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.

Previous Showing 11-20 of 36 results. Next

A047341 Numbers that are congruent to {3, 4} mod 7.

Original entry on oeis.org

3, 4, 10, 11, 17, 18, 24, 25, 31, 32, 38, 39, 45, 46, 52, 53, 59, 60, 66, 67, 73, 74, 80, 81, 87, 88, 94, 95, 101, 102, 108, 109, 115, 116, 122, 123, 129, 130, 136, 137, 143, 144, 150, 151, 157, 158, 164, 165, 171
Offset: 1

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Author

Keywords

Comments

Numbers m such that m^2 == 2 (mod 7). - Vincenzo Librandi, Aug 05 2010
Numbers k such that A056107(k)/7 is an integer. - Bruno Berselli, Feb 14 2017

Crossrefs

Programs

Formula

a(n)^2 = 7*A056834(a(n)) + 2. - Bruno Berselli, Nov 28 2010
G.f.: x*(3 + x + 3*x^2)/((1 + x)*(1 - x)^2). - R. J. Mathar, Oct 08 2011
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi*tan(Pi/14)/7. - Amiram Eldar, Dec 12 2021
E.g.f.: 3 + ((14*x - 7)*exp(x) - 5*exp(-x))/4. - David Lovler, Sep 01 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 1.
Product_{n>=1} (1 + (-1)^n/a(n)) = 2*cos(Pi/7) - 1 (A160389 - 1). (End)

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

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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

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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 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

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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

Views

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

A070234 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); then a(n) = v(n).

Original entry on oeis.org

1, 3, 15, 303, 325023, 2896797882687, 10689080432835089614170716799, 1051462916692114532403603811392745230616355871287492722818364671
Offset: 1

Views

Author

Benoit Cloitre, May 08 2002

Keywords

Crossrefs

Programs

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) ~ gv^((C + 1)^n), where C is defined above and gv = 1.250231610564761084... The relation between constants gu (see A070231), gv and gw (see A070233) is gu^(1 + C) = gv^C = gw. - Vaclav Kotesovec, May 11 2020

A220605 Decimal expansion of Gamma(2/7).

Original entry on oeis.org

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

Views

Author

Bruno Berselli, Dec 17 2012

Keywords

Comments

See the second comment of A220086.

Examples

			3.1491151177599365909701136646807688922297786117662526...
		

Crossrefs

Cf. A160389 (see the first comment of A220086).

Programs

  • Mathematica
    RealDigits[Gamma[2/7], 10, 90][[1]]
  • Maxima
    fpprec:90; ev(bfloat(gamma(2/7)));
    
  • PARI
    gamma(2/7) \\ Charles R Greathouse IV, Feb 11 2025

Formula

Equals Pi*sec(3*Pi/14)/A220606.

A220607 Decimal expansion of Gamma(6/7).

Original entry on oeis.org

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

Views

Author

Bruno Berselli, Dec 17 2012

Keywords

Comments

See the second comment of A220086.

Crossrefs

Cf. A160389 (see the first comment of A220086).

Programs

  • Mathematica
    RealDigits[Gamma[6/7], 10, 90][[1]]
  • Maxima
    fpprec:90; ev(bfloat(gamma(6/7)));

Formula

Equals Pi*csc(Pi/7)/A220086, where csc is the cosecant function.

A039921 Continued fraction expansion of w = 2*cos(Pi/7).

Original entry on oeis.org

1, 1, 4, 20, 2, 3, 1, 6, 10, 5, 2, 2, 1, 2, 2, 1, 18, 1, 1, 3, 2, 1, 2, 1, 2, 1, 39, 2, 1, 1, 1, 13, 1, 2, 1, 30, 1, 1, 1, 3, 2, 5, 4, 1, 5, 1, 5, 1, 2, 1, 1, 94, 6, 2, 19, 11, 1, 60, 1, 1, 50, 2, 1, 1, 8, 53, 1, 3, 1, 6, 3, 2, 1, 5, 1, 1, 3, 4, 636, 1, 2, 1, 3, 3, 7, 9, 1, 2, 10, 3, 1, 22, 1, 119, 3
Offset: 0

Views

Author

Keywords

Comments

Arises in the approximation of 14-fold quasipatterns by 14 Fourier modes.

Examples

			w = 1.80193773580483825247220463901489010233183832426371430010712484639886...
Equals 1 + 1/(1 + 1/(4 + 1/(20 + 1/(2 + ...)))). - _Harry J. Smith_, May 31 2009
		

References

  • A. M. Rucklidge & W. J. Rucklidge (preprint) 2002.

Crossrefs

Cf. A160389 (Decimal expansion). - Harry J. Smith, May 31 2009

Programs

  • Mathematica
    ContinuedFraction[2*Cos[Pi/7], 100]
  • PARI
    { allocatemem(932245000); default(realprecision, 21000); x=contfrac(2*cos(Pi/7)); for (n=0, 20000, write("b039921.txt", n, " ", x[n+1])); } \\ Harry J. Smith, May 31 2009

Formula

w satisfies w^3 - w^2 - 2w + 1 = 0 and so is algebraic.
The other two roots are 2*cos(3 Pi/7) and 2*cos(5 Pi/7); their continued fraction expansions also end in 20, 2, 3, 1, 6, 10, 5, 2, 2, 1, ... which is a(n) for n >= 3. - Greg Dresden, Jul 01 2018
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