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A003010 A Lucas-Lehmer sequence: a(0) = 4; for n>0, a(n) = a(n-1)^2 - 2.
(Formerly M3494)
37
4, 14, 194, 37634, 1416317954, 2005956546822746114, 4023861667741036022825635656102100994, 16191462721115671781777559070120513664958590125499158514329308740975788034 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
COMMENTS
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
REFERENCES
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.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437, alternative link.
Larry Ericksen, Primality Testing and Prime Constellations, Šiauliai Mathematical Seminar, Vol. 3 (11), 2008.
J. 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.
P. Liardet and P. Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
J. Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
P. Vellucci and A. M. Bersani, The class of Lucas-Lehmer polynomials, arXiv preprint arXiv:1603.01989 [math.CA], 2016.
Pierluigi Vellucci and A. M. 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
FORMULA
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)
MAPLE
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);
MATHEMATICA
seqLucasLehmer[0] = 4; seqLucasLehmer[n_] := seqLucasLehmer[n - 1]^2 - 2; Array[seqLucasLehmer, 8, 0] (* Robert G. Wilson v, Jun 28 2012 *)
PROG
(PARI)
a(n)=if(n, a(n-1)^2-2, 4)
vector(10, i, a(i-1)) \\ M. F. Hasler, Mar 09 2007
(Magma) [n le 1 select 4 else Self(n-1)^2-2: n in [1..10]]; // Vincenzo Librandi, Aug 24 2015
(Python)
from itertools import accumulate
def f(anm1, _): return anm1**2 - 2
print(list(accumulate([4]*8, f))) # Michael S. Branicky, Apr 14 2021
CROSSREFS
Cf. A001566 (starting with 3), A003423 (starting with 6), A003487 (starting with 5).
Sequence in context: A129224 A129225 A129226 * A118770 A226943 A292708
KEYWORD
nonn
AUTHOR
EXTENSIONS
One more term from Thomas A. Rockwell (LlewkcoRAT(AT)aol.com), Jan 18 2005
STATUS
approved

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Last modified March 18 22:56 EDT 2024. Contains 370952 sequences. (Running on oeis4.)