A003423 -id:A003423 - 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

A001601 a(n) = 2*a(n-1)^2 - 1, if n>1. a(0)=1, a(1)=3.

Original entry on oeis.org

1, 3, 17, 577, 665857, 886731088897, 1572584048032918633353217, 4946041176255201878775086487573351061418968498177

Offset: 0

N. J. A. Sloane

Reduced numerators of Newton's iteration for sqrt(2). - Eric W. Weisstein
An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
Evaluation of the 2^n - 1 degree interpolating polynomial of 1/x at Chebyshev nodes in the interval (1,2): v = 1.0; for(i = 1, n, v *= 4*(a(i) - x*v)); v *= 2/a(n+1). - Jose Hortal, Apr 07 2012
Smallest positive integer x satisfying the Pell equation x^2 - 2^(2*n+1) * y^2 = 1, for n > 0. - A.H.M. Smeets, Sep 29 2017

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, 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).

Dennis Martin, Table of n, a(n) for n = 0..11
Dov Jarden, Recurring Sequences, Riveon Lematematika, Jerusalem, 1966. [Annotated scanned copy]
M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.
Jeffrey Shallit, Rational numbers with non-terminating, non-periodic modified Engel-type expansions, Fib. Quart., 31 (1993), 37-40.
Eric Weisstein's World of Mathematics, Newtons Iteration.
Eric Weisstein's World of Mathematics, Square Root.
Eric Weisstein's World of Mathematics, Pythagoras's Constant.
H. S. Wilf, D. C. B. Marsh and J. V. Whittaker, Problem E1093, Amer. Math. Monthly, 61 (1954), 424-425.
Index entries for sequences related to Engel expansions

Cf. A051009, A003423.

a(n) = A001333(2^n).

Table[Simplify[Expand[(1/2) ((1 + Sqrt[2])^(2^n) + (1 - Sqrt[2])^(2^n))]], {n, 0, 7}]  (* Artur Jasinski, Oct 10 2008 *)
Join[{1},NestList[2#^2-1&,3,7]]  (* Harvey P. Dale, Mar 24 2011 *)

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

From Mario Catalani (mario.catalani(AT)unito.it), May 27 2003, May 30 2003: (Start)
a(n) = a(n-1)^2 + 2*A051009(n)^2 for n > 0.
a(n)^2 = 2*A051009(n+1)^2 + 1.
a(n) = Sum_{r=0..2^(n-1)} binomial(2^n, 2*r)*2^r. (End)
Expansion of 1/sqrt(2) as an infinite product: 1/sqrt(2) = Product_{k>=1} (1 - 1/(a(n)+1)). a(1)=3; a(n) = floor(1/(1-1/(sqrt(2)*Product_{k=1..n-1} 1-1/(a(k)+1)))). - Thomas Baruchel, Nov 06 2003
2*a(n+1) = A003423(n).
a(n) = (1/2)*((1 + sqrt(2))^(2^n) + (1 - sqrt(2))^(2^n)). - Artur Jasinski, Oct 10 2008
For n > 1: a(n) - 1 = 4^n * Product_{i=1..n-2} a(i)^2. - Jose Hortal, Apr 13 2012
From Peter Bala, Nov 11 2012: (Start)
4*sqrt(2)/7 = Product_{n>=1} (1 - 1/(2*a(n))).
sqrt(2) = Product_{n>=1} (1 + 1/a(n)).
a(n) = (1/2)*A003423(n-1). (End)
a(n) = cos(2^(n-1) * arccos(3)) = cosh(2^(n-1) * log(3 + 2*sqrt(2))) for n >= 1. - Daniel Suteu, Jul 28 2016
a(n+1) = T(2^n,3), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Feb 01 2017
a(n) = A001541(2^(n-1)). - A.H.M. Smeets, May 28 2017

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

A228931 Optimal ascending continued fraction expansion of sqrt(2)-1.

Original entry on oeis.org

2, -6, 34, 1154, 1331714, 1773462177794, 3145168096065837266706434, 9892082352510403757550172975146702122837936996354

Offset: 1

Giovanni Artico, Sep 09 2013

See A228929 for the definition of "optimal ascending continued fraction".
Conjecture: The terms from a(3) are all positive and can be generated by the recurrence relation a(k+1) = a(k)^2 - 2.
This relation was studied by Lucas with reference to Engel expansion.
This recurrence is not peculiar of sqrt(2) but is present in the expansion of the square root of many other numbers, starting from some term onward, but not for all numbers. Here is a list of the numbers in range 1..200 having the recurrence: 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 45, 47, 48, 50, 51, 52, 54, 55, 56, 57, 59, 60, 62, 63, 65, 66, 68, 69, 70, 71, 72, 74, 75, 77, 78, 79, 80, 82, 83, 84, 87, 88, 90, 92, 93, 95, 96, 98, 99, 101, 102, 104, 105, 107, 110, 111, 112, 114, 117, 119, 120, 122, 123, 124, 126, 128, 130, 132, 133, 135, 136, 138, 140, 141, 142, 143, 145, 146, 147, 148, 150, 152, 155, 156, 158, 162, 164, 165, 167, 168, 170, 171, 174, 175, 178, 180, 182, 183, 185, 187, 188, 189, 192, 194, 195, 197, 198, 200
Essentially the same as A003423. - R. J. Mathar, Sep 21 2013

			sqrt(2)=1+1/2*(1-1/6*(1+1/34*(1+1/1154*(1+1/1331714*(1+1/1773462177794*(1+.....))))))

P. Bala, A modified Engel expansion for certain quadratic irrationals
Giovanni Artico, Proof of the conjecture

Cf. A228929, A220335.

ArticoExp := proc (n, q::posint)::list; local L, i, z; Digits := 50000; L := []; z := frac(evalf(n)); for i to q+1 do if z = 0 then break end if; L := [op(L), round(1/abs(z))*sign(z)]; z := abs(z)*round(1/abs(z))-1 end do; return L end proc
# List the first 8 terms of the expansion of sqrt(2)-1
ArticoExp(sqrt(2),8)

Flatten[{2, RecurrenceTable[{a[n] == a[n-1]^2 - 2, a[2] == -6}, a, {n, 2, 10}]}] (* Vaclav Kotesovec, Sep 20 2013 *)

a(n) = a(n-1)^2 - 2, for n > 2.

For n>2, a(n) = (sqrt(2)+1)^(2^(n-1)) + (sqrt(2)-1)^(2^(n-1)). - Vaclav Kotesovec, Sep 20 2013

Added a pdf file with a proof of the conjecture by Giovanni Artico

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)

A219165 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 6.

Original entry on oeis.org

6, 1154, 1773462177794, 9892082352510403757550172975146702122837936996354

Offset: 0

Peter Bala, Nov 13 2012

Bisection of A003423.

Cf. A001999, A003423, A003499, A219162, A219163, A219164.

Let alpha = 3 + 2*sqrt(2). Then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
a(n) = A003423(2*n) = A003499(4^n).
Product {n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(2).
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2*T(4^n,3), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
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)

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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A001601 a(n) = 2*a(n-1)^2 - 1, if n>1. a(0)=1, a(1)=3.

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

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A228931 Optimal ascending continued fraction expansion of sqrt(2)-1.

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

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