A003010 -id:A003010 - cp's OEIS frontend

A003500 a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.

2, 4, 14, 52, 194, 724, 2702, 10084, 37634, 140452, 524174, 1956244, 7300802, 27246964, 101687054, 379501252, 1416317954, 5285770564, 19726764302, 73621286644, 274758382274, 1025412242452, 3826890587534, 14282150107684, 53301709843202, 198924689265124

Offset: 0

N. J. A. Sloane

a(n) gives values of x satisfying x^2 - 3*y^2 = 4; corresponding y values are given by 2*A001353(n).
If M is any given term of the sequence, then the next one is 2*M + sqrt(3*M^2 - 12). - Lekraj Beedassy, Feb 18 2002
For n > 0, the three numbers a(n) - 1, a(n), and a(n) + 1 form a Fleenor-Heronian triangle, i.e., a Heronian triangle with consecutive sides, whose area A(n) may be obtained from the relation [4*A(n)]^2 = 3([a(2n)]^2 - 4); or A(n) = 3*A001353(2*n)/2 and whose semiperimeter is 3*a[n]/2. The sequence is symmetrical about a[0], i.e., a[-n] = a[n].
For n > 0, a(n) + 2 is the number of dimer tilings of a 2*n X 2 Klein bottle (cf. A103999).
Tsumura shows that, for prime p, a(p) is composite (contrary to a conjecture of Juricevic). - Charles R Greathouse IV, Apr 13 2010
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 12 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 192 = 0. - Colin Barker, Feb 16 2014
A268281(n) - 1 is a member of this sequence iff A268281(n) is prime. - Frank M Jackson, Feb 27 2016
a(n) gives values of x satisfying 3*x^2 - 4*y^2 = 12; corresponding y values are given by A005320. - Sture Sjöstedt, Dec 19 2017
Middle side lengths of almost-equilateral Heronian triangles. - Wesley Ivan Hurt, May 20 2020
For all elements k of the sequence, 3*(k-2)*(k+2) is a square. - Davide Rotondo, Oct 25 2020

B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 82.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p.91.
Michael P. Cohen, Generating Heronian Triangles With Consecutive Integer Sides. Journal of Recreational Mathematics, vol. 30 no. 2 1999-2000 p. 123.
L. E. Dickson, History of The Theory of Numbers, Vol. 2 pp. 197;198;200;201. Chelsea NY.
Charles R. Fleenor, Heronian Triangles with Consecutive Integer Sides, Journal of Recreational Mathematics, Volume 28, no. 2 (1996-7) 113-115.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.
V. D. To, "Finding All Fleenor-Heronian Triangles", Journal of Recreational Mathematics vol. 32 no.4 2003-4 pp. 298-301 Baywood NY.

T. D. Noe, Table of n, a(n) for n=0..200
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
R. A. Beauregard and E. R. Suryanarayan, The Brahmagupta Triangles, The College Mathematics Journal 29(1) 13-7 1998 MAA.
Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Daniel Birmajer, Juan B. Gil and Michael D. Weiner, Linear recurrence sequences with indices in arithmetic progression and their sums, arXiv preprint arXiv:1505.06339 [math.NT], 2015.
K. S. Brown's Mathpages, Some Properties of the Lucas Sequence(2, 4, 14, 52, 194, ...)
H. W. Gould, A triangle with integral sides and area, Fib. Quart., 11 (1973), 27-39.
Tanya Khovanova, Recursive Sequences
E. Keith Lloyd, The Standard Deviation of 1, 2, ..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
Hideyuki Ohtskua, proposer, Problem B-1351, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 3 (2024), p. 258.
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
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Yu Tsumura, On compositeness of special types of integers, arXiv:1004.1244 [math.NT], 2010.
Eric Weisstein's World of Mathematics, Heronian Triangle
Wikipedia, Heronian triangle
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Cf. A001075, A001353, A001835.
Cf. A011945 (areas), A334277 (perimeters).
Cf. this sequence (middle side lengths), A016064 (smallest side lengths), A335025 (largest side lengths).
Cf. A001570, A002530, A005320, A006051, A048788, A174500, A268281.
Cf. A011943, A102344, A019673, A105531.

a003500 n = a003500_list !! n
a003500_list = 2 : 4 : zipWith (-)
   (map (* 4) $ tail a003500_list) a003500_list
-- Reinhard Zumkeller, Dec 17 2011

I:=[2,4]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2018

A003500 := proc(n) option remember; if n <= 1 then 2*n+2 else 4*procname(n-1)-procname(n-2); fi;
end proc;

a[0]=2; a[1]=4; a[n_]:= a[n]= 4a[n-1] -a[n-2]; Table[a[n], {n, 0, 23}]
LinearRecurrence[{4,-1},{2,4},30] (* Harvey P. Dale, Aug 20 2011 *)
Table[Round@LucasL[2n, Sqrt[2]], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)

x='x+O('x^99); Vec(-2*(-1+2*x)/(1-4*x+x^2)) \\ Altug Alkan, Apr 04 2016

[lucas_number2(n,4,1) for n in range(0, 24)] # Zerinvary Lajos, May 14 2009

a(n) = ( 2 + sqrt(3) )^n + ( 2 - sqrt(3) )^n.
a(n) = 2*A001075(n).
G.f.: 2*(1 - 2*x)/(1 - 4*x + x^2). Simon Plouffe in his 1992 dissertation.
a(n) = A001835(n) + A001835(n+1).
a(n) = trace of n-th power of the 2 X 2 matrix [1 2 / 1 3]. - Gary W. Adamson, Jun 30 2003 [corrected by Joerg Arndt, Jun 18 2020]
From the addition formula, a(n+m) = a(n)*a(m) - a(m-n), it is easy to derive multiplication formulas, such as: a(2*n) = (a(n))^2 - 2, a(3*n) = (a(n))^3 - 3*(a(n)), a(4*n) = (a(n))^4 - 4*(a(n))^2 + 2, a(5*n) = (a(n))^5 - 5*(a(n))^3 + 5*(a(n)), a(6*n) = (a(n))^6 - 6*(a(n))^4 + 9*(a(n))^2 - 2, etc. The absolute values of the coefficients in the expansions are given by the triangle A034807. - John Blythe Dobson, Nov 04 2007
a(n) = 2*A001353(n+1) - 4*A001353(n). - R. J. Mathar, Nov 16 2007
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n=0..infinity} (1 + x^(4*n + 1))/(1 + x^(4*n + 3)). Let alpha = 2 - sqrt(3). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.24561 99455 06551 88869 ... = 2 + 1/(4 + 1/(14 + 1/(52 + ...))). Cf. A174500.
Also F(-alpha) = 0.74544 81786 39692 68884 ... has the continued fraction representation 1 - 1/(4 - 1/(14 - 1/(52 - ...))) and the simple continued fraction expansion 1/(1 + 1/((4 - 2) + 1/(1 + 1/((14 - 2) + 1/(1 + 1/((52 - 2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((4^2 - 4) + 1/(1 + 1/((14^2 - 4) + 1/(1 + 1/((52^2 - 4) + 1/(1 + ...))))))).
(End)
a(2^n) = A003010(n). - John Blythe Dobson, Mar 10 2014
a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 12*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
E.g.f.: 2*exp(2*x)*cosh(sqrt(3)*x). - Ilya Gutkovskiy, Apr 27 2016
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*n*(n - k - 1)!/(k!*(n - 2*k)!)*4^(n - 2*k) for n >= 1. - Peter Luschny, May 10 2016
From Peter Bala, Oct 15 2019: (Start)
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; -1, 4].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
2*Sum_{n >= 1} 1/( a(n) - 6/a(n) ) = 1.
6*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 2/a(n) ) = 1.
8*Sum_{n >= 1} 1/( a(n) + 24/(a(n) - 12/(a(n))) ) = 1.
8*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 8/(a(n) + 4/(a(n))) ) = 1.
Series acceleration formulas for sums of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/2 - 6*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 6)),
Sum_{n >= 1} 1/a(n) = 1/8 + 24*Sum_{n >= 1} 1/(a(n)*(a(n)^2 + 12)),
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/6 + 2*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 2)) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/8 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 12)).
Sum_{n >= 1} 1/a(n) = ( theta_3(2-sqrt(3))^2 - 1 )/4 = 0.34770 07561 66992 06261 .... See Borwein and Borwein, Proposition 3.5 (i), p.91.
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - theta_3(sqrt(3)-2)^2 )/4. Cf. A003499 and A153415. (End)
a(n) = tan(Pi/12)^n + tan(5*Pi/12)^n. - Greg Dresden, Oct 01 2020
From Wolfdieter Lang, Sep 06 2021: (Start)
a(n) = S(n, 4) - S(n-2, 4) = 2*T(n, 2), for n >= 0, with S and T Chebyshev polynomials, with S(-1, x) = 0 and S(-2, x) = -1. S(n, 4) = A001353(n+1), for n >= -1, and T(n, 2) = A001075(n).
a(2*k) = A067902(k), a(2*k+1) = 4*A001570(k+1), for k >= 0. (End)
a(n) = sqrt(2 + 2*A011943(n+1)) = sqrt(2 + 2*A102344(n+1)), n>0. - Ralf Steiner, Sep 23 2021
Sum_{n>=1} arctan(3/a(n)^2) = Pi/6 - arctan(1/3) = A019673 - A105531 (Ohtskua, 2024). - Amiram Eldar, Aug 29 2024

More terms from James Sellers, May 03 2000

Additional comments from Lekraj Beedassy, Feb 14 2002

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

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

Original entry on oeis.org

2, 7, 97, 18817, 708158977, 1002978273411373057, 2011930833870518011412817828051050497, 8095731360557835890888779535060256832479295062749579257164654370487894017

Offset: 0

N. J. A. Sloane

An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
2^p-1 is prime iff it divides a(p-2), since a(n) = A003010(n)/2, where A003010 is the Lucas-Lehmer sequence used for Mersenne number primality testing. - M. F. Hasler, Mar 09 2007
From Cino Hilliard, Sep 28 2008: (Start)
Also numerators of the convergents to the square root of 3 using the following recursion for initial x = 1: x1=x, x=3/x, x=(x+x1)/2.
This recursion was derived by experimenting with polynomial recursions of the form x = -a(0)/(a(n-1)x^(n-1) + ... + a(1)) in an effort to find a root for the polynomial a(n)x^n + a(n-1)x^(n-1) + ... + a(0). The process was hit-and-miss until I introduced the averaging step described above. This method is equivalent to Newton's Method although derived somewhat differently. (End)
The sequence satisfies the Pell equation a(n)^2 - 3*A071579(n)^2 = 1. - Vincenzo Librandi, Dec 19 2011
From Peter Bala, Nov 2012: (Start)
The present sequence corresponds to the case x = 2 of the following general remarks. Let x > 1 and let alpha := {x + sqrt(x^2 - 1)}. Define a sequence a(n) (which depends on x) by setting a(n) = (1/2)*(alpha^(2^n) + (1/alpha)^(2^n)) for n >= 0. It is easy to verify that a(n) is a solution to the recurrence equation a(n+1) = 2*a(n)^2 - 1 with the initial condition a(0) = x.
The following algebraic identity is valid for x > 1:
sqrt(4*x^2 - 4)/(2*x + 1) = (1 - 1/(2*x))*sqrt(4*y^2 - 4)/(2*y + 1), where y = 2*x^2 - 1. Iterating the identity yields the product expansion sqrt(4*x^2 - 4)/(2*x + 1) = Product_{n >= 0} (1 - 1/(2*a(n))). A second expansion is Product_{n >= 0} (1 + 1/a(n)) = sqrt((x + 1)/(x - 1)). See Mendes-France and van der Poorten. (End)

			G.f. = 2 + 7*x + 97*x^2 + 18817*x^3 + 708158977*x^4 + ...

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.
E. Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
W. Sierpiński, Sur les développements systématiques des nombres en produits infinis, in Œuvres choisies, tome 1, PWN Editions scientifiques de Pologne, 1974.
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..10
Georg Cantor, Zwei Sätze über eine gewisse Zerlegung der Zahlen in unendliche Producte, Zeitschrift für Mathematik und Physik. Band 14, 1869, S. 152-158. (See page 155, II, error in the fourth term.)
E. Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy)
M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.
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, Newton's Iteration
Index entries for sequences related to Engel expansions

Cf. A001075, A002531, A003010, A071579.

Cf. A177879 (lpf).

a:=[2];; for n in [2..10] do a[n]:=2*a[n-1]^2-1; od; a; # Muniru A Asiru, Aug 12 2018

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

a:=n->((2+sqrt(3))^(2^n)+(2-sqrt(3))^(2^n))/2: seq(floor(a(n)),n=0..10); # Muniru A Asiru, Aug 12 2018

Table[((2 + Sqrt[3])^2^n + (2 - Sqrt[3])^2^n)/2, {n, 0, 7}] (* Bruno Berselli, Dec 20 2011 *)
NestList[2#^2-1&,2,10] (* Harvey P. Dale, May 04 2013 *)
a[ n_] := If[ n < 0, 0, ChebyshevT[2^n, 2]]; (* Michael Somos, Dec 06 2016 *)

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

/* Roots by recursion. Find first root of ax^2 + b^x + c */
rroot2(a,b,c,p) = { local(x=1,x1=1,j); for(j=1,p, x1=x; x=-c/(a*x+b); x=(x1+x)/2; /* Let x be the average of the last 2 values */ print1(numerator(x)","); ); } \\ Cino Hilliard, Sep 28 2008

a(n) = A001075(2^n).
a(n) = ((2+sqrt(3))^(2^n) + (2-sqrt(3))^(2^n))/2. - Bruno Berselli, Dec 20 2011
From Peter Bala, Nov 11 2012: (Start)
2*sqrt(3)/5 = Product_{n >= 0} (1 - 1/(2*a(n))).
sqrt(3) = Product_{n >= 0} (1 + 1/a(n)).
a(n) = (1/2)*A003010(n). (End)
a(n) = cos(2^n*arccos(2)). - Peter Luschny, Oct 12 2022
a(n) = A002531(2^(n+1)). - Robert FERREOL, Apr 13 2023

A095847 Lucas-Lehmer residues for Mersenne numbers with prime indices.

Original entry on oeis.org

1, 0, 0, 0, 1736, 0, 0, 0, 6107895, 458738443, 0, 117093979072, 856605019673, 5774401272921, 96699253829728, 5810550306096509, 450529175803834166, 0, 44350645312365507266, 271761692158955752596, 2941647823169311845731

Offset: 1

Eric W. Weisstein, Jun 08 2004

nonn

If a(n) = 0, then 2^prime(n) - 1 is a prime greater than 3. - Alonso del Arte, May 09 2014

For n > 1, 2^prime(n) - 1 is prime if and only if a(n) = 0. - Thomas Ordowski, Aug 12 2018

			The first term is 1 since 4 mod 3 = 1. - Zvi Mendlowitz (zvi113(AT)zahav.net.il), May 10 2006

Dennis Martin, Table of n, a(n) for n = 1..100
Eric Weisstein's World of Mathematics, Lucas-Lehmer Test

Cf. A003010.

(* First run the program for A003010 to define seqLucasLehmer *) Table[Mod[seqLucasLehmer[Prime[n] - 2], 2^Prime[n] - 1], {n, 20}] (* Alonso del Arte, May 09 2014 *)

First, s(0) = 4, s(i) = s(i - 1)^2 - 2. Then, a(n) = s(prime(n) - 2) mod 2^prime(n) - 1. - Alonso del Arte, May 09 2014

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)

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

cp's OEIS Frontend

A003500 a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.

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

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

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A095847 Lucas-Lehmer residues for Mersenne numbers with prime indices.

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

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

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