A145503 -id:A145503 - cp's OEIS frontend

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

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

A145502 a(n+1) = a(n)^2+2*a(n)-2 and a(1)=2.

Original entry on oeis.org

2, 6, 46, 2206, 4870846, 23725150497406, 562882766124611619513723646, 316837008400094222150776738483768236006420971486980606

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))).
From Peter Bala, Nov 12 2012: (Start)
The present sequence corresponds to the case x = 3 of the following general remarks. Sequences A145503 through A145510 correspond to the cases x = 4 through x = 11 respectively.
Let x > 2 and let alpha := {x + sqrt(x^2 - 4)}/2. Define a sequence a(n) (which depends on x) by setting a(n) = alpha^(2^(n-1)) + (1/alpha)^(2^(n-1)) - 1. Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^2 + 2*a(n) - 2 with the initial condition a(1) = x - 1.
A second recurrence is a(n) = (a(1) + 2)*{Product_{k = 1..n-1} a(k)} - 2.
The following algebraic identity is valid for x > 2:
(x + 1)/sqrt(x^2 - 4) = (1 + 1/(x - 1))*(y + 1)/sqrt(y^2 - 4), where y - 1 = (x - 1)^2 + 2*(x - 1) - 2. Iterating the identity yields the product expansion (x + 1)/sqrt(x^2 - 4) = Product_{n >= 1} (1 + 1/a(n)).
A second expansion is Product_{n >= 1} (1 + 2/(a(n) + 1)) = sqrt((x + 2)/(x - 2)). For an alternative approach to these identities see the Bala link.
(End)

Peter Bala, Notes on A145502-A145510.
Daniel Duverney, Irrationality of Fast Converging Series of Rational Numbers, Journal of Mathematical Sciences-University of Tokyo, Vol. 8, No. 2 (2001), pp. 275-316.
Daniel Duverney and Takeshi Kurosawa, Transcendence of infinite products involving Fibonacci and Lucas numbers, Research in Number Theory, Vol. 8 (2002), Article 68.

Cf. A145503, A145504, A145505, A145506, A145507, A145508, A145509, A145510.

Cf. A000324, A001566.

aa = {}; k = 2; Do[AppendTo[aa, k]; k = k^2 + 2 k - 2, {n, 1, 10}]; aa
(* or *)
k = 1; Table[Floor[((k + Sqrt[k^2 + 4])/2)^(2^(n - 1))], {n, 2, 7}]
NestList[#^2+2#-2&,2,10] (* Harvey P. Dale, Dec 14 2021 *)

From Peter Bala, Nov 12 2012: (Start)
a(n) = phi^(2^n) + (1/phi)^(2^n) - 1, where phi := (1 + sqrt(5))/2 is the golden ratio.
a(n) = A001566(n-1) - 1.
Recurrence: a(n) = 4*(Product_{k = 1..n-1} a(k)) - 2 with a(1) = 2.
Product_{n >= 1} (1 + 1/a(n)) = 4/sqrt(5).
Product_{n >= 1} (1 + 2/(a(n) + 1)) = sqrt(5). (End)
From Amiram Eldar, Sep 10 2022: (Start)
a(n) = A000324(n) - 3.
Sum_{n>=1} (-2)^n/a(n) = -1/2 (Duverney, 2001). (End)
Product_{n>=1} (1 + 3/a(n)) = 4 (Duverney and Kurosawa, 2022). - Amiram Eldar, Jan 07 2023

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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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A145502 a(n+1) = a(n)^2+2*a(n)-2 and a(1)=2.

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A145510 a(n+1) = a(n)^2 + 2*a(n) - 2 and a(1)=10.

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