A084764 -id:A084764 - 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

A005828 a(n) = 2*a(n-1)^2 - 1, a(0) = 4, a(1) = 31.

Original entry on oeis.org

4, 31, 1921, 7380481, 108942999582721, 23737154316161495960243527681, 1126904990058528673830897031906808442930637286502826475521

Offset: 0

Jeffrey Shallit

An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004

The next term has 115 digits. - Harvey P. Dale, May 25 2018

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

G. C. Greubel, Table of n, a(n) for n = 0..10
Anonymous, Fermat's rule for 3-fold perfect numbers [Broken link]

Cf. A001091, A001601, A002812, A084764 (essentially the same).

[n le 2 select 2^(3*n-1)-n+1 else 2*Self(n-1)^2 - 1: n in [1..10]]; // G. C. Greubel, May 17 2023

NestList[2#^2-1&,4,10] (* Harvey P. Dale, May 25 2018 *)

PARI
```
a(n)=if(n<1,4*(n==0),2*a(n-1)^2-1)
```

a(n)=if(n<0,0,subst(poltchebi(2^n),x,4))

[chebyshev_T(2^n, 4) for n in range(11)] # G. C. Greubel, May 17 2023

a(n) = A001091(2^n).
From Peter Bala, Nov 11 2012, (Start)
a(n) = (1/2)*((4 + sqrt(15))^(2^n) + (4 - sqrt(15))^(2^n)).
2*sqrt(15)/9 = Product_{n>=0} (1 - 1/(2*a(n))).
sqrt(5/3) = Product_{n>=0} (1 + 1/a(n)).
See A002812 for general properties of the recurrence a(n+1) = 2*a(n)^2 - 1.
(End)
a(n) = T(2^n,4), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Feb 01 2017
a(n) = cos(2^n*arccos(4)). - Peter Luschny, Oct 12 2022

A084765 a(n) = 2*a(n-1)^2 - 1, a(0)=1, a(1)=5.

Original entry on oeis.org

1, 5, 49, 4801, 46099201, 4250272665676801, 36129635465198759610694779187201, 2610701117696295981568349760414651575095962187244375364404428801

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Jun 04 2003

Product_{k>=1} (1 + 1/a(k)) = sqrt(3/2) (see A010527).
A subsequence of A001079 (cf. formula), which must contain any prime occurring in A001079. The initial term a(0)=1 seems rather unnatural; using the recurrence relation it would yield the constant sequence 1,1,1,... Note that this sequence corresponds to sequence b(n) in Shallit's paper, which starts only at offset n=1. - M. F. Hasler, Sep 27 2009
Since if x is even (x^2-2)/2 = 2*y^2-1 and 10 is even from a(1) onward this is a reduced version of the LL sequence starting with 10 (A135927) as it is reduced by dividing by 2 it is also the difference between two possible LL sequences. - Roderick MacPhee, May 31 2015
For n >= 3, a(n) == 201 (mod 1000) if n is even, a(n) == 801 (mod 1000) if n is odd. - Robert Israel, Jun 01 2015
The next term -- a(8) -- has 128 digits. - Harvey P. Dale, Mar 28 2020

G. C. Greubel, Table of n, a(n) for n = 0..10
Jeffrey Shallit, Rational numbers with non-terminating, non-periodic modified Engel-type expansions, Fib. Quart., 31 (1993), 37-40.
H. S. Wilf, Limit of a sequence, Elementary Problem E 1093, Amer. Math. Monthly 61 (1954), 424-425.

Cf. A001079, A002812, A010527, A084764, A135927.

[n le 2 select 5^(n-1) else 2*Self(n-1)^2-1: n in [1..10]]; // Vincenzo Librandi, Jun 02 2015

1,seq(expand((5+2*sqrt(6))^(2^n)+(5-2*sqrt(6))^(2^n))/2, n=0..10); # Robert Israel, Jun 01 2015

a[n_]:= a[n]= If[n<2, 5^n, 2 a[n-1]^2 -1]; Table[a[n], {n,0,10}]
Join[{1}, NestList[2 #^2 - 1 &, 5, 10]] (* Harvey P. Dale, Mar 28 2020 *)

first(m)={my(v=[1,5]);for(i=3,m,v=concat(v, 2*v[i-1]^2 - 1));v;} \\ Anders Hellström, Aug 22 2015

def A084765(n): return 1 if n==0 else chebyshev_T(2^(n-1), 5)
[A084765(n) for n in range(11)] # G. C. Greubel, May 17 2023

a(n+1) = (x^(2^n) + y^(2^n))/2, with x = 5 + 2*sqrt(6), y = 5 - 2*sqrt(6).
a(n) = A001079(2^(n-1)) with a(0) = 1. - M. F. Hasler, Sep 27 2009
4*sqrt(6)/11 = Product_{n >= 1} (1 - 1/(2*a(n))). See A002812 for some general properties of the recurrence a(n+1) = 2*a(n)^2 - 1. - Peter Bala, Nov 11 2012
a(n) = cos(2^(n-1)*arccos(5)) for n >= 1. - Peter Luschny, Oct 12 2022

cp's OEIS Frontend

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

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A005828 a(n) = 2*a(n-1)^2 - 1, a(0) = 4, a(1) = 31.

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A084765 a(n) = 2*a(n-1)^2 - 1, a(0)=1, a(1)=5.

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