A001999 -id:A001999 - cp's OEIS frontend

A045529 a(n+1) = 5a(n)^3 - 3a(n), a(0) = 1.

1, 2, 34, 196418, 37889062373143906, 271964099255182923543922814194423915162591622175362

Offset: 0

Jose Eduardo Blazek, Dec 11 1999

The next term, a(6), has 153 digits. - Harvey P. Dale, Oct 24 2011

Seiichi Manyama, Table of n, a(n) for n = 0..7
Daniel Duverney and Takeshi Kurosawa, Transcendence of infinite products involving Fibonacci and Lucas numbers, Research in Number Theory, Vol. 8 (2002), Article 68.
Zalman Usiskin, Problem B-265, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 11, No. 3 (1973), p. 333; Fibonacci Numbers for Powers of 3, Solution to Problem B-265 by Ralph Garfield and David Zeitlin, ibid., Vol. 12, No. 3 (1974), p. 315.

Cf. (k^n)-th Fibonacci number: A058635 (k=2), this sequence (k=3), A145231 (k=4), A145232 (k=5), A145233 (k=6), A145234 (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).

Cf. A000032, A000045, A001622, A001999, A002814, A006267, A094214.

a := proc(n) option remember; if n = 0 then 1 else 5*a(n-1)^3 - 3*a(n-1) end if; end:
seq(a(n), n = 0..5); # Peter Bala, Nov 24 2022

G = (1 + Sqrt[5])/2; Table[Expand[(G^(3^n) - (1 - G)^(3^n))/Sqrt[5]], {n, 1, 7}] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[(4/5)^(1/2)*Cosh[3^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 4}] (* Artur Jasinski, Oct 05 2008 *)
RecurrenceTable[{a[0]==1,a[n]==5a[n-1]^3-3a[n-1]},a[n],{n,6}] (* Harvey P. Dale, Oct 24 2011 *)
NestList[5#^3-3#&,1,5] (* Harvey P. Dale, Dec 21 2014 *)

A045529(n):=fib(3^n)$
makelist(A045529(n),n,0,10); /* Martin Ettl, Nov 12 2012 */

The first example I know in which a(n) can be expressed as (4/5)^(1/2)*cosh(3^n*arccosh((5/4)^(1/2))).
a(n) = Fibonacci(3^n). - Leroy Quet, Mar 17 2002
a(n+1) = a(n)*A002814(n+1). - Lekraj Beedassy, Jun 16 2003
a(n) = (phi^(3^n) - (1 - phi)^(3^n))/sqrt(5), where phi is the golden ratio (A001622). - Artur Jasinski, Oct 05 2008
a(n) = Product_{k=0..n-1} (Lucas(2*3^k) - 1) (Usiskin, 1973). - Amiram Eldar, Jan 29 2022
From Peter Bala, Nov 24 2022: (Start)
a(2*n+2) == a(2*n) (mod 3^(2*n+1)); a(2*n+3) == a(2*n+1) (mod 3^(2*n+2));
a(2*n+1) + a(2*n) == 0 (mod 3^(2*n+1)).
a(2*n) == 1 (mod 3) and a(2*n+1) == 2 (mod 3).
5*a(n)^2 == 2 (mod 3^(n+1)).
In the ring of 3-adic integers, the sequences {a(2*n)} and {a(2*n+1)} are both Cauchy sequences and converge to the pair of 3-adic roots of the quadratic equation 5*x^2 - 2 = 0. (End)
From Amiram Eldar, Jan 07 2023: (Start)
Product_{n>=1} (1 + 2/(sqrt(5)*a(n)-1)) = phi (A001622).
Product_{n>=1} (1 - 2/(sqrt(5)*a(n)+1)) = 1/phi (A094214).
Both formulas are from Duverney and Kurosawa (2022). (End)

A006267 Continued cotangent for the golden ratio.

Original entry on oeis.org

1, 4, 76, 439204, 84722519070079276, 608130213374088941214747405817720942127490792974404

Offset: 0

N. J. A. Sloane

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Harry J. Smith, Table of n, a(n) for n = 0..7
Mohammad K. Azarian, Problem 123, Missouri Journal of Mathematical Sciences, Vol. 10, No. 3 (Fall 1998), p. 176; Solution, ibid., Vol. 12, No. 1 (Winter 2000), pp. 61-62.
Jeffrey Shallit, Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B, Vol. 80B, No. 2 (1976), pp. 285-290.
Zalman Usiskin, Problem B-266, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 11, No. 3 (1973), p. 334; Lucas Numbers for Powers of 3, Solution to Problem B-266 by David Zeitlin, ibid., Vol. 12, No. 3 (1974), p. 315-316.
Eric Weisstein's World of Mathematics, Lehmer Cotangent Expansion.

Cf. A000032, A000204, A001622, A001999, A002666, A002667, A002668, A006266, A002813, A045529, A271223, A268924.

a := proc(n) option remember; if n = 1 then 4 else a(n-1)^3 + 3*a(n-1) end if; end: seq(a(n), n = 1..5); # Peter Bala, Nov 15 2022

c = N[GoldenRatio, 1000]; Table[Round[c^(3^n)], {n, 1, 8}] (* Artur Jasinski, Sep 22 2008 *)
a = {}; x = 4; Do[AppendTo[a, x]; x = x^3 + 3 x, {n, 1, 10}]; a (* Artur Jasinski, Sep 24 2008 *)

a(n)=fibonacci(3^n+1) + fibonacci(3^n-1) \\ Andrew Howroyd, Dec 30 2024

a(n)={my(t=1); for(i=1, n, t = t^3 + 3*t); t} \\ Andrew Howroyd, Dec 30 2024

(1+sqrt(5))/2 = cot(Sum_{n>=0} (-1)^n*acot(a(n))); let b(0) = (1+sqrt(5))/2, b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1))) then a(n) = floor(b(n)). - Benoit Cloitre, Apr 10 2003
a(n) = A000204(3^n). - Benoit Cloitre, Sep 18 2005
a(n) = round(c^(3^n)) where c = GoldenRatio = 1.6180339887498948482... = (sqrt(5)+1)/2 (A001622). - Artur Jasinski, Sep 22 2008
a(n) = a(n-1)^3 + 3*a(n-1), a(0) = 1. - Artur Jasinski, Sep 24 2008
a(n+1) = Product_{k = 0..n} A002813(k). Thus a(n) divides a(n+1). - Peter Bala, Nov 22 2012
Sum_{n>=0} a(n)^2/A045529(n+1) = 1. - Amiram Eldar, Jan 12 2022
a(n) = Product_{k=0..n-1} (Lucas(2*3^k) + 1) (Usiskin, 1973). - Amiram Eldar, Jan 29 2022
From Peter Bala, Nov 15 2022: (Start)
a(n) = Lucas(3^n) for n >= 1.
a(n) == 1 (mod 3) for n >= 1.
a(n+1) == a(n) (mod 3^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).
The smallest positive residue of a(n) mod 3^n = A268924(n).
In the ring of 3-adic integers the limit_{n -> oo} a(n) exists and is equal to A271223. Cf. A006266. (End)

The next term is too large to include.

A002814 For n > 1: a(n) = a(n-1)^3 + 3a(n-1)^2 - 3; a(0) = 1, a(1) = 2.

Original entry on oeis.org

1, 2, 17, 5777, 192900153617, 7177905237579946589743592924684177

Offset: 0

N. J. A. Sloane

An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
Next terms have 102 and 305 digits. - Harvey P. Dale, Jun 06 2011
From Peter Bala, Nov 15 2012: (Start)
The present sequence is the case x = 3 of the following general remarks. The recurrence equation a(n+1) = a(n)^3 + 3*a(n)^2 - 3 with the initial condition a(1) = x - 1 > 1 has the explicit solution a(n+1) = alpha^(3^n) + (1/alpha)^(3^n) - 1 for n >= 0, where alpha := {x + sqrt(x^2 - 4)}/2.
Two other recurrences satisfied by the sequence are a(n+1) = (a(1) + 3)*(Product_{k = 1..n} a(k)^2) - 3 and a(n+1) = 1 + (a(1) - 1)*Product_{k = 1..n} (a(k) + 2)^2, both with a(1) = x - 1.
The associated sequence b(n) := a(n) + 1 satisfies the recurrence equation b(n+1) = b(n)^3 - 3*b(n) with the initial condition b(1) = x. See A001999 for the case x = 3. The sequence c(n) := a(n) + 2 satisfies the recurrence equation c(n+1) = c(n)^3 - 3*c(n)^2 + 3 with the initial condition c(1) = x + 1.
The sequences a(n) and b(n) have been considered by Fine and Escott in connection with a product expansion for quadratic irrationals. We have the following identity, valid for x > 2: sqrt((x + 2)/(x - 2)) = (1 + 2/(x-1))*sqrt((y + 2)/(y - 2)), where y = x^3 - 3*x or, equivalently, y - 1 = (x - 1)^3 + 3*(x - 1)^2 - 3. Iterating the identity produces the product expansion sqrt((x+2)/(x-2)) = Product_{n >= 1} (1 + 2/a(n)), with a(1) = x - 1 and a(n+1) = a(n)^3 + 3*a(n)^2 - 3.
For similar results to the above see A145502. See also A219162. (End)
Conjecture: The sequence {a(n) - 2: n >= 1} is a strong divisibility sequence, that is, gcd(a(n) - 2, a(m) - 2) = a(gcd(n, m)) - 2 for n, m >= 1. - Peter Bala, Dec 08 2022

L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 397.
E. Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 0..8
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8, Oct. 1977.
E. Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy)
M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.
J. Shallit, Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.

Cf. A000045, A001566.
Cf. A045529. A001999, A145502, A219162.
Cf. A000244.

a002814 n = a002814_list !! n
a002814_list = 1 : zipWith div (tail xs) xs
   where xs = map a000045 a000244_list
-- Reinhard Zumkeller, Nov 24 2012

Join[{1}, NestList[#^3+3#^2-3&,2,5]] (* Harvey P. Dale, Apr 01 2011 *)

a[0]:1$ a[1]:2$ a[n]:=a[n-1]^3 + 3*a[n-1]^2-3$ A002814(n):=a[n]$
makelist(A002814(n),n,0,6); /* Martin Ettl, Nov 12 2012 */

a(n)=if(n<2,max(0,n+1),a(n-1)^3+3*a(n-1)^2-3)

a(n) = Fibonacci(3^n)/Fibonacci(3^(n-1)). - Henry Bottomley, Jul 10 2001
a(n+1) = 5*(f(n))^2 - 3, where f(n) = Fibonacci(3^n) = product of first n entries. - Lekraj Beedassy, Jun 16 2003
From Artur Jasinski, Oct 05 2008: (Start)
a(n+2) = (G^(3^(n + 1)) - (1 - G)^(3^(n + 1)))/((G^(3^n)) - (1 - G)^(3^n)) where G = (1 + sqrt(5))/2;
a(n+2) = A045529(n+1)/A045529(n). (End)
From Peter Bala, Nov 15 2012: (Start)
a(n+1) = (1/2*(3 + sqrt(5)))^(3^n) + (1/2*(3 - sqrt(5)))^(3^n) - 1.
The sequence b(n):= a(n) + 2 is a solution to the recurrence b(n+1) = b(n)^3 - 3*b(n)^2 + 3 with b(1) = 4.
Other recurrence equations:
a(n+1) = -3 + 5*(Product_{k = 1..n} a(k)^2) with a(1) = 2.
a(n+1) = 1 + Product_{k = 1..n} (a(k) + 2)^2 with a(1) = 2.
Thus Y := Product_{k = 1..n} a(k) and X := Product_{k = 1..n} (a(k) + 2) give a solution to the Diophantine equation X^2 - 5*Y^2 = -4.
sqrt(5) = Product_{n >= 1} (1 + 2/a(n)). The rate of convergence is cubic. Fine remarks that twelve factors of the product would give well over 300,000 correct decimal digits for sqrt(5).
5 - {Product_{n = 1..N} (1 + 2/a(n))}^2 = 20/(a(N+1) + 3). (End)
a(n) = 2*T(3^(n-1),3/2) - 1 for n >= 1, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Dec 06 2022

Definition improved by Reinhard Zumkeller, Feb 29 2012

A006276 Pierce expansion of (3 - sqrt(5))/2.

Original entry on oeis.org

2, 4, 17, 19, 5777, 5779, 192900153617, 192900153619, 7177905237579946589743592924684177, 7177905237579946589743592924684179, 369822356418414944143680173221426891716916679027557977938929258031490127514207143830378340325399155217

Offset: 0

N. J. A. Sloane

From Peter Bala, Nov 22 2012: (Start)
For x in the open interval (0,1) define the map f(x) = 1 - x*floor(1/x). The n-th term (n >= 0) in the Pierce expansion of x is given by floor(1/f^(n)(x)), where f^(n)(x) denotes the n-th iterate of the map f, with the convention that f^(0)(x) = x.
Let x = (sqrt(5) - 1)/2, the reciprocal of the golden ratio, and let X = (3 - sqrt(5))/2 so that X = x^2. The Pierce expansion of X^(3^n) is [a(2*n), a(2*n+1), a(2*n+2), ...]. The Pierce expansion of x is A118242 = [1, a(0), a(1), a(2), ...]. The Pierce expansion of x^3 is [a(1), a(2), a(3), ...]. In general, the Pierce expansion of x^(3^n) for n >= 1 is [a(1)*a(3)*...*a(2*n-1), a(2*n), a(2*n+1), a(2*n+2), ...] = [sqrt(a(2*n) - 1), a(2*n), a(2*n+1), a(2*n+2), ...]. Some examples of the associated alternating series are given below.
(End)

			From _Peter Bala_, Nov 22 2012: (Start)
Let x = (sqrt(5) - 1)/2. We have the alternating series expansions
x = 1 - 1/2 + 1/(2*4) - 1/(2*4*17) + 1/(2*4*17*19) - ...
x^2 = 1/2 - 1/(2*4) + 1/(2*4*17) - 1/(2*4*17*19) + ...
x^6 = 1/17 - 1/(17*19) + 1/(17*19*5777) - ...
as well as
x^3 = 1/4 - 1/(4*17) + 1/(4*17*19) - 1/(4*17*19*5777) + ...
4*x^9 = 1/19 - 1/(19*5777) + 1/(19*5777*5779) - ...
4*19*x^27 = 1/5779 - 1/(5779*192900153617) + ....
(End)

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..15
T. A. Pierce, On an algorithm and its use in approximating roots of algebraic equations, Amer. Math. Monthly, Vol. 36 No. 10, (1929) p.523-525.
Jeffrey Shallit, Some predictable Pierce expansions, Fib. Quart., 22 (1984), 332-335.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A118242.

Table[c=2*3^Floor[n/2]; 2*Fibonacci[c+1]-Fibonacci[c]-(-1)^n, {n,0,10}] (* Harvey P. Dale, Oct 22 2013 *)
PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[(3 - Sqrt[5])/2, 7!], 10] (* G. C. Greubel, Nov 14 2016 *)

r=(1+sqrt(5))/2; for(n=1,10, r=r/(r-floor(r)) print1(floor(r),","))

Let c(0)=3, c(n+1) = c(n)^3-3*c(n) [A001999]; then this sequence is c(0)-1, c(0)+1, c(1)-1, c(1)+1, c(2)-1, c(2)+1, ......
a(n) = 2*F(2*3^floor(n/2)+1)-F(2*3^floor(n/2))-(-1)^n where F(k) denotes the k-th Fibonacci number A000045(k)
Let u(0)=(1+sqrt(5))/2 and u(n+1)=u(n)/frac(u(n)) where frac(x) is the fractional part of x, then a(n)=floor(u(n)). - Benoit Cloitre, Mar 09 2004
a(2*n) = ((3 + sqrt(5))/2)^(3^n) + ((3 - sqrt(5))/2)^(3^n) - 1.
a(2*n+1) = ((3 + sqrt(5))/2)^(3^n) + ((3 - sqrt(5))/2)^(3^n) + 1. - Peter Bala, Nov 22 2012

More terms from James Sellers, May 19 2000

A112845 Recurrence a(n) = a(n-1)^3 - 3*a(n-1) with a(0) = 6.

Original entry on oeis.org

6, 198, 7761798, 467613464999866416198, 102249460387306384473056172738577521087843948916391508591105798

Offset: 0

Eric W. Weisstein, Sep 21 2005

Identical to A006243 apart from the initial term. For some general remarks on this recurrence see A001999. - Peter Bala, Nov 13 2012

G. C. Greubel, Table of n, a(n) for n = 0..6
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8, Oct. 1977, 629-630.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A006275, A006276.
Cf. A006243. - R. J. Mathar, Aug 15 2008
Cf. A001999, A219160, A219161.

RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 6}, a, {n,
  0, 5}] (* G. C. Greubel, Dec 30 2016 *)
NestList[#^3-3#&,6,5] (* Harvey P. Dale, Jul 23 2025 *)

a(n) = -2*cos(3^n*arccos(-3)).
From Peter Bala, Nov 13 2012: (Start)
a(n) = (3 + 2*sqrt(2))^(3^n) + (3 - 2*sqrt(2))^(3^n).
Product {n = 0..inf} (1 + 2/(a(n) - 1)) = sqrt(2).
(End)

A219160 Recurrence equation a(n+1) = a(n)^3 - 3*a(n) with a(0) = 4.

Original entry on oeis.org

4, 52, 140452, 2770663499604052, 21269209556953516583554114034636483645584976452

Offset: 0

Peter Bala, Nov 13 2012

For some general remarks on this recurrence see A001999.

G. C. Greubel, Table of n, a(n) for n = 0..6
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8, Oct. 1977.

Cf. A001999, A112845, A219161.

RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 4}, a, {n,
  0, 5}] (* G. C. Greubel, Dec 30 2016 *)

a(n) = (2 + sqrt(3))^(3^n) + (2 - sqrt(3))^(3^n).

Product {n = 0..inf} (1 + 2/(a(n) - 1)) = sqrt(3). The rate of convergence is cubic. Fine remarks that taking the first twelve factors of the product would give well over 300,000 correct decimals for sqrt(3).

A219161 Recurrence equation a(n+1) = a(n)^3 - 3*a(n) with a(0) = 5.

Original entry on oeis.org

5, 110, 1330670, 2356194280407770990, 13080769480548649962914459850235688797656360638877986030

Offset: 0

Peter Bala, Nov 13 2012

For some general remarks on this recurrence see A001999.

The next term (a(5)) has 166 digits. - Harvey P. Dale, Apr 23 2019

G. C. Greubel, Table of n, a(n) for n = 0..6
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8, Oct. 1977, 629-630.

Cf. A001999, A112845, A219160.

RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 5}, a, {n,
  0, 5}] (* G. C. Greubel, Dec 30 2016 *)
NestList[#^3-3#&,5,5] (* Harvey P. Dale, Apr 23 2019 *)

a(n) = (1/2*(5 + sqrt(21)))^(3^n) + (1/2*(5 - sqrt(21)))^(3^n).
Product_{n = 0..inf} (1 + 2/(a(n) - 1)) = sqrt(7/3).
a(n) = 2*T(3^n,5/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. Cf. A001999. - Peter Bala, Feb 01 2017

A002813 a(0) = 4; for n > 0, a(n) = a(n-1)^3 - 3*a(n-1)^2 + 3.

Original entry on oeis.org

4, 19, 5779, 192900153619, 7177905237579946589743592924684179, 369822356418414944143680173221426891716916679027557977938929258031490127514207143830378340325399155219

Offset: 0

N. J. A. Sloane

An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
The next term, a(7), has 305 digits. - Harvey P. Dale, Jul 19 2011
From Peter Bala, Nov 22 2012: (Start)
The present sequence is the case x = 1 of the following general remarks about the recurrence a(n+1) = a(n)^3 - 3*a(n-1)^2 + 3. Cf. A002814.
Define a sequence of polynomials P(n,x) inductively by setting P(0,x) = x^2 + 3 and P(n+1,x) = P(n,x^3 + 3*x) for n >= 0. Then P(n,x) satisfies the cubic recurrence P(n+1,x) = P(n,x)^3 - 3*P(n-1,x)^2 + 3 with the initial condition P(0,x) = x^2 + 3.
An explicit formula is P(n,x) = Q(3^(n+1),x)/Q(3^n,x), where Q(n,x) = ((x + sqrt(x^2 + 4))/2)^n + ((x - sqrt(x^2 + 4))/2)^n.
Alternatively, P(n,x) = ((x^2 + 2 + sqrt(x^4 + 4*x^2))/2)^(3^n) + ((x^2 + 2 - sqrt(x^4 + 4*x^2))/2)^(3^n) + 1.
Iterating the algebraic identity x/sqrt(x^2 + 4) = (1 - 2/(x^2 + 3))*y/sqrt(y^2 + 4), where y = x^3 + 3*x, leads to the product expansion x/sqrt(x^2 + 4) = Product_{n = 0..oo} (1 - 2/P(n,x)). See Escott and also Fine.
The sequence A(n,x) := x*Product_{k = 0..n} P(k,x) satisfies the recurrence A(n+1,x) = A(n,x)^3 + 3*A(n,x). These sequences occur in the continued cotangent expansions of Lehmer. Cases currently in the database are A006267 (x = 1), A006266 (x = 2), A006268 (x = 3), A006269 (x = 5) and A145180 through A145189 (x = 6 through x = 15).
(End)

L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 397.
E. Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..8
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8, Oct. 1977.
E. Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy)
J. Shallit Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.
Eric Weisstein's World of Mathematics, Lehmer Cotangent Expansion

Cf. A002814, A006267.

[Lucas(2*3^n)+1: n in [0..5]]; // Vincenzo Librandi, Jul 20 2011

NestList[#^3-3#^2+3&,4,6] (* Harvey P. Dale, Jul 19 2011 *)

a(n)=if(n<1,4*(n==0),a(n-1)^3-3*a(n-1)^2+3)

a(n)=if(n<0,0,n=2*3^n;fibonacci(n+1)+fibonacci(n-1)+1)

a(n) = L(2*3^n)+1 where L=Lucas numbers.
a(n) = L(3^(n+1))/L(3^n). - Benoit Cloitre, Sep 18 2005
a(n) = A001999(n)+1. - R. J. Mathar, Apr 22 2007
From Peter Bala, Nov 22 2012: (Start)
a(n) = ((3 + sqrt(5))/2)^(3^n) + ((3 - sqrt(5))/2)^(3^n) + 1.
(1/5)*sqrt(5) = Product_{n = 0..oo} (1 - 2/a(n)).
A006267(n+1) = Product_{k = 0..n} a(k).
A002814(n+1) = a(n) - 2. (End)

A219162 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 3.

Original entry on oeis.org

3, 47, 4870847, 562882766124611619513723647

Offset: 0

Peter Bala, Nov 13 2012

Bisection of A001566. Compare the following remarks with A001999.
The present sequence is the case x = 3 of the following general remarks. For other cases see A219163 (x = 4), A219164 (x = 5) and A219165 (x = 6).
Let x > 2 and let alpha := {x + sqrt(x^2 - 4)}/2. Define a sequence a(n) (which depends on x) by setting a(n) = alpha^(4^n) + (1/alpha)^(4^n). Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^4 + 4*a(n)^2 - 2 with the initial condition a(0) = x.
We have the product expansion sqrt((x + 2)/(x - 2)) = Product_{n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)).

Cf. A000032, A001566, A001999, A002814, A145502, A219163, A219164, A219165.

a(n)={if(n==0,3,a(n-1)^4-4*a(n-1)^2+2)} \\ Edward Jiang, Sep 11 2014

Let alpha = 1/2*(3 + sqrt(5)) then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
a(n) = A001566(2*n) = A000032(2*4^n).
Product {n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(5).
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2*T(4^n,3/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
Let b(n) = a(n) - 3. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. (End)

A219163 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 4.

Original entry on oeis.org

4, 194, 1416317954, 4023861667741036022825635656102100994

Offset: 0

Peter Bala, Nov 13 2012

Bisection of A003010.

a(4) has 147 digits and a(5) has 586 digits. - Harvey P. Dale, Mar 03 2020

Cf. A001999, A003010, A003500, A219162, A219164, A219165.

NestList[#^4-4#^2+2&,4,5] (* Harvey P. Dale, Mar 03 2020 *)

a(n)={if(n==0,4,a(n-1)^4-4*a(n-1)^2+2)} \\ Edward Jiang, Sep 11 2014

Let alpha = 2 + sqrt(3). Then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
a(n) = A003010(2*n) = A003500(4^n).
Product_{n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(3).
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2*T(4^n,2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
Let b(n) = a(n) - 4. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. (End)

cp's OEIS Frontend

A045529 a(n+1) = 5*a(n)^3 - 3*a(n), a(0) = 1.

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Maple

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A006267 Continued cotangent for the golden ratio.

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A002814 For n > 1: a(n) = a(n-1)^3 + 3a(n-1)^2 - 3; a(0) = 1, a(1) = 2.

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A006276 Pierce expansion of (3 - sqrt(5))/2.

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A112845 Recurrence a(n) = a(n-1)^3 - 3*a(n-1) with a(0) = 6.

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A219160 Recurrence equation a(n+1) = a(n)^3 - 3*a(n) with a(0) = 4.

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A219161 Recurrence equation a(n+1) = a(n)^3 - 3*a(n) with a(0) = 5.

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A045529 a(n+1) = 5a(n)^3 - 3a(n), a(0) = 1.