^2 and (2*cos(5*Pi/7))^2 = (-A255249)^2. - Wolfdieter Lang, Mar 30 2020

Examples

			3.246979603717467061...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 5.25 Tutte-Beraha Constants, p. 417.

Crossrefs

2 + 2*cos(2*Pi/n): A104457 (n = 5), A332438 (n = 9), A296184 (n = 10), A019973 (n = 12).

Programs

Formula

Equals (2*cos(Pi/7))^2 = (A160389)^2.
Equals 2 + i^(4/7) - i^(10/7). - Peter Luschny, Apr 04 2020
Let c = 2 + 2*cos(2*Pi/7). The linear fractional transformation z -> c - c/z has order 7, that is, z = c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/z)))))). - Peter Bala, May 09 2024

A231187 Decimal expansion of the length ratio (largest diagonal)/side in the regular 7-gon (or heptagon).

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, Nov 21 2013

Keywords

Comments

The length ratio (largest diagonal)/side in the regular 7-gon (heptagon) is sigma(7) = S(2, rho(7)) = -1 + rho(7)^2, with rho(7) = 2*cos(Pi/7), which is approx. 1.8019377358 (see A160389 for its decimal expansion, and A049310 for the Chebyshev S-polynomials). sigma(7), approx. 2.2469796, is also the reciprocal of one of the solutions of the minimal polynomial C(7, x) = x^3 - x^2 - 2*x + 1 of rho(7) (see A187360), namely 1/(2*cos(3*Pi/7)).
sigma(7) is the limit of a(n+1)/a(n) for n->infinity for the sequences A006054 and A077998 which can be considered as analogs of the Fibonacci sequence in the pentagon. Thus sigma(7) plays in the heptagon the role of the golden section in the pentagon.
See the Steinbach link.

Examples

			2.24697960371746706105000976800847962126454946179280421073109887819...
		

Crossrefs

Programs

Formula

sigma(7) = -1 + (2*cos(Pi/7))^2 = 1/(2*cos(3*Pi/7)).
Equals A116425 -1.
From Geoffrey Caveney, Apr 23 2014: (Start)
sigma(7) = exp(asinh(cos(Pi/7))).
cos(Pi/7) + sqrt(1+cos(Pi/7)^2). (End)
From Peter Bala, Oct 12 2021: (Start)
Minimal polynomial x^3 - 2*x^2 - x + 1.
Equals 2*(cos(3*Pi/7) - cos(6*Pi/7)). The other zeros of the minimal polynomial are 2*(cos(Pi/7) - cos(2*Pi/7)) = A255240 and 2*(cos(5*Pi/7) - cos(10*Pi/7)) = 1 - A160389.
The quadratic mapping z -> z^2 - 2*z cyclically permutes the zeros of the minimal polynomial. The inverse cyclic permutation is given by the mapping z -> 2 + z - z^2.
Equals Product_{n >= 0} (7*n+3)*(7*n+4)/((7*n+1)*(7*n+6)) = 1 + Product_{n >= 0} (7*n+3)*(7*n+4)/((7*n+2)*(7*n+5)) = 1 + A255249 = 1/A255241. (End)
Equals 1/(2*sin(Pi/14)) = 1 + 2*sin(3*Pi/14). - Gary W. Adamson, Jun 25 2022
Equals (2*cos(Pi/7)) * (2*cos(2*Pi/7)) = (i^(2/7) + i^(-2/7)) * (i^(4/7) + i^(-4/7)) = 1 + i^(4/7) + i^(-4/7). - Gary W. Adamson, Jul 16 2022
Equals 2F1(1/7,2/7;1/2;1) [Zucker] - R. J. Mathar, Jun 24 2024

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

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

Original entry on oeis.org

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

Views

Author

Peter Bala, Nov 06 2021

Keywords

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 ...
		

Crossrefs

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....

A070231 Let u(k), v(k), w(k) satisfy the recursions u(1) = v(1) = w(1) = 1, u(k+1) = u(k) + v(k) + w(k), v(k+1) = u(k)*v(k) + v(k)*w(k) + w(k)*u(k), and w(k+1) = u(k)*v(k)*w(k) for k >= 1; then a(n) = u(n).

Original entry on oeis.org

1, 3, 7, 31, 1279, 9202687, 3692849258577919, 98367959484921734629696721986125823, 3882894052327310957045599009145809243674851356642054390303168725061781159935999
Offset: 1

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Author

Benoit Cloitre, May 08 2002

Keywords

Crossrefs

Programs

  • Mathematica
    a[1] = 1; v[1] = 1; w[1] = 1; a[k_] := a[k] = a[k - 1] + v[k - 1] + w[k - 1]; v[k_] := v[k] = a[k - 1]*v[k - 1] + v[k - 1]*w[k - 1] + w[k - 1]*a[k - 1]; w[k_] := w[k] = a[k - 1]*v[k - 1]*w[k - 1]; Table[a[n], {n, 1, 9}] (* Vaclav Kotesovec, May 11 2020 *)
  • PARI
    lista(nn) = {my(u = vector(nn)); my(v = vector(nn)); my(w = vector(nn)); u[1] = 1; v[1] = 1; w[1] = 1; for (n=2, nn, u[n] = u[n-1] + v[n-1] + w[n-1]; v[n] = u[n-1]*v[n-1] + v[n-1]*w[n-1] + w[n-1]*u[n-1]; w[n] = u[n-1]*v[n-1]*w[n-1];); u; } \\ Petros Hadjicostas, May 11 2020

Formula

Let C be the positive root of x^3 + x^2 - 2*x - 1 = 0; that is, C = 1.246979603717... = A255249. Then Lim_{n -> infinity} u(n)^(C+1)/w(n)= Lim_{n -> infinity} v(n)^C/w(n) = Lim_{n -> infinity} u(n)^B/v(n) = 1 with B = C + 1 - 1/(1 + C) = 1.8019377... = A160389. [corrected by Vaclav Kotesovec, May 11 2020]
a(n) ~ gu^((1 + C)^n), where C is defined above and gu = 1.131945853718244297... The relation between constants gu, gv (see A070234) and gw (see A070233) is gu^(1 + C) = gv^C = gw. - Vaclav Kotesovec, May 11 2020

A070233 Let u(k), v(k), w(k) satisfy the recursions u(1) = v(1) = w(1) = 1, u(k+1) = u(k) + v(k) + w(k), v(k+1) = u(k)*v(k) + v(k)*w(k) + w(k)*u(k), and w(k+1) = u(k)*v(k)*w(k) for k >= 1; then a(n) = w(n).

Original entry on oeis.org

1, 1, 9, 945, 8876385, 3689952451492545, 98367948795841301790914258556831105, 3882894052327309905582682317031276840071039865528864289025562807872336355445505
Offset: 1

Views

Author

Benoit Cloitre, May 08 2002

Keywords

Comments

Next term is too large to include.

Crossrefs

Programs

  • Mathematica
    u[1] = 1; v[1] = 1; a[1] = 1; u[k_] := u[k] = u[k - 1] + v[k - 1] + a[k - 1]; v[k_] := v[k] = u[k - 1]*v[k - 1] + v[k - 1]*a[k - 1] + a[k - 1]*u[k - 1]; a[k_] := a[k] = u[k - 1]*v[k - 1]*a[k - 1]; Table[a[n], {n, 1, 9}] (* Vaclav Kotesovec, May 11 2020 *)
  • PARI
    lista(nn) = {my(u = vector(nn)); my(v = vector(nn)); my(w = vector(nn)); u[1] = 1; v[1] = 1; w[1] = 1; for (n=2, nn, u[n] = u[n-1] + v[n-1] + w[n-1]; v[n] = u[n-1]*v[n-1] + v[n-1]*w[n-1] + w[n-1]*u[n-1]; w[n] = u[n-1]*v[n-1]*w[n-1]; ); w; } \\ Petros Hadjicostas, May 11 2020

Formula

Let C be the positive root of x^3 + x^2 - 2*x - 1 = 0: that is, C = 1.246979603717... = A255249. Then Lim_{n -> infinity} u(n)^(C+1)/w(n)= Lim_{n -> infinity} v(n)^C/w(n) = Lim_{n -> infinity} u(n)^B/v(n) = 1 with B = C + 1 - 1/(1 + C) = 1.8019377... = A160389. [corrected by Vaclav Kotesovec, May 11 2020]
a(n) ~ gw^((C + 1)^n), where C is defined above and gw = 1.321128752475732548... The relation between constants gu (see A070231), gv (see A070234) and gw is gu^(1 + C) = gv^C = gw. - Vaclav Kotesovec, May 11 2020
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