A003487 -id:A003487 - 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

A003010 A Lucas-Lehmer sequence: a(0) = 4; for n>0, a(n) = a(n-1)^2 - 2.

Original entry on oeis.org

4, 14, 194, 37634, 1416317954, 2005956546822746114, 4023861667741036022825635656102100994, 16191462721115671781777559070120513664958590125499158514329308740975788034

Offset: 0

N. J. A. Sloane

nonn

Albert Beiler states (page 228 of Recreations in the Theory of Numbers): D. H. Lehmer modified Lucas's test to the relatively simple form: If and only if 2^n-1 divides a(n-2) then 2^n-1 is a prime, otherwise it is composite. Since 2^3 - 1 is a factor of a(1) = 14, 2^3 - 1 = 7 is a prime. - Gary W. Adamson, Jun 07 2003

a(n) - a(n-1) divides a(n+1) - a(n). - Thomas Ordowski, Dec 24 2016

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 228.
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. 399.
R. K. Guy, Unsolved Problems in Number Theory, Section A3.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 78.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 205.

N. J. A. Sloane, Table of n, a(n) for n = 0..10
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!)
Larry Ericksen, Primality Testing and Prime Constellations, Šiauliai Mathematical Seminar, Vol. 3 (11), 2008.
James S. Hall, A remark on the primeness of Mersenne numbers, J. London Math. Soc. 28, (1953). 285-287.
D. H. Lehmer, On Lucas's test for the primality of Mersenne's numbers, Journal of the London Mathematical Society 1.3 (1935): 162-165. See page 162.
D. H. Lehmer. Mathematical Reviews. Review of: Hall, James S. A remark on the primeness of Mersenne numbers. (1953) [Annotated scanned copy]
Pierre Liardet and Pierre Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Mario Raso, Integer Sequences in Cryptography: A New Generalized Family and its Application, Ph. D. Thesis, Sapienza University of Rome (Italy 2025). See p. 25.
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.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Pierluigi Vellucci and Alberto Maria Bersani, The class of Lucas-Lehmer polynomials, arXiv preprint arXiv:1603.01989 [math.CA], 2016.
Pierluigi Vellucci and Alberto Maria Bersani, New formulas for pi involving infinite nested square roots and Gray code, arXiv preprint arXiv:1606.09597 [math.NT], 2016 (The OEIS is cited in version 1, but has been dropped from version 4.)
Eric Weisstein's World of Mathematics, Lucas-Lehmer Test.
Wikipedia, Engel Expansion
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Cf. A001566 (starting with 3), A003423 (starting with 6), A003487 (starting with 5).

Cf. A002812, A145503.

[n le 1 select 4 else Self(n-1)^2-2: n in [1..10]]; // Vincenzo Librandi, Aug 24 2015

a := n-> if n>0 then a(n-1)^2-2 else 4 fi: 'a(i)' $ i=0..9; # M. F. Hasler, Mar 09 2007
a := n-> simplify(2*ChebyshevT(2^n, 2), 'ChebyshevT'): seq(a(n), n=0..7);

seqLucasLehmer[0] = 4; seqLucasLehmer[n_] := seqLucasLehmer[n - 1]^2 - 2; Array[seqLucasLehmer, 8, 0] (* Robert G. Wilson v, Jun 28 2012 *)

a(n)=if(n,a(n-1)^2-2,4)
vector(10,i,a(i-1)) \\ M. F. Hasler, Mar 09 2007

from itertools import accumulate
def f(anm1, _): return anm1**2 - 2
print(list(accumulate([4]*8, f))) # Michael S. Branicky, Apr 14 2021

a(n) = ceiling((2 + sqrt(3))^(2^n)). - Benoit Cloitre, Nov 30 2002
More generally, if u(0) = z, integer > 2 and u(n) = a(n-1)^2 - 2 then u(n) = ceiling(c^(2^n)) where c = (1/2)*(z+sqrt(z^2-4)) is the largest root of x^2 - zx + 1 = 0. - Benoit Cloitre, Dec 03 2002
a(n) = (2+sqrt(3))^(2^n) + (2-sqrt(3))^(2^n). - John Sillcox (johnsillcox(AT)hotmail.com), Sep 20 2003
a(n) = ceiling(tan(5*Pi/12)^(2^n)). Note: 5*Pi/12 radians is 75 degrees. - Jason M. Follas (jasonfollas(AT)hotmail.com), Jan 16 2004
Sum_{n >= 0} 1/( Product_{k = 0..n} a(k) ) = 2 - sqrt(3). - Paul D. Hanna, Aug 11 2004
From Ulrich Sondermann, Sep 04 2006: (Start)
To generate the n-th number in the sequence: let x = 2^(n-1), a = 2, b = sqrt(3). Take every other term of the binomial expansion (a+b)^x times 2.
E.g., for the 4th term: x = 2^(4-1) = 8, the binomial expansion is: a^8 + 7a^7 b + 28a^6 b^2 + 56a^5 b^3 + 70a^4 b^4 + 56a^3 b^5 + 28a^2 b^6 + 7a b^7 + b^8, every other term times 2: 2(a^8 + 28a^6 b^2 + 70a^4 b^4 + 28a^2 b^6 + b^8) = 2(256 + (28)(64)(3) + (70)(16)(9) + (28)(4)(27) + 81) = 2(18817) = 37634. (End)
a(n) = 2*cosh( 2^(n-1)*log(sqrt(3)+2) ) For n > 0, a(n) = 2 + 3 * 4^n * (Product_{k=0..n-2} (a(k)/2))^2, where a(k)/2 = A002812(k) is a coprime sequence. - M. F. Hasler, Mar 09 2007
a(n) = A003500(2^n). - John Blythe Dobson, Oct 28 2007
a(n) = 2*T(2^n,2) where T(n,x) is the Chebyshev polynomial of the first kind. - Leonid Bedratyuk, Mar 17 2011
Engel expansion of 2 - sqrt(3). Thus 2 - sqrt(3) = 1/4 + 1/(4*14) + 1/(4*14*194) + ... as noted by Hanna above. See Liardet and Stambul. Cf. A001566, A003423 and A003487. - Peter Bala, Oct 31 2012
From Peter Bala, Nov 11 2012: (Start)
2*sqrt(3)/5 = Product_{n = 0..oo} (1 - 1/a(n)).
sqrt(3) = Product_{n = 0..oo} (1 + 2/a(n)).
a(n) - 1 = A145503(n+1).
a(n) = 2*A002812(n). (End)
a(n+1) - a(n) = a(n)^2 - a(n-1)^2. - Thomas Ordowski, Dec 24 2016
a(n) = 2*cos(2^n * arccos(2)). - Ryan Brooks, Oct 27 2020
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2 + 2*Product_{k = 0..n-1} (a(k) + 2) for n >= 1.
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)

One more term from Thomas A. Rockwell (LlewkcoRAT(AT)aol.com), Jan 18 2005

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)

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

A219164 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 5.

Original entry on oeis.org

5, 527, 77132286527, 35395236908668169265765137996816180039862527

Offset: 0

Peter Bala, Nov 13 2012

Bisection of A003487.

The next term -- a(4) -- has 175 digits. - Harvey P. Dale, Jun 09 2017

Cf. A001999, A003487, A003501, A219162, A219163, A219165.

NestList[#^4-4#^2+2&,5,5] (* Harvey P. Dale, Jun 09 2017 *)

Let alpha = 1/2*(5 + sqrt(21)). Then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
a(n) = A003487(2*n) = A003501(4^n).
Product_{n>=0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(7/3).
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2*T(4^n,5/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
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)

A230338 Recurrence equation: a(0) = 1 and a(n) = a(n-1)sqrt(21a(n-1)^2 + 4) for n >= 1.

Original entry on oeis.org

1, 5, 115, 60605, 16831644835, 1298263252133919638045, 7723873922612696850892381990249713732303715, 273388347343560518533856033712658350781293745092679040607342582493129736504927611387805

Offset: 0

Peter Bala, Oct 30 2013

For integer N, the recurrence equation a(n) = a(n-1)*sqrt((N^2 - 4)*a(n-1)^2 + 4) for n >= 1, with starting value a(0) = 1, produces an integer sequence. The present sequence is the case N = 5. Cf. A000079 (case N = 2), A058635 (case N = 3) and A071579 (case N = 4).
Sequence of numerators in the Engel series representation of 1/2*(7 - sqrt(21)) = 1 + 1/5 + 1 /115 + 1/60605 + .... The corresponding Engel expansion is A003487.
The sequence also has a description as a Pierce expansion of the quadratic irrational 1/2*(5 - sqrt(21)) to the base b := 1/sqrt(21) (see A058635 for a definition of this term).
The associated Pierce series representation of 1/2*(5 - sqrt(21)) to the base b begins 1/2*(5 - sqrt(21)) = b/1 - b^2/(1*5) + b^3/(1*5*115) - b^4/(1*5*115*60605) + ....
More generally, for n >= 0, the sequence [a(n), a(n+1), a(n+2), ...] gives a Pierce expansion of ( 1/2*(5 - sqrt(21)) )^(2^n) to the base b = 1/sqrt(21). Some examples are given below.

			Let b = 1/sqrt(21) and x = 1/2*(5 - sqrt(21)). We have the following Pierce expansions to base b:
x = b/1 - b^2/(1*5) + b^3/(1*5*115) - b^4/(1*5*115*60605) + b^5/(1*5*115*60605*16831644835) - ....
x^2 = b/5 - b^2/(5*115) + b^3/(5*115*60605) - b^4/(5*115*60605*16831644835) + ....
x^4 = b/115 - b^2/(115*60605) + b^3/(115*60605*16831644835) - ....
x^8 = b/60605 - b^2/(60605*16831644835) + ....

Seiichi Manyama, Table of n, a(n) for n = 0..10

Cf. A003487, A004254, A058635, A071579.

a[n_] := a[n - 1]*Sqrt[21 a[n - 1]^2 + 4]; a[0] = 1; Array[a, 8, 0] (* Robert G. Wilson v, Mar 19 2014 *)
a[ n_] := If[ n < 0, 0, ChebyshevU[2^n - 1, 5/2]]; (* Michael Somos, Dec 06 2016 *)

a(n) = if( n<0, 0, imag( (5 + quadgen(84))^2^n) / 2^(2^n - 1)); /* Michael Somos, Dec 06 2016 */

a(n) = 1/sqrt(21)*( alpha^(2^n) - (1/alpha)^(2^n) ), where alpha = 1/2*(5 + sqrt(21)).
a(n) = product {k = 0..n-1} A003487(k).
Defining recurrence equation:
a(0) = 1 and a(n) = a(n-1)*sqrt(21*a(n-1)^2 + 4) for n >= 1.
Other recurrence equations:
a(0) = 1, a(1) = 5 and a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 2.
a(0) = 1, a(1) = 5 and a(n)/a(n-1) = 21*a(n-2)^2 + 2 for n >= 2.
a(n) = A004254(2^n). - Michael Somos, Dec 06 2016

A099175 a(1) = 1, a(2) = 8; for n > 2: a(n) = a(n-1)(3a(n-1)-2).

Original entry on oeis.org

1, 8, 176, 92576, 25710762176, 1983129874961075240576, 11798412302889389755255045998938726679954176, 417607598606915139792416465578985202696175989567316713156694229302007260054247019608576

Offset: 1

Jonathan Vos Post, Nov 15 2004

nonn

n*(3n-2) is the n-th octagonal number (A000567).

Cf. A000567, A003487.

Join[{1}, NestList[ #*(3*# - 2) &, 8, 7]]

a(n) = (A003487(n-1)+1)/3 for n >= 2. - Ondrej Kutal, Sep 08 2023

Edited by Stefan Steinerberger, Jul 25 2007

A319749 a(n) is the numerator of the Heron sequence with h(0)=3.

Original entry on oeis.org

3, 11, 119, 14159, 200477279, 40191139395243839, 1615327685887921300502934267457919, 2609283532796026943395592527806764363779539144932833602430435810559

Offset: 0

Paul Weisenhorn, Sep 27 2018

The denominator of the Heron sequence is in A319750.
The following relationship holds between the numerator of the Heron sequence and the numerator of the continued fraction A041018(n)/A041019(n) convergent to sqrt(13).
n even: a(n)=A041018((5*2^n-5)/3).
n  odd: a(n)=A041018((5*2^n-1)/3).
More generally, all numbers c(n)=A078370(n)=(2n+1)^2+4 have the same relationship between the numerator of the Heron sequence and the numerator of the continued fraction convergent to 2n+1.
sqrt(c(n)) has the continued fraction 2n+1; n,1,1,n,4n+2.
hn(n)^2-c(n)*hd(n)^2=4 for n>1.
From Peter Bala, Mar 29 2022: (Start)
Applying Heron's method (sometimes called the Babylonian method) to approximate the square root of the function x^2 + 4, starting with a guess equal to x, produces the sequence of rational functions [x, 2*T(1,(x^2+2)/2)/x, 2*T(2,(x^2+2)/2)/( 2*x*T(1,(x^2+2)/2) ), 2*T(4,(x^2+2)/2)/( 4*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2) ), 2*T(8,(x^2+2)/2)/( 8*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2)*T(4,(x^2+2)/2) ), ...], where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. The present sequence is the case x = 3. Cf. A001566 and A058635 (case x = 1), A081459 and A081460 (essentially the case x = 4). (End)

			A078370(2)=29.
hn(0)=A041046(0)=5; hn(1)=A041046(3)=27; hn(2)=A041046(5)=727;
hn(3)=A041046(13)=528527.

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
Wikipedia, Methods of computing square roots
Index entries for sequences of form a(n+1)=a(n)^2 + ...

2*T(2^n,x/2) modulo differences of offset: A001566 (x = 3 and x = 7), A003010 (x = 4), A003487 (x = 5), A003423 (x = 6), A346625 (x = 8), A135927 (x = 10), A228933 (x = 18).

Cf. A041018, A041019, A057076, A078370, A081459, A010122, A319750, A041046.

hn[0]:=3:  hd[0]:=1:
for n from 1 to 6 do
hn[n]:=(hn[n-1]^2+13*hd[n-1]^2)/2:
hd[n]:=hn[n-1]*hd[n-1]:
   printf("%5d%40d%40d\n", n, hn[n], hd[n]):
end do:
#alternative program
a := n -> if n = 0 then 3 else simplify( 2*ChebyshevT(2^(n-1), 11/2) ) end if:
seq(a(n), n = 0..7); # Peter Bala, Mar 16 2022

def aupton(nn):
    hn, hd, alst = 3, 1, [3]
    for n in range(nn):
        hn, hd = (hn**2 + 13*hd**2)//2, hn*hd
        alst.append(hn)
    return alst
print(aupton(7)) # Michael S. Branicky, Mar 16 2022

h(n) = hn(n)/hd(n); hn(0)=3; hd(0)=1.
hn(n+1) = (hn(n)^2+13*hd(n)^2)/2.
hd(n+1) = hn(n)*hd(n).
A041018(n) = A010122(n)*A041018(n-1) + A041018(n-2).
A041019(n) = A010122(n)*A041019(n-1) + A041019(n-2).
From Peter Bala, Mar 16 2022: (Start)
a(n) = 2*T(2^(n-1),11/2) for n >= 1, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(n) = 2*T(2^n, 3*sqrt(-1)/2) for n >= 2.
a(n) = ((11 + 3*sqrt(13))/2)^(2^(n-1)) + ((11 - 3*sqrt(13))/2)^(2^(n-1)) for n >= 1.
a(n+1) = a(n)^2 - 2 for n >= 1.
a(n) = A057076(2^(n-1)) for n >= 1.
Engel expansion of (1/6)*(13 - 3*sqrt(13)); that is, (1/6)*(13 - 3*sqrt(13)) = 1/3 + 1/(3*11) + 1/(3*11*119) + .... (Define L(n) = (1/2)*(n - sqrt(n^2 - 4)) for n >= 2 and show L(n) = 1/n + L(n^2-2)/n. Iterate this relation with n = 11. See also Liardet and Stambul, Section 4.)
sqrt(13) = 6*Product_{n >= 0} (1 - 1/a(n)).
sqrt(13) = (9/5)*Product_{n >= 0} (1 + 2/a(n)). See A001566. (End)

a(6) and a(7) added by Peter Bala, Mar 16 2022

cp's OEIS Frontend

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

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A003010 A Lucas-Lehmer sequence: a(0) = 4; for n>0, a(n) = a(n-1)^2 - 2.

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

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

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A219164 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 5.

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A230338 Recurrence equation: a(0) = 1 and a(n) = a(n-1)*sqrt(21*a(n-1)^2 + 4) for n >= 1.

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A099175 a(1) = 1, a(2) = 8; for n > 2: a(n) = a(n-1)*(3*a(n-1)-2).

A230338 Recurrence equation: a(0) = 1 and a(n) = a(n-1)sqrt(21a(n-1)^2 + 4) for n >= 1.

A099175 a(1) = 1, a(2) = 8; for n > 2: a(n) = a(n-1)(3a(n-1)-2).