A145502 -id:A145502 - cp's OEIS frontend

A001566 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2.

3, 7, 47, 2207, 4870847, 23725150497407, 562882766124611619513723647, 316837008400094222150776738483768236006420971486980607

Offset: 0

N. J. A. Sloane

Expansion of 1/phi: 1/phi = (1-1/3)*(1-1/((3-1)*7))*(1-1/(((3-1)*7-1)*47))*(1-1/((((3-1)*7-1)*47-1)*2207))... (phi being the golden ration (1+sqrt(5))/2). - Thomas Baruchel, Nov 06 2003
An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
Starting with 7, the terms end with 7,47,07,47,07,..., of the form 8a+7 where a = 0,1,55,121771,... Conjecture: Every a is squarefree, every other a is divisible by 55, the a's are a subset of A046194, the heptagonal triangular numbers (the first, 2nd, 3rd, 6th, 11th, ?, ... terms). - Gerald McGarvey, Aug 08 2004
Also the reduced numerator of the convergents to sqrt(5) using Newton's recursion x = (5/x+x)/2. - Cino Hilliard, Sep 28 2008
The subsequence of primes begins a(n) for n = 0, 1, 2, 3. - Jonathan Vos Post, Feb 26 2011
We have Sum_{n=0..N} a(n)^2 = 2*(N+1) + Sum_{n=1..N+1} a(n), Sum_{n=0..N} a(n)^4 = 5*(Sum_{n=1..N+1} a(n)) + a(N+1)^2 + 6*N -3, etc. which is very interesting with respect to the fact that a(n) = Lucas(2^(n+1)); see W. Webb's problem in Witula-Slota's paper. - Roman Witula, Nov 02 2012
From Peter Bala, Nov 11 2012: (Start)
The present sequence corresponds to the case x = 3 of the following general remarks.
The recurrence a(n+1) = a(n)^2 - 2 with initial condition a(0) = x > 2 has the solution a(n) = ((x + sqrt(x^2 - 4))/2)^(2^n) + ((x - sqrt(x^2 - 4))/2)^(2^n).
We have the product expansion sqrt(x + 2)/sqrt(x - 2) = Product_{n>=0} (1 + 2/a(n)) (essentially due to Euler - see Mendes-France and van der Poorten). Another expansion is sqrt(x^2 - 4)/(x + 1) = Product_{n>=0} (1 - 1/a(n)), which follows by iterating the identity sqrt(x^2 - 4)/(x + 1) = (1 - 1/x)*sqrt(y^2 - 4)/(y + 1), where y = x^2 - 2.
The sequence b(n) := a(n) - 1 satisfies b(n+1) = b(n)^2 + 2*b(n) - 2. Cases currently in the database are A145502 through A145510. The sequence c(n) := a(n)/2 satisfies c(n+1) = 2*c(n)^2 - 1. Cases currently in the database are A002812, A001601, A005828, A084764 and A084765.
(End)
E. Lucas in Section XIX of "The Theory of Simply Periodic Numerical Functions" (page 56 of English translation) equation "(127) (1-sqrt(5))/2 = -1/1 + 1/3 + 1/(3*7) + 1/(3*7*47) + 1/(3*7*47*2207) + ..." - Michael Somos, Oct 11 2022
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. - Peter Bala, Dec 08 2022
The number of digits of a(n) is given by A094057(n+1). - Hans J. H. Tuenter, Jul 29 2025

			From _Cino Hilliard_, Sep 28 2008: (Start)
Init x=1;
x = (5/1 + 1)/2 = 3/1;
x = (5/3 + 3)/2 = 7/3;
x = ((5/7)/3 + 7/3)/2 = 47/21;
x = ((5/47)/21 + 47/21)/2 = 2207/987;
(2207/987)^2 = 5.000004106... (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.-B. Escott, Note #1741, L'Intermédiaire des Mathématiciens, 8 (1901), page 13. - N. J. A. Sloane, Mar 02 2022
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 223.
Édouard Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 7.
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..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!)
Pierre Liardet and Pierre Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Édouard Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy)
Édouard Lucas, The Theory of Simply Periodic Numerical Functions, Fibonacci Association, 1969. English translation of article "Théorie des Fonctions Numériques Simplement Périodiques, I", Amer. J. Math., 1 (1878), 184-240.
M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.
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. 2.
Chance Sanford, Infinite Series Involving Fibonacci Numbers Via Apéry-Like Formulae, arXiv:1603.03765 [math.NT], 2016.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Wikipedia, Engel Expansion.
Roman Wituła and Damian Słota, delta-Fibonacci numbers, Applicable Analysis and Discrete Mathematics, Vol. 3, No. 2 (2009), pp. 310-329.
Index entries for sequences of form a(n+1)=a(n)^2 + ....

Lucas numbers (A000032) with subscripts that are powers of 2 greater than 1 (Herbert S. Wilf). Cf. A000045.
Cf. A003010 (starting with 4), A003423 (starting with 6), A003487 (starting with 5).
Cf. A058635. - Artur Jasinski, Oct 05 2008
Cf. A002812, A094874, A145502, A145274, A050614, A088334, A181393, A181419, A186750, A186751, A338304.

a:= n-> simplify(2*ChebyshevT(2^n, 3/2), 'ChebyshevT'):
seq(a(n), n=0..8);

NestList[#^2-2&,3,10] (* Harvey P. Dale, Dec 17 2014 *)
Table[LucasL[2^n], {n, 1, 8}] (* Amiram Eldar, Oct 22 2020 *)

a[0]:3$
a[n]:=a[n-1]^2-2$
A001566(n):=a[n]$
makelist(A001566(n),n,0,7); /* Martin Ettl, Nov 12 2012 */

{a(n) = if( n<1, 3*(n==0), a(n-1)^2 - 2)}; /* Michael Somos, Mar 14 2004 */

g(n,p) = x=1;for(j=1,p,x=(n/x+x)/2;print1(numerator(x)","));
g(5,8) \\ Cino Hilliard, Sep 28 2008

{a(n) = my(w = quadgen(5)); if( n<0, 0, n++; imag( (2*w - 1) * w^2^n ))}; /* Michael Somos, Nov 30 2014 */

{a(n) = my(y = x^2-x-1); if( n<0, 0, n++; for(i=1, n, y = polgraeffe(y)); -polcoeff(y, 1))}; /* Michael Somos, Nov 30 2014 */

a(n) = Fibonacci(2^(n+2))/Fibonacci(2^(n+1)) = A058635(n+2)/A058635(n+1). - Len Smiley, May 08 2000, and Artur Jasinski, Oct 05 2008
a(n) = ceiling(c^(2^n)) where c = (3+sqrt(5))/2 = tau^2 is the largest root of x^2-3*x+1=0. - Benoit Cloitre, Dec 03 2002
a(n) = round(G^(2^n)) where G is the golden ratio (A001622). - Artur Jasinski, Sep 22 2008
a(n) = (G^(2^(n+1))-(1-G)^(2^(n+1)))/((G^(2^n))-(1-G)^(2^n)) = G^(2^n)+(1-G)^(2^n) = G^(2^n)+(-G)^(-2^n) where G is the golden ratio. - Artur Jasinski, Oct 05 2008
a(n) = 2*cosh(2^(n+1)*arccosh(sqrt(5)/2)). - Artur Jasinski, Oct 09 2008
a(n) = Fibonacci(2^(n+1)-1) + Fibonacci(2^(n+1)+1). (3-sqrt(5))/2 = 1/3 + 1/(3*7) + 1/(3*7*47) + 1/(3*7*47*2207) + ... (E. Lucas). - Philippe Deléham, Apr 21 2009
a(n)*(a(n+1)-1)/2 = A023039(2^n). - M. F. Hasler, Sep 27 2009
For n >= 1, a(n) = 2 + Product_{i=0..n-1} (a(i) + 2). - Vladimir Shevelev, Nov 28 2010
a(n) = 2*T(2^n,3/2) where T(n,x) is the Chebyshev polynomial of the first kind. - Leonid Bedratyuk, Mar 17 2011
From Peter Bala, Oct 31 2012: (Start)
Engel expansion of 1/2*(3 - sqrt(5)). Thus 1/2*(3 - sqrt(5)) = 1/3 + 1/(3*7) + 1/(3*7*47) + ... as noted above by Deleham. See Liardet and Stambul.
sqrt(5)/4 = Product_{n>=0} (1 - 1/a(n)).
sqrt(5) = Product_{n>=0} (1 + 2/a(n)). (End)
a(n) - 1 = A145502(n+1). - Peter Bala, Nov 11 2012
a(n) == 2 (mod 9), for n > 1. - Ivan N. Ianakiev, Dec 25 2013
From Amiram Eldar, Oct 22 2020: (Start)
a(n) = A000032(2^(n+1)).
Sum_{k>=0} 1/a(k) = -1 + A338304. (End)
a(n) = (A000045(m+2^(n+2))+A000045(m))/A000045(m+2^(n+1)) for any m>=0. - Alexander Burstein, Apr 10 2021
a(n) = 2*cos(2^n*arccos(3/2)). - Peter Luschny, Oct 12 2022
a(n) == -1 ( mod 2^(n+2) ). - Peter Bala, Nov 07 2022
a(n) = 5*Fibonacci(2^n)^2+2 = 5*A058635(n)^2+2, for n>0. - Jianglin Luo, Sep 21 2023
Sum_{n>=0} a(n)/Fibonacci(2^(n+2)) = A094874 (Sanford, 2016). - Amiram Eldar, Mar 01 2024

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

A145510 a(n+1) = a(n)^2 + 2*a(n) - 2 and a(1)=10.

Original entry on oeis.org

10, 118, 14158, 200477278, 40191139395243838, 1615327685887921300502934267457918, 2609283532796026943395592527806764363779539144932833602430435810558

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))

Cf. A145502, A145503, A145504, A145505, A145506, A145507, A145508, A145509.

aa = {}; k = 10; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
(* or *)
k =9; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)

From Peter Bala, Nov 12 2012: (Start)
a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 1/2*(11 + sqrt(117)). a(n) = 1 (mod 9).
Recurrence: a(n) = 12*{Product_{k = 1..n-1} a(k)} - 2 with a(1) = 10.
Product {n = 1..inf} (1 + 1/a(n)) = 12/sqrt(117).
Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(13/9).
(End)

A145503 a(n+1) = a(n)^2+2*a(n)-2 and a(1)=3.

Original entry on oeis.org

3, 13, 193, 37633, 1416317953, 2005956546822746113, 4023861667741036022825635656102100993

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)).

Essentially the same as A110407. [R. J. Mathar, Mar 18 2009]

Indranil Ghosh, Table of n, a(n) for n = 1..11

Cf. A145502-A145510. A003010, A002812.

aa = {}; k = 3; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
(* or *)
k = 2; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}]
NestList[#^2+2#-2&,3,10] (* Harvey P. Dale, Feb 01 2018 *)

From Peter Bala, Nov 12 2012: (Start)
a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 2 + sqrt(3).
a(n) = A003010(n-1) - 1. a(n) = 2*A002812(n-1) - 1.
Recurrence: a(n) = 5*(Product {k = 1..n-1} a(k)) - 2 with a(1) = 3.
Product_{n >= 1} (1 + 1/a(n)) = 5/6*sqrt(3).
Product_{n >= 1} (1 + 2/(a(n) + 1)) = sqrt(3).
(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)

A145504 a(n+1)=a(n)^2+2*a(n)-2 and a(1)=4.

Original entry on oeis.org

4, 22, 526, 277726, 77132286526, 5949389624883225721726, 35395236908668169265765137996816180039862526, 1252822795820745419377249396736955608088527968701950139470082687906021780162741058825726

Offset: 1

Artur Jasinski, Oct 11 2008

nonn

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))

The next term (a(9)) has 175 digits. - Harvey P. Dale, Nov 16 2013

A145502-A145510. A003487.

NestList[#^2+2#-2&,4,7] (* Harvey P. Dale, Nov 16 2013 *)

From Peter Bala, Nov 12 2012: (Start)
a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 1/2*(5 + sqrt(21)).
a(n) = A003487(n-1) - 1.
Recurrence: a(n) = 6*{product {k = 1..n-1} a(k)} - 2 with a(1) = 4.
Product {n = 1..inf} (1 + 1/a(n)) = 2/7*sqrt(21).
Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(7/3).
(End)

One additional term (a(8)) from Harvey P. Dale, Nov 16 2013

A145505 a(n+1)=a(n)^2+2*a(n)-2 and a(1)=5.

Original entry on oeis.org

5, 33, 1153, 1331713, 1773462177793, 3145168096065837266706433, 9892082352510403757550172975146702122837936996353

Offset: 1

Artur Jasinski, Oct 11 2008

nonn

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))

A145502-A145510. A003423, A001601, A190840.

aa = {}; k = 5; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
or
k =4; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)
NestList[#^2+2#-2&,5,7]  (* Harvey P. Dale, Mar 19 2011 *)

From Peter Bala, Nov 12 2012: (Start)
a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 3 + 2*sqrt(2).
a(n) = (1 + sqrt(2))^(2^n) + (sqrt(2) - 1)^(2^n) - 1.
a(n) = A003423(n-1) - 1. a(n) = 2*A001601(n) - 1. a(n) = 4*A190840(n-1) + 1.
Recurrence: a(n) = 7*{product {k = 1..n-1} a(k)} - 2 with a(1) = 5.
Product {n = 1..inf} (1 + 1/a(n)) = 7/8*sqrt(2).
Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(2).
(End)

A145506 a(n+1) = a(n)^2+2*a(n)-2 and a(1)=6.

Original entry on oeis.org

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

Offset: 1

Artur Jasinski, Oct 11 2008

nonn

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))).

Essentially the same as A145502. - R. J. Mathar, Mar 18 2009

Cf. A145502-A145511.

aa = {}; k = 6; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
or
k = 5; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (* Artur Jasinski *)

A145507 a(n+1)=a(n)^2+2*a(n)-2 and a(1)=7.

Original entry on oeis.org

7, 61, 3841, 14760961, 217885999165441, 47474308632322991920487055361, 2253809980117057347661794063813616885861274573005652951041

Offset: 1

Artur Jasinski, Oct 11 2008

nonn

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))

A145502-A145510. A005828.

aa = {}; k = 7; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
or
k =6; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)

From Peter Bala, Nov 12 2012: (Start)
a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 4 + sqrt(15).
a(n) = 2*A005828(n) - 1.
Recurrence: a(n) = 9*{product {k = 1..n-1} a(k)} - 2 with a(1) = 7.
Product {n = 1..inf} (1 + 1/a(n)) = 3/10*sqrt(15).
Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(5/3).
(End)

A145508 a(n+1)=a(n)^2+2*a(n)-2 and a(1)=8.

Original entry on oeis.org

8, 78, 6238, 38925118, 1515164889164158, 2295724641355838227053650177278, 5270351628928392053240255925779522360603268430188068127843838

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))

A145502-A145510.

aa = {}; k = 8; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
or
k =7; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}] (*Artur Jasinski*)
NestList[#^2+2#-2&,8,8] (* Harvey P. Dale, Sep 20 2013 *)

From Peter Bala, Nov 12 2012: (Start)
a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1, where alpha := 1/2*(9 + sqrt(77)). a(n) = 1 (mod 7).
Recurrence: a(n) = 10*{product {k = 1..n-1} a(k)} - 2 with a(1) = 8.
Product {n = 1..inf} (1 + 1/a(n)) = 10/sqrt(77).
Product {n = 1..inf} (1 + 2/(a(n) + 1)) = sqrt(11/7).
(End)

cp's OEIS Frontend

A001566 a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2.

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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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A145510 a(n+1) = a(n)^2 + 2*a(n) - 2 and a(1)=10.

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A145503 a(n+1) = a(n)^2+2*a(n)-2 and a(1)=3.

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

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A145504 a(n+1)=a(n)^2+2*a(n)-2 and a(1)=4.

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A145505 a(n+1)=a(n)^2+2*a(n)-2 and a(1)=5.

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