A002812 -id:A002812 - cp's OEIS frontend

A071579 a(n) = 2a(n-1)A002812(n-1), starting a(0)=1.

1, 4, 56, 10864, 408855776, 579069776145402304, 1161588808526051807570761628582646656, 4674072680304961790168962360144614650442718636276775741658113370728376064

Offset: 0

Joe Keane (jgk(AT)jgk.org), May 31 2002

Also the denominators of the convergents to sqrt(3) using Newton's recursion x = (3/x+x)/2. - Cino Hilliard, Sep 28 2008
For n>1, Egyptian fraction of 2-sqrt(3): 2-sqrt(3) = 1/4 + 1/56 + 1/10864 + 1/408855776 + ... - Simon Plouffe, Feb 20 2011
The sequence satisfies the Pell equation A002812(n)^2-3*a(n)^2 = 1. - Vincenzo Librandi, Dec 19 2011
From Peter Bala, Oct 30 2013: (Start)
Apart from giving the numerators in the Engel series representation of 2 - sqrt(3), as stated above by Plouffe, this sequence is also a Pierce expansion of the real number x = 2 - sqrt(3) to the base b := 1/sqrt(12) (see A058635 for a definition of this term).
The associated series representation begins 2 - sqrt(3) = b/1 - b^2/(1*4) + b^3/(1*4*56) - b^4/(1*4*56*10864) + .... Cf. A230338.
More generally, for n >= 0, the sequence [a(n), a(n+1), a(n+2), ...] gives a Pierce expansion of (2 - sqrt(3))^(2^n) to the base b = 1/sqrt(12). Some examples are given below. (End)

			Let b = 1/sqrt(12) and x = 2 - sqrt(3). We have the following Pierce expansions to base b:
x = b/1 - b^2/(1*4) + b^3/(1*4*56) - b^4/(1*4*56*10864) + b^5/(1*4*56*10864*408855776) - ....
x^2 = b/4 - b^2/(4*56) + b^3/(4*56*10864) - b^4/(4*56*10864*408855776) + ....
x^4 = b/56 - b^2/(56*10864) + b^3/(56*10864*408855776) - ....
x^8 = b/10864 - b^2/(10864*408855776) + .... - _Peter Bala_, Oct 30 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..10
Eric Weisstein's World of Mathematics, Newton's Iteration.
Doron Zeilberger, Another Book of Somos-Like Miracles, Prop. Number, 2; Local copy.

Cf. A002812, A001353. A058635, A230338.

I:=[1,4]; [n le 2 select I[n] else 2*Self(n-1)*(6*Self(n-2)^2+1): n in [1..8]]; // Vincenzo Librandi, Dec 19 2011

a[ n_] := If[n<0, 0, Coefficient[PolynomialMod[x^2^n, x^2 - 4*x + 1], x]]; (* Michael Somos, Jul 29 2024 *)
a[ n_] := If[n<1, Boole[n==0], a[n] = a[n-1]*Sqrt[12*a[n-1]^2 + 4] ]; (* Michael Somos, Jul 29 2024 *)

g(n,p) = x=1;for(j=1,p,x=(n/x+x)/2;print1(denominator(x)","))

{a(n) = if(n<0, 0, imag((2 + quadgen(12))^2^n))}; /* Michael Somos, Jul 29 2024 */

{a(n) = if(n<0, 0, polcoef(lift(Mod(x^2^n, x^2 - 4*x + 1)), 1))}; /* Michael Somos, Jul 29 2024 */
g(3,8) \\ Cino Hilliard, Sep 28 2008

a(n) = 1/sqrt(12)*( (2 + sqrt(3))^2^n - (2 - sqrt(3))^2^n ) = A001353(2^n).
a(n) = 2*a(n-1)*(6*a(n-2)^2+1). - Max Alekseyev, Apr 19 2006
Recurrence equations:
a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 2.
a(n) = a(n-1)*sqrt(12*a(n-1)^2 + 4) for n >= 1. - Peter Bala, Oct 30 2013
0 = 6*a(n)^2*a(n+2) - 6*a(n+1)^3 - 2*a(n+1) + a(n+2) for n>=1. - Michael Somos, Dec 05 2016
0 = a(n)^2*(2*a(n+1) + a(n+2)) - a(n+1)^3 for n>=1. - Michael Somos, Dec 05 2016
a(n) = A001353(2^n). - Michael Somos, Jul 29 2024

A001541 a(0) = 1, a(1) = 3; for n > 1, a(n) = 6*a(n-1) - a(n-2).

Original entry on oeis.org

1, 3, 17, 99, 577, 3363, 19601, 114243, 665857, 3880899, 22619537, 131836323, 768398401, 4478554083, 26102926097, 152139002499, 886731088897, 5168247530883, 30122754096401, 175568277047523, 1023286908188737, 5964153172084899, 34761632124320657

Offset: 0

N. J. A. Sloane

Chebyshev polynomials of the first kind evaluated at 3.
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 8*y^2 = 1, the corresponding values of y are in A001109. For n > 0, the ratios a(n)/A001090(n) may be obtained as convergents to sqrt(8): either successive convergents of [3; -6] or odd convergents of [2; 1, 4]. - Lekraj Beedassy, Sep 09 2003 [edited by Jon E. Schoenfield, May 04 2014]
Also gives solutions to the equation x^2 - 1 = floor(x*r*floor(x/r)) where r = sqrt(8). - Benoit Cloitre, Feb 14 2004
Appears to give all solutions greater than 1 to the equation: x^2 = ceiling(x*r*floor(x/r)) where r = sqrt(2). - Benoit Cloitre, Feb 24 2004
This sequence give numbers n such that (n-1)*(n+1)/2 is a perfect square. Remark: (i-1)*(i+1)/2 = (i^2-1)/2 = -1 = i^2 with i = sqrt(-1) so i is also in the sequence. - Pierre CAMI, Apr 20 2005
a(n) is prime for n = {1, 2, 4, 8}. Prime a(n) are {3, 17, 577, 665857}, which belong to A001601(n). a(2k-1) is divisible by a(1) = 3. a(4k-2) is divisible by a(2) = 17. a(8k-4) is divisible by a(4) = 577. a(16k-8) is divisible by a(8) = 665857. - Alexander Adamchuk, Nov 24 2006
The upper principal convergents to 2^(1/2), beginning with 3/2, 17/12, 99/70, 577/408, comprise a strictly decreasing sequence; essentially, numerators=A001541 and denominators=A001542. - Clark Kimberling, Aug 26 2008
Also index of sequence A082532 for which A082532(n) = 1. - Carmine Suriano, Sep 07 2010
Numbers n such that sigma(n-1) and sigma(n+1) are both odd numbers. - Juri-Stepan Gerasimov, Mar 28 2011
Also, numbers such that floor(a(n)^2/2) is a square: base 2 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001075. - M. F. Hasler, Jan 15 2012
Numbers such that 2n^2 - 2 is a square. Also integer square roots of the expression 2*n^2 + 1, at values of n given by A001542. Also see A228405 regarding 2n^2 -+ 2^k generally for k >= 0. - Richard R. Forberg, Aug 20 2013
Values of x (or y) in the solutions to x^2 - 6xy + y^2 + 8 = 0. - Colin Barker, Feb 04 2014
Panda and Ray call the numbers in this sequence the Lucas-balancing numbers C_n (see references and links).
Partial sums of X or X+1 of Pythagorean triples (X,X+1,Z). - Peter M. Chema, Feb 03 2017
a(n)/A001542(n) is the closest rational approximation to sqrt(2) with a numerator not larger than a(n), and 2*A001542(n)/a(n) is the closest rational approximation to sqrt(2) with a denominator not larger than a(n). These rational approximations together with those obtained from the sequences A001653 and A002315 give a complete set of closest rational approximations to sqrt(2) with restricted numerator or denominator. a(n)/A001542(n) > sqrt(2) > 2*A001542(n)/a(n). - A.H.M. Smeets, May 28 2017
x = a(n), y = A001542(n) are solutions of the Diophantine equation x^2 - 2y^2 = 1 (Pell equation). x = 2*A001542(n), y = a(n) are solutions of the Diophantine equation x^2 - 2y^2 = -2. Both together give the set of fractional approximations for sqrt(2) obtained from limited fractions obtained from continued fraction representation to sqrt(2). - A.H.M. Smeets, Jun 22 2017
a(n) is the radius of the n-th circle among the sequence of circles generated as follows: Starting with a unit circle centered at the origin, every subsequent circle touches the previous circle as well as the two limbs of hyperbola x^2 - y^2 = 1, and lies in the region y > 0. - Kaushal Agrawal, Nov 10 2018
All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0Michael Somos, Jun 26 2022

			99^2 + 99^2 = 140^2 + 2. - _Carmine Suriano_, Jan 05 2015
G.f. = 1 + 3*x + 17*x^2 + 99*x^3 + 577*x^4 + 3363*x^5 + 19601*x^6 + 114243*x^7 + ...

Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009)
J. W. L. Glaisher, On Eulerian numbers (formulas, residues, end-figures), with the values of the first twenty-seven, Quarterly Journal of Mathematics, vol. 45, 1914, pp. 1-51.
G. K. Panda, Some fascinating properties of balancing numbers, In Proc. of Eleventh Internat. Conference on Fibonacci Numbers and Their Applications, Cong. Numerantium 194 (2009), 185-189.
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).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-258.
P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022

T. D. Noe, Table of n, a(n) for n = 0..200
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Jean-Paul Allouche, Zeta-regularization of arithmetic sequences, EPJ Web of Conferences (2020) Vol. 244, 01008.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
H. Brocard, Notes élémentaires sur le problème de Peel, Nouvelle Correspondance Mathématique, 4 (1878), 161-169.
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
P. Catarino, H. Campos, and P. Vasco, On some identities for balancing and cobalancing numbers, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24.
Kwang-Wu Chen and Yu-Ren Pan, Greatest Common Divisors of Shifted Horadam Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.5.8.
S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Morgan Fiebig, aBa Mbirika, and Jürgen Spilker, Period patterns, entry points, and orders in the Lucas sequences: theory and applications, arXiv:2408.14632 [math.NT], 2024. See p. 5.
Robert Frontczak, A Note on Hybrid Convolutions Involving Balancing and Lucas-Balancing Numbers, Applied Mathematical Sciences, Vol. 12, 2018, No. 25, 1201-1208.
Robert Frontczak, Sums of Balancing and Lucas-Balancing Numbers with Binomial Coefficients, International Journal of Mathematical Analysis (2018) Vol. 12, No. 12, 585-594.
Robert Frontczak, Powers of Balancing Polynomials and Some Consequences for Fibonacci Sums, International Journal of Mathematical Analysis (2019) Vol. 13, No. 3, 109-115.
Robert Frontczak, Identities for generalized balancing numbers, Notes on Number Theory and Discrete Mathematics (2019) Vol. 25, No. 2, 169-180.
Robert Frontczak and Taras Goy, Additional close links between balancing and Lucas-balancing polynomials, arXiv:2007.14048 [math.NT], 2020.
Robert Frontczak and Taras Goy, More Fibonacci-Bernoulli relations with and without balancing polynomials, arXiv:2007.14618 [math.NT], 2020.
Robert Frontczak and Taras Goy, Lucas-Euler relations using balancing and Lucas-balancing polynomials, arXiv:2009.09409 [math.NT], 2020.
O. Khadir, K. Liptai, and L. Szalay, On the Shifted Product of Binary Recurrences, J. Int. Seq. 13 (2010), 10.6.1.
J. M. Katri and D. R. Byrkit, Problem E1976, Amer. Math. Monthly, 75 (1968), 683-684.
Tanya Khovanova, Recursive Sequences
D. H. Lehmer, A cotangent analogue of continued fractions, Duke Math. J., 4 (1935), 323-340.
D. H. Lehmer, A cotangent analogue of continued fractions, Duke Math. J., 4 (1935), 323-340. [Annotated scanned copy]
D. H. Lehmer, Lacunary recurrence formulas for the numbers of Bernoulli and Euler, Annals Math., 36 (1935), 637-649.
Dino Lorenzini and Z. Xiang, Integral points on variable separated curves, Preprint 2016.
aBa Mbirika, Janeè Schrader, and Jürgen Spilker, Pell and associated Pell braid sequences as GCDs of sums of k consecutive Pell, balancing, and related numbers, arXiv:2301.05758 [math.NT], 2023. See also J. Int. Seq. (2023) Vol. 26, Art. 23.6.4.
A. Patra, G. K. Panda, and T. Khemaratchatakumthorn, Exact divisibility by powers of the balancing and Lucas-balancing numbers, Fibonacci Quart., 59:1 (2021), 57-64; see C(n).
Robert Phillips, Polynomials of the form 1+4ke+4ke^2, 2008.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Prasanta K. Ray, Curious congruences for balancing numbers, Int. J. Contemp. Math. Sci. 7 (2012), 881-889.
Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
Bibhu Prasad Tripathy and Bijan Kumar Patel, On balancing and Lucas-balancing numbers expressible as product of two k-Fibonacci numbers, arXiv:2508.12238 [math.NT], 2025.
N. J. Wildberger, Pell's equation without irrational numbers, J. Int. Seq. 13 (2010), 10.4.3, Section 4.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Bisection of A001333. A003499(n) = 2a(n).
Cf. A055997 = numbers n such that n(n-1)/2 is a square.
Row 1 of array A188645.
Cf. A046090, A001109, A053142, A084130, A001601, A056771, A077420, A005319, A082532, A001542.
Cf. A055792 (terms squared), A132592.

a001541 n = a001541_list !! (n-1)
a001541_list =
1 : 3 : zipWith (-) (map (* 6) $ tail a001541_list) a001541_list
-- Reinhard Zumkeller, Oct 06 2011
(Scheme, with memoization-macro definec)
(definec (A001541 n) (cond ((zero? n) 1) ((= 1 n) 3) (else (- (* 6 (A001541 (- n 1))) (A001541 (- n 2))))))
;; Antti Karttunen, Oct 04 2016

[n: n in [1..10000000] |IsSquare(8*(n^2-1))]; // Vincenzo Librandi, Nov 18 2010

a[0]:=1: a[1]:=3: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006
A001541:=-(-1+3*z)/(1-6*z+z**2); # Simon Plouffe in his 1992 dissertation

Table[Simplify[(1/2) (3 + 2 Sqrt[2])^n + (1/2) (3 - 2 Sqrt[2])^n], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
a[ n_] := If[n == 0, 1, With[{m = Abs @ n}, m Sum[4^i Binomial[m + i, 2 i]/(m + i), {i, 0, m}]]]; (* Michael Somos, Jul 11 2011 *)
a[ n_] := ChebyshevT[ n, 3]; (* Michael Somos, Jul 11 2011 *)
LinearRecurrence[{6, -1}, {1, 3}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)

{a(n) = real((3 + quadgen(32))^n)}; /* Michael Somos, Apr 07 2003 */

{a(n) = subst( poltchebi( abs(n)), x, 3)}; /* Michael Somos, Apr 07 2003 */

{a(n) = if( n<0, a(-n), polsym(1 - 6*x + x^2, n) [n+1] / 2)}; /* Michael Somos, Apr 07 2003 */

{a(n) = polchebyshev( n, 1, 3)}; /* Michael Somos, Jul 11 2011 */

a(n)=([1,2,2;2,1,2;2,2,3]^n)[3,3] \\ Vim Wenders, Mar 28 2007

G.f.: (1-3*x)/(1-6*x+x^2). - Barry E. Williams and Wolfdieter Lang, May 05 2000
E.g.f.: exp(3*x)*cosh(2*sqrt(2)*x). Binomial transform of A084128. - Paul Barry, May 16 2003
From N. J. A. Sloane, May 16 2003: (Start)
a(n) = sqrt(8*((A001109(n))^2) + 1).
a(n) = T(n, 3), with Chebyshev's T-polynomials A053120. (End)
a(n) = ((3+2*sqrt(2))^n + (3-2*sqrt(2))^n)/2.
a(n) = cosh(2*n*arcsinh(1)). - Herbert Kociemba, Apr 24 2008
a(n) ~ (1/2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all elements x of the sequence, 2*x^2 - 2 is a square. Limit_{n -> infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 10 2002 [corrected by Peter Pein, Mar 09 2009]
a(n) = 3*A001109(n) - A001109(n-1), n >= 1. - Barry E. Williams and Wolfdieter Lang, May 05 2000
For n >= 1, a(n) = A001652(n) - A001652(n-1). - Charlie Marion, Jul 01 2003
From Paul Barry, Sep 18 2003: (Start)
a(n) = ((-1+sqrt(2))^n + (1+sqrt(2))^n + (1-sqrt(2))^n + (-1-sqrt(2))^n)/4 (with interpolated zeros).
E.g.f.: cosh(x)*cosh(sqrt(2)x) (with interpolated zeros). (End)
For n > 0, a(n)^2 + 1 = 2*A001653(n-1)*A001653(n). - Charlie Marion, Dec 21 2003
a(n)^2 + a(n+1)^2 = 2*(A001653(2*n+1) - A001652(2*n)). - Charlie Marion, Mar 17 2003
a(n) = Sum_{k >= 0} binomial(2*n, 2*k)*2^k = Sum_{k >= 0} A086645(n, k)*2^k. - Philippe Deléham, Feb 29 2004
a(n)*A002315(n+k) = A001652(2*n+k) + A001652(k) + 1; for k > 0, a(n+k)*A002315(n) = A001652(2*n+k) - A001652(k-1). - Charlie Marion, Mar 17 2003
For n > k, a(n)*A001653(k) = A011900(n+k) + A053141(n-k-1). For n <= k, a(n)*A001653(k) = A011900(n+k) + A053141(k-n). - Charlie Marion, Oct 18 2004
A053141(n+1) + A055997(n+1) = a(n+1) + A001109(n+1). - Creighton Dement, Sep 16 2004
a(n+1) - A001542(n+1) = A090390(n+1) - A046729(n) = A001653(n); a(n+1) - 4*A079291(n+1) = (-1)^(n+1). Formula generated by the floretion - .5'i + .5'j - .5i' + .5j' - 'ii' + 'jj' - 2'kk' + 'ij' + .5'ik' + 'ji' + .5'jk' + .5'ki' + .5'kj' + e. - Creighton Dement, Nov 16 2004
a(n) = sqrt( A055997(2*n) ). - Alexander Adamchuk, Nov 24 2006
a(2n) = A056771(n). a(2*n+1) = 3*A077420(n). - Alexander Adamchuk, Feb 01 2007
a(n) = (A000129(n)^2)*4 + (-1)^n. - Vim Wenders, Mar 28 2007
2*a(k)*A001653(n)*A001653(n+k) = A001653(n)^2 + A001653(n+k)^2 + A001542(k)^2. - Charlie Marion, Oct 12 2007
a(n) = A001333(2*n). - Ctibor O. Zizka, Aug 13 2008
A028982(a(n)-1) + 2 = A028982(a(n)+1). - Juri-Stepan Gerasimov, Mar 28 2011
a(n) = 2*A001108(n) + 1. - Paul Weisenhorn, Dec 17 2011
a(n) = sqrt(2*x^2 + 1) with x being A001542(n). - Zak Seidov, Jan 30 2013
a(2n) = 2*a(n)^2 - 1 = a(n)^2 + 2*A001542(n)^2. a(2*n+1) = 1 + 2*A002315(n)^2. - Steven J. Haker, Dec 04 2013
a(n) = 3*a(n-1) + 4*A001542(n-1); e.g., a(4) = 99 = 3*17 + 4*12. - Zak Seidov, Dec 19 2013
a(n) = cos(n * arccos(3)) = cosh(n * log(3 + 2*sqrt(2))). - Daniel Suteu, Jul 28 2016
From Ilya Gutkovskiy, Jul 28 2016: (Start)
Inverse binomial transform of A084130.
Exponential convolution of A000079 and A084058.
Sum_{n>=0} (-1)^n*a(n)/n! = cosh(2*sqrt(2))/exp(3) = 0.4226407909842764637... (End)
a(2*n+1) = 2*a(n)*a(n+1) - 3. - Timothy L. Tiffin, Oct 12 2016
a(n) = a(-n) for all n in Z. - Michael Somos, Jan 20 2017
a(2^n) = A001601(n+1). - A.H.M. Smeets, May 28 2017
a(A298210(n)) = A002350(2*n^2). - A.H.M. Smeets, Jan 25 2018
a(n) = S(n, 6) - 3*S(n-1, 6), for n >= 0, with S(n, 6) = A001109(n+1), (Chebyshev S of A049310). See the first comment and the formula a(n) = T(n, 3). - Wolfdieter Lang, Nov 22 2020
From Peter Bala, Dec 31 2021: (Start)
a(n) = [x^n] (3*x + sqrt(1 + 8*x^2))^n.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) hold for all prime p and positive integers n and k.
O.g.f. A(x) = 1 + x*d/dx(log(B(x))), where B(x) = 1/sqrt(1 - 6*x + x^2) is the o.g.f. of A001850. (End)
From Peter Bala, Aug 17 2022: (Start)
Sum_{n >= 1} 1/(a(n) - 2/a(n)) = 1/2.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 1/a(n)) = 1/4.
Sum_{n >= 1} 1/(a(n)^2 - 2) = 1/2 - 1/sqrt(8). (End)
From  Peter Bala, Jun 23 2025: (Start)
Product_{n >= 0} (1 + 1/a(2^n)) = sqrt(2).
Product_{n >= 0} (1 - 1/(2*a(2^n))) = (4/7)*sqrt(2). See A002812. (End)

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

Original entry on oeis.org

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

A220335 A modified Engel expansion for sqrt(3) - 1.

Original entry on oeis.org

2, 3, 4, 2, 8, 14, 2, 98, 194, 2, 18818, 37634, 2, 708158978, 1416317954, 2, 1002978273411373058, 2005956546822746114, 2, 2011930833870518011412817828051050498, 4023861667741036022825635656102100994

Offset: 1

Peter Bala, Dec 12 2012

This is the case p = 2 of a family of quadratic irrationals of the form (p - 1)*sqrt(p^2 - 1) - (p^2 - p - 1) whose modified Engel expansion, as defined below, has a predictable form. For other cases see A220336 (p = 3), A220337 (p = 4) and A220338 (p = 5).
The Engel expansion of a positive real number x in the half-open interval (0,1] is the unique nondecreasing sequence {e(1), e(2), e(3), ...} of positive integers such that x = 1/e(1) + 1/(e(1)*e(2)) + 1/(e(1)*e(2)*e(3)) + .... The terms in the Engel expansion of x are obtained from the iterates of the map g(x) = x*(1 + floor(1/x)) - 1 by means of the formula e(n) = 1 + floor(1/g^(n-1)(x)). Here g^(n)(x) = g(g^(n-1)(x)) denotes the n-th iterate of g(x) with the convention g^(0)(x) = x.
In a similar way, the modified Engel expansion of x belonging to (0,1] is a sequence {E(1), E(2), E(3), ...} of positive integers such that x = 1/E(1) + 1/(E(1)*E(2)) + 1/(E(1)*E(2)*E(3)) + ... whose terms are obtained from the iterates of the harmonic sawtooth map h(x) = floor(1/x)*g(x). The general formula is E(1) = 1 + floor(1/x) and for n >= 1, E(n) = floor(1/h^(n-2)(x))*(1 + floor(1/h^(n-1)(x))). For further details see the Bala link.

Peter Bala, A modified Engel expansion for certain quadratic irrationals
Wikipedia, Engel Expansion

Cf. A002812, A028257, A220336 (p = 3), A220337 (p = 4), A220338 (p = 5).

Let x = sqrt(3) - 1. Then a(1) = ceiling(1/x) and for n >= 2, a(n) = floor(1/h^(n-2)(x))*ceiling(1/h^(n-1)(x)), where h^(n)(x) denotes the n-th iterate of the harmonic sawtooth map h(x), with the convention h^(0)(x) = x.
a(3*n+2) = 1/2*{2 + (2 + sqrt(3))^(2^n) + (2 - sqrt(3))^(2^n)} and
a(3*n+3) = (2 + sqrt(3))^(2^n) + (2 - sqrt(3))^(2^n), both for n >= 0.
For n >= 0, a(3*n+1) = 2. For n >= 1, a(3*n+2) = 2*(A002812(n-1))^2 and a(3*n+3) = 4*(A002812(n-1))^2 - 2.
Recurrence equations:
For n >= 1, a(3*n+2) = 2*{a(3*n-1)^2 - 2*a(3*n-1) + 1} and
a(3*n+3) = 2*a(3*n+2) - 2.
Put P(n) = Product_{k=1..n} a(k). Then we have the infinite Egyptian fraction representation sqrt(3) - 1 = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*3) + 1/(2*3*4) + 1/(2*3*4*2) + ....

A081459 Consider the mapping f(r) = (1/2)*(r + N/r) from rationals to rationals where N = 5. Starting with r = 2 and applying the mapping to each new (reduced) rational number gives 2, 9/4, 161/72, 51841/23184, ..., tending to N^(1/2). Sequence gives values of the numerators.

Original entry on oeis.org

2, 9, 161, 51841, 5374978561, 57780789062419261441, 6677239169351578707225356193679818792961, 89171045849445921581733341920411050611581102638589828325078491812417901966295041

Offset: 1

Amarnath Murthy, Mar 22 2003

Related sequence pairs (numerator, denominator) can be obtained by choosing N = 2, 3, 6 etc.

The sequence satisfies the Pell equation a(n+1)^2 - 5*A081460(n+1)^2 = 1. - Vincenzo Librandi, Dec 20 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..11

Cf. A000032, A000129, A001333, A081460.

m:=8; f:=[ n eq 1 select 2 else (Self(n-1)+5/Self(n-1))/2: n in [1..m] ]; [ Numerator(f[n]): n in [1..m] ]; // Bruno Berselli, Dec 20 2011

k = 4; Table[Simplify[Expand[(1/2) (((k + Sqrt[k^2 + 4])/2)^(2^(n - 1)) + ((k - Sqrt[k^2 + 4])/2)^(2^(n - 1)))]], {n, 1, 6}] (* Artur Jasinski, Oct 12 2008 *)
aa = {}; k = 9; Do[AppendTo[aa, k]; k = 2 k^2 - 1, {n, 1, 5}]; aa (* Artur Jasinski, Oct 12 2008 *)

{r=2; N=5; for(n=1,8,a=numerator(r); b=denominator(r); print1(a,","); r=(1/2)*(r + N/r) )}

a(n) = a(n-1)^2 + 5*A081460(n-1)^2. - Mario Catalani (mario.catalani(AT)unito.it), May 21 2003
a(n) = (1/2)*(((4+2*sqrt(5))/2)^(2^(n-1)) + ((4-2*sqrt(5))/2)^(2^(n-1))). a(n+1) = 2*a(n)^2 - 1 for n > 1. - Artur Jasinski, Oct 12 2008
a(n) = A000032(3*2^(n-1))/2. - Ehren Metcalfe, Oct 05 2017
a(n) = A001077(2^(n-1)). - Robert FERREOL, Apr 16 2023
From  Peter Bala, Jun 22 2025: (Start)
Product_{n >= 1} (1 + 1/a(n)) = (3/4)*sqrt(5).
Product_{n >= 1} (1 - 1/(2*a(n))) = (6/19)*sqrt(5). See A002812. (End)

Edited and extended by Klaus Brockhaus and Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 06 2003

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

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)

A219510 Pierce expansion of 84 - 48*sqrt(3).

Original entry on oeis.org

1, 7, 32, 97, 392, 18817, 75272, 708158977, 2832635912, 1002978273411373057, 4011913093645492232, 2011930833870518011412817828051050497, 8047723335482072045651271312204201992

Offset: 0

Peter Bala, Nov 23 2012

Paradis et al. have determined the Pierce expansion of the quadratic irrationality 2*(p - 1)*(p - sqrt(p^2 - 1)), p a positive integer greater than or equal to 3. This is the case p = 7. For other cases see A219508 (p = 3), A219509 (p = 5) and A219511 (p = 9).

G. C. Greubel, Table of n, a(n) for n = 0..20
J. Paradis, P. Viader, and L. Bibiloni Approximation to quadratic irrationals and their Pierce expansions, The Fibonacci Quarterly, Vol.36 No. 2 (1998) 146-153.
T. A. Pierce, On an algorithm and its use in approximating roots of algebraic equations, Amer. Math. Monthly, Vol. 36 No. 10, (1929) p.523-525.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A002812, A219508 (p = 3), A219509 (p = 5), A219511 (p = 9).

PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[84 - 48*Sqrt[3] , 7!], 10] (* G. C. Greubel, Nov 15 2016 *)

r=(7 + 4*sqrt(3))/12; for(n=1, 10, print(floor(r), ", "); r=r/(r-floor(r))) \\ G. C. Greubel, Nov 15 2016

a(2*n) = 2*{(2 + sqrt(3))^(2^n) + (2 - sqrt(3))^(2^n) + 2} for n >= 1.
a(2*n-1) = 1/2*{(2 + sqrt(3))^(2^n) + (2 - sqrt(3))^(2^n)} for n >= 1.
Recurrence equations: a(0) = 1, a(1) = 7 and for n >= 1 a(2*n) = 4*(a(2*n-1) + 1) and a(2*n+1) = 2*(a(2*n-1))^2 - 1.
84 - 48*sqrt(3) = 1 - 1/7 + 1/(7*32) - 1/(7*32*97) + 1/(7*32*97*392) - ....
a(2*n) = 8*A002812(n-1)^2 for n >= 1.
a(2*n+1) = A002812(n+1) for n >= 0.

cp's OEIS Frontend

A071579 a(n) = 2*a(n-1)*A002812(n-1), starting a(0)=1.

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A001541 a(0) = 1, a(1) = 3; for n > 1, a(n) = 6*a(n-1) - a(n-2).

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Haskell

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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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A220335 A modified Engel expansion for sqrt(3) - 1.

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A081459 Consider the mapping f(r) = (1/2)*(r + N/r) from rationals to rationals where N = 5. Starting with r = 2 and applying the mapping to each new (reduced) rational number gives 2, 9/4, 161/72, 51841/23184, ..., tending to N^(1/2). Sequence gives values of the numerators.

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A071579 a(n) = 2a(n-1)A002812(n-1), starting a(0)=1.