A001566 -id:A001566 - cp's OEIS frontend

A145502 a(n+1) = a(n)^2+2*a(n)-2 and a(1)=2.

2, 6, 46, 2206, 4870846, 23725150497406, 562882766124611619513723646, 316837008400094222150776738483768236006420971486980606

Offset: 1

Artur Jasinski, Oct 11 2008

General formula for a(n+1) = a(n)^2+2*a(n)-2 and a(1) = k+1 is a(n) = floor(((k + sqrt(k^2 + 4))/2)^(2^((n+1) - 1))).
From Peter Bala, Nov 12 2012: (Start)
The present sequence corresponds to the case x = 3 of the following general remarks. Sequences A145503 through A145510 correspond to the cases x = 4 through x = 11 respectively.
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^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1. Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^2 + 2*a(n) - 2 with the initial condition a(1) = x - 1.
A second recurrence is a(n) = (a(1) + 2)*{Product_{k = 1..n-1} a(k)} - 2.
The following algebraic identity is valid for x > 2:
(x + 1)/sqrt(x^2 - 4) = (1 + 1/(x - 1))*(y + 1)/sqrt(y^2 - 4), where y - 1 = (x - 1)^2 + 2*(x - 1) - 2. Iterating the identity yields the product expansion (x + 1)/sqrt(x^2 - 4) = Product_{n >= 1} (1 + 1/a(n)).
A second expansion is Product_{n >= 1} (1 + 2/(a(n) + 1)) = sqrt((x + 2)/(x - 2)). For an alternative approach to these identities see the Bala link.
(End)

Peter Bala, Notes on A145502-A145510.
Daniel Duverney, Irrationality of Fast Converging Series of Rational Numbers, Journal of Mathematical Sciences-University of Tokyo, Vol. 8, No. 2 (2001), pp. 275-316.
Daniel Duverney and Takeshi Kurosawa, Transcendence of infinite products involving Fibonacci and Lucas numbers, Research in Number Theory, Vol. 8 (2002), Article 68.

Cf. A145503, A145504, A145505, A145506, A145507, A145508, A145509, A145510.

Cf. A000324, A001566.

aa = {}; k = 2; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
(* or *)
k = 1; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}]
NestList[#^2+2#-2&,2,10] (* Harvey P. Dale, Dec 14 2021 *)

From Peter Bala, Nov 12 2012: (Start)
a(n) = phi^(2^n) + (1/phi)^(2^n) - 1, where phi := (1 + sqrt(5))/2 is the golden ratio.
a(n) = A001566(n-1) - 1.
Recurrence: a(n) = 4*(Product_{k = 1..n-1} a(k)) - 2 with a(1) = 2.
Product_{n >= 1} (1 + 1/a(n)) = 4/sqrt(5).
Product_{n >= 1} (1 + 2/(a(n) + 1)) = sqrt(5). (End)
From Amiram Eldar, Sep 10 2022: (Start)
a(n) = A000324(n) - 3.
Sum_{n>=1} (-2)^n/a(n) = -1/2 (Duverney, 2001). (End)
Product_{n>=1} (1 + 3/a(n)) = 4 (Duverney and Kurosawa, 2022). - Amiram Eldar, Jan 07 2023

A000324 A nonlinear recurrence: a(0) = 1, a(1) = 5, a(n) = a(n-1)^2 - 4*a(n-1) + 4 for n>1.

Original entry on oeis.org

1, 5, 9, 49, 2209, 4870849, 23725150497409, 562882766124611619513723649, 316837008400094222150776738483768236006420971486980609

Offset: 0

N. J. A. Sloane

An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
This is the special case k=4 of sequences with exact mutual k-residues. In general, a(1)=k+1 and a(n)=min{m | m>a(n-1), mod(m,a(i))=k, i=1,...,n-1}. k=1 gives Sylvester's sequence A000058 and k=2 Fermat sequence A000215. - Seppo Mustonen, Sep 04 2005
A000058, A000215, A000289 and this sequence here can be represented as values of polynomials defined via P_0(z)= 1+z, P_{n+1}(z) = z+ prod_{i=0..n} P_i(z), with recurrences P_{n+1}(z) = (P_n(z))^2 -z*P_n(z) +z, n>=0. - Vladimir Shevelev, Dec 08 2010

Derek Jennings, Some reciprocal summation identities with applications to the Fibonacci and Lucas numbers, in: G. E. Bergum, Applications of Fibonacci Numbers, Vol. 7, Bergum G. E. et al. (eds.), Kluwer Academic Publishers, 1998, pp. 197-200.
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).

T. D. Noe, Table of n, a(n) for n = 0..12
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
Daniel Duverney, Irrationality of Fast Converging Series of Rational Numbers, Journal of Mathematical Sciences-University of Tokyo, Vol. 8, No. 2 (2001), pp. 275-316.
Daniel Duverney and Takeshi Kurosawa, Transcendence of infinite products involving Fibonacci and Lucas numbers, Research in Number Theory, Vol. 8 (2002), Article 68.
Solomon W. Golomb, On certain nonlinear recurring sequences, Amer. Math. Monthly, Vol. 70, No. 4 (1963), 403-405.
Romeo Meštrović, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv preprint arXiv:1202.3670 [math.HO], 2012. - _N. J. A. Sloane_, Jun 13 2012
Seppo Mustonen, On integer sequences with mutual k-residues, 2005.
Seppo Mustonen, On integer sequences with mutual k-residues, 2005. [Local copy]
Index entries for sequences of form a(n+1)=a(n)^2 + ....

a(n) = A001566(n-1)+2 (for n>0).

Cf. A000032, A000058, A000215, A000289.

t = {1, 5}; Do[AppendTo[t, t[[-1]]^2 - 4*t[[-1]] + 4], {n, 11}] (* T. D. Noe, Jun 19 2012 *)
Join[{1}, RecurrenceTable[{a[n] == a[n-1]^2 - 4*a[n-1] + 4, a[1] == 5}, a, {n, 1, 8}]] (* Jean-François Alcover, Feb 07 2016 *)
Join[{1},NestList[#^2-4#+4&,5,10]] (* Harvey P. Dale, Dec 11 2023 *)

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

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

a(n) = L(2^n)+2, if n>0 where L() is Lucas sequence.
For n>=1, a(n) = 4 + Product_{i=0..n-1} a(i). - Vladimir Shevelev, Dec 08 2010
From Amiram Eldar, Sep 10 2022: (Start)
a(n) = Lucas(2^(n-1))^2 for n > 1.
Sum_{n>=1} 4^n/a(n) = 4 (Jennings, 1998; Duverney, 2001). (End)
Product_{n>=1} (1 - 3/a(n)) = 1/4 (Duverney and Kurosawa, 2022). - Amiram Eldar, Jan 07 2023

A003423 a(n) = a(n-1)^2 - 2, with a(0) = 6.

Original entry on oeis.org

6, 34, 1154, 1331714, 1773462177794, 3145168096065837266706434, 9892082352510403757550172975146702122837936996354

Offset: 0

N. J. A. Sloane

If x is either of the roots of x^2 - 6*x + 1 = 0 (i.e., x = 3 +- 2*sqrt(2)), then x^(2^n) + 1 = a(n)*x^(2^(n-1)). For example, x^8 + 1 = 1154*x^4. - James East, Oct 05 2018

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. 376.
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..10
P. Liardet and P. Stambul, Series d'Engel et fractions continuees, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
E. Lucas, Théorie des Fonctions Numériques Simplement Périodiques, II, Amer. J. Math., 1 (1878), 289-321.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.
J. Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
J. Shallit & N. J. A. Sloane, Correspondence 1974-1975
Wikipedia, Engel Expansion
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Cf. A001566 (starting with 3), A003010 (starting with 4), A003487 (starting with 5).

Cf. A145505.

a:= n-> simplify(2*ChebyshevT(2^n, 3), 'ChebyshevT'):
seq(a(n), n=0..7);

a[1] := 6; a[n_] := a[n - 1]^2 - 2; Table[a[n], {n, 1, 8}] (* Stefan Steinerberger, Apr 11 2006 *)
Table[Round[(1 + Sqrt[2])^(2^n)], {n, 1, 7}] (* Artur Jasinski, Sep 25 2008 *)
NestList[#^2-2&,6,10] (* Harvey P. Dale, Nov 11 2011 *)

PARI
```
a(n)=if(n<1, 6*(n==0), a(n-1)^2-2)
```

a(n) = ceiling(c^(2^n)) where c = 3 + 2*sqrt(2) is the largest root of x^2 - 6x + 1 = 0. - Benoit Cloitre, Dec 03 2002
From Paul D. Hanna, Aug 11 2004: (Start)
a(n) = (3+sqrt(8))^(2^n) + (3-sqrt(8))^(2^n).
Sum_{n>=0} 1/(Product_{k=0..n} a(k) ) = 3 - sqrt(8). (End)
a(n) = 2*A001601(n+1).
a(n-1) = Round((1 + sqrt(2))^(2^n)). - Artur Jasinski, Sep 25 2008
a(n) = 2*T(2^n,3) where T(n,x) is the Chebyshev polynomial of the first kind. - Leonid Bedratyuk, Mar 17 2011
Engel expansion of 3 - 2*sqrt(2). Thus 3 - 2*sqrt(2) = 1/6 + 1/(6*34) + 1/(6*34*1154) + .... See Liardet and Stambul. - Peter Bala, Oct 31 2012
From Peter Bala, Nov 11 2012: (Start)
4*sqrt(2)/7 = Product_{n >= 0} (1 - 1/a(n))
sqrt(2) = Product_{n >= 0} (1 + 2/a(n)).
a(n) - 1 = A145505(n+1). (End)
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2 + 4*Product_{k = 0 ..n-1} (a(k) + 2) for n >= 1.
Let b(n) = a(n) - 6. 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)

A003487 a(n) = a(n-1)^2 - 2, with a(0) = 5.

Original entry on oeis.org

5, 23, 527, 277727, 77132286527, 5949389624883225721727, 35395236908668169265765137996816180039862527, 1252822795820745419377249396736955608088527968701950139470082687906021780162741058825727

Offset: 0

N. J. A. Sloane

The next term has 175 digits. - Harvey P. Dale, Feb 19 2015

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

Seiichi Manyama, Table of n, a(n) for n = 0..10
P. Liardet and P. Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Wikipedia, Engel Expansion
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Cf. A001566 (starting with 3), A003010 (starting with 4), A003423 (starting with 6). A001601, A145504.

a:= n-> simplify(2*ChebyshevT(2^n, 1/2*5), 'ChebyshevT'):
seq(a(n), n=0..7);

NestList[#^2-2&,5,10] (* Harvey P. Dale, Feb 19 2015 *)
a[ n_] := If[ n < 0, 0, 2 ChebyshevT[2^n, 5/2]]; (* Michael Somos, Dec 06 2016 *)

{a(n) = if( n<0, 0, polchebyshev(2^n, 1, 5/2) * 2)}; /* Michael Somos, Dec 06 2016 */

a(n) = ceiling(c^(2^n)) where c=(5+sqrt(21))/2 is the largest root of x^2-5x+1=0. - Benoit Cloitre, Dec 03 2002
a(n) = 2*T(2^n,5/2) where T(n,x) is the Chebyshev polynomial of the first kind. - Leonid Bedratyuk, Mar 17 2011
Engel expansion of 1/2*(5 - sqrt(21)). Thus 1/2*(5 - sqrt(21)) = 1/5 + 1/(5*23) + 1/(5*23*527) + .... See Liardet and Stambul. Cf. A001566, A003010 and A003423. - Peter Bala, Oct 31 2012
From Peter Bala, Nov 11 2012: (Start)
a(n) = ((5 + sqrt(21))/2)^(2^n) + ((5 - sqrt(21))/2)^(2^n).
sqrt(21)/6 = Product_{n = 0..oo} (1 - 1/a(n)).
sqrt(7/3) = Product_{n = 0..oo} (1 + 2/a(n)).
a(n) - 1 = A145504(n+1). (End)
a(n) = A003501(2^n). - Michael Somos, Dec 06 2016
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2 + 3*Product_{k = 0 ..n-1} (a(k) + 2) for n >= 1.
Let b(n) = a(n) - 5. 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)

One more term from Harvey P. Dale, Feb 19 2015

A048757 Sum_{i=0..2n} (C(2n,i) mod 2)Fibonacci(i+2) = Sum_{i=0..n} (C(n,i) mod 2)Fibonacci(2i+2).

Original entry on oeis.org

1, 4, 9, 33, 56, 203, 441, 1596, 2585, 9353, 20304, 73461, 124033, 448756, 974169, 3524577, 5702888, 20633243, 44791065, 162055596, 273617239, 989956471, 2149017696, 7775219067, 12591974497, 45558191716, 98898651657

Offset: 0

Antti Karttunen, Jul 13 1999

The history of 1-D CA Rule 90 starting from the seed pattern 1 interpreted as Zeckendorffian expansion.

Also, product of distinct terms of A001566 and appropriate Fibonacci or Lucas numbers: a(n) = FL(n+2)Product(L(2^i)^bit(n,i),i=0..) Here L(2^i) = A001566 and FL(n) = n-th Fibonacci number if n even, n-th Lucas number if n odd. bit(n,i) is the i-th digit (0 or 1) in the binary expansion of n, with the least significant digit being bit(n,0).

			1 = Fib(2) = 1;
101 = Fib(4) + Fib(2) = 3 + 1 = 4;
10001 = Fib(6) + Fib(2) = 8 + 1 = 9;
1010101 = Fib(8) + Fib(6) + Fib(4) + Fib(2) = 21 + 8 + 3 + 1 = 33; etc.

Antti Karttunen, On Pascal's Triangle Modulo 2 in Fibonacci Representation, Fibonacci Quarterly, 42 (2004), 38-46.

a(n) = A022290(A038183(n)) = A022290(A048723(5, n)) = A003622(A051656(n)) = A075148(n, 2)*A050613(n). Third row of A050609, third column of A050610.

Cf. A054433.

Table[Sum[Mod[Binomial[2n, i], 2] Fibonacci[i + 2], {i, 0, 2n}], {n, 0, 19}] (* Alonso del Arte, Apr 27 2014 *)

A144837 a(n) = Lucas(5^n).

Original entry on oeis.org

11, 167761, 132878596168524201724674011

Offset: 1

Artur Jasinski, Sep 22 2008

Previous name was: a(n) = round(phi^(5^n)) where phi = 1.6180339887... = (sqrt(5) + 1)/2 = A001622.

a(4), a 131-digit number, is too large to show here.

			The base 5 representation of a(3) = 132878596168524201724674011 begins 1 + 2*5 + 0*(5^2) + 2*(5^3) + 3*(5^4) + 0*(5^5) + 4*(5^6) + O(5^7) so A269591 begins [1, 2, 0, 2, 3, 0, 4, ...]. - _Peter Bala_, Nov 14 2022

Amiram Eldar, Table of n, a(n) for n = 1..5

Cf. A000032, A000351, A001622, A001566, A006267, A144836, A144838, A144839, A268922, A269591.

a := proc(n) option remember; if n = 1 then 11 else a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1) end if; end;
seq(a(n), n = 1..5); # Peter Bala, Nov 14 2022

Table[Round[GoldenRatio^(5^n)], {n, 1, 5}]
c = (1 + Sqrt[5])/2; Table[Expand[c^(5^n) + (1 - c)^(5^n)], {n, 1, 5}] (* Artur Jasinski, Oct 05 2008 *)
LucasL[5^Range[5]] (* Harvey P. Dale, Apr 01 2023 *)

a(n) = phi^(5^n) + (1-phi)^(5^n) = phi^(5^n) + (-phi)^(-5^n). - Artur Jasinski, Oct 05 2008
From Peter Bala, Nov 14 2022: (Start)
a(n) = A000032(5^n).
a(n) = a(n-1)^5 + 5*a(n-1)^3 + 5*a(n-1) with a(1) = 11.
a(n) == 1 (mod 5).
a(n+1) == a(n) (mod 5^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).
Conjecture: a(n+1) == a(n) (mod 5^(n+r+1)) for n >= r.
The smallest positive residue of a(n) mod(5^n) = A268922(n).
In the ring of 5-adic integers the limit_{n -> oo} a(n) exists and is equal to A269591. An example is given below. (End)

New name from Peter Bala, Nov 10 2022

A135927 a(n) = a(n-1)^2 - 2 with a(1) = 10.

Original entry on oeis.org

10, 98, 9602, 92198402, 8500545331353602, 72259270930397519221389558374402, 5221402235392591963136699520829303150191924374488750728808857602

Offset: 1

Ant King, Dec 07 2007

This is the Lucas-Lehmer sequence with starting value u(1) = 10 and the position of the zeros when it is reduced mod(2^p - 1) also gives the position of the Mersenne primes. As we have started with n = 1, these will occupy the (p - 1)th positions in the sequence. For example, the first 12 terms mod(2^13 - 1) are 10, 98, 1411, 506, 2113, 672, 1077, 4996, 2037, 4721, 128, 0 and hence 8191 is a Mersenne prime. The radicals in the above closed forms are the solutions to x^2 - 10x + 1 = 0.

			a(4) = 2*cosh(2^3*log(5 + 2*sqrt(6))) = 92198402.

Gabriel Klambauer, Summation of Series, Amer. Math. Monthly, Vol. 87, No. 2 (Feb., 1980), pp. 128-130.
Raphael M. Robinson, Mersenne and Fermat Numbers, Proceedings of the American Mathematical Society, Vol. 5, No. 5. (October 1954), pp. 842-846.
E. L. Roettger and H. C. Williams, Some Remarks Concerning the Lucas-Lehmer Primality Test, Journal of Integer Sequences, Vol. 28 (2025), Article 25.2.5. See p. 3.
Wikipedia, Engel expansion

Cf. A000668, A000043, A003010, A095847, A001566, A135928, A246723.

a[1] = 10; a[n_] := a[n] = a[n - 1]^2 - 2; a[#]&/@Range[7]

A135927 = [10]
for n in range(1, 8): A135927.append(A135927[-1]**2-2)
print(A135927) # Karl-Heinz Hofmann, Feb 01 2022

a(n) = 2*cosh(2^(n-1)*log(5 + 2*sqrt(6))) = exp(2^(n-1)*log(5 + 2*sqrt(6))) + exp(2^(n-1)*log(5 - 2*sqrt(6))) = (5 + 2*sqrt(6))^(2^(n-1)) + (5 - 2*sqrt(6))^(2^(n-1)) = ceiling(exp(2^(n-1)*log(5 + 2*sqrt(6)))) = ceiling((5 + 2*sqrt(6))^(2^(n-1))).
From Peter Bala, Feb 01 2022: (Start)
Product_{n >= 1} (1 + 2/a(n)) = (1/2)*sqrt(6); Product_{n >= 1} (1 - 1/a(n)) = (4/11)*sqrt(6).
Engel expansion of 5 - sqrt(24) = 1/a(1) + 1/(a(1)*a(2)) + 1/(a(1)*a(2)*a(3)) + .... See Klambauer, p. 130. (End)

A145275 a(n) = A145232(n+1)/A145232(n).

Original entry on oeis.org

15005, 792070839820228500005, 311759807762174781605301007201736860141952393239819056256447450170889021063181630442743411596527196875005

Offset: 1

Artur Jasinski, Oct 06 2008

nonn

A member of the family of sequences of type:
(G^(k^(n + 1)) - (1 - G)^(k^(n + 1)))/(G^(k^n) - (1 - G)^(k^n)) where G = (1 + sqrt(5))/2.
For k=2 see A001566.
For k=3 see A002814(n+2).
For k=4 see A145274.
For k=5 see this sequence.
For k=6 see A145276.
For k=7 see A145277.

Amiram Eldar, Table of n, a(n) for n = 1..4

Cf. A001566, A001622, A002814, A145232, A145274, A145276, A145277.

G = (1 + Sqrt[5])/2; Table[Expand[(G^(5^(n + 1)) - (1 - G)^(5^(n + 1)))/Sqrt[5]]/Expand[(G^(5^n) - (1 - G)^(5^n))/Sqrt[5]], {n, 1, 5}]

a(n) = (G^(5^(n + 1)) - (1 - G)^(5^(n + 1)))/(G^(5^n) - (1 - G)^(5^n)) where G = (1 + sqrt(5))/2.

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)

A145274 a(n) = A145231(n+1)/A145231(n).

Original entry on oeis.org

329, 10749959329, 13354478338703157414450712411084788083329

Offset: 1

Artur Jasinski, Oct 06 2008

nonn

A member of the family of sequences of type:
(G^(k^(n + 1)) - (1 - G)^(k^(n + 1)))/(G^(k^n) - (1 - G)^(k^n)) where G = (1 + sqrt(5))/2.
For k=2 see A001566.
For k=3 see A002814(n+2).
For k=4 see this sequence.
For k=5 see A145275.
For k=6 see A145276.
For k=7 see A145277.

Amiram Eldar, Table of n, a(n) for n = 1..5

Cf. A001566, A002814, A145231, A145274, A145275, A145276, A145277.

G = (1 + Sqrt[5])/2; Table[Expand[(G^(4^(n + 1)) - (1 - G)^(4^(n + 1)))/Sqrt[5]]/Expand[(G^(4^n) - (1 - G)^(4^n))/Sqrt[5]], {n, 1, 5}]

a(n) = (G^(4^(n + 1)) - (1 - G)^(4^(n + 1)))/(G^(4^n) - (1 - G)^(4^n)) where G = (1 + sqrt(5))/2.

cp's OEIS Frontend

A145502 a(n+1) = a(n)^2+2*a(n)-2 and a(1)=2.

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A000324 A nonlinear recurrence: a(0) = 1, a(1) = 5, a(n) = a(n-1)^2 - 4*a(n-1) + 4 for n>1.

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A003423 a(n) = a(n-1)^2 - 2, with a(0) = 6.

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A003487 a(n) = a(n-1)^2 - 2, with a(0) = 5.

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A048757 Sum_{i=0..2n} (C(2n,i) mod 2)*Fibonacci(i+2) = Sum_{i=0..n} (C(n,i) mod 2)*Fibonacci(2i+2).

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A144837 a(n) = Lucas(5^n).

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A135927 a(n) = a(n-1)^2 - 2 with a(1) = 10.

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A048757 Sum_{i=0..2n} (C(2n,i) mod 2)Fibonacci(i+2) = Sum_{i=0..n} (C(n,i) mod 2)Fibonacci(2i+2).