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A003010
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A Lucas-Lehmer sequence: a(0) = 4; for n>0, a(n) = a(n-1)^2 - 2.
(Formerly M3494)
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30
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4, 14, 194, 37634, 1416317954, 2005956546822746114, 4023861667741036022825635656102100994, 16191462721115671781777559070120513664958590125499158514329308740975788034
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OFFSET
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0,1
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COMMENTS
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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
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REFERENCES
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Albert Beiler, Recreations in the Theory of Numbers, page 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.
J. S. Hall, A remark on the primeness of Mersenne numbers, J. London Math. Soc. 28, (1953). 285-287.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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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, Fib. Quart., 11 (1973), 429-437.
P. Liardet and P. Stambul, Series d'Engel et fractions continuees, Jour. de Theorie des Nombres de Bordeaux 12 (2000), 37-68.
Eric Weisstein's World of Mathematics, Lucas-Lehmer Test.
Wikipedia, Engel Expansion
Index entries for sequences of form a(n+1)=a(n)^2 + ...
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FORMULA
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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/( prod_{k=0..n} a(k) ) = 2-sqrt(3). - Paul D. Hanna, Aug 11 2004
Comment from Ulrich Sondermann, Sep 04 2006: 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
a(n) = 2*cosh( 2^(n-1)*log(sqrt(3)+2) ) For n>0, a(n) = 2 + 3 * 4^n * product( a(k)/2, k=0..n-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 (j.dobson(AT)uwinnipeg.ca), 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
2*sqrt(3)/5 = product {n = 0..inf} (1 - 1/a(n)). sqrt(3) = product {n = 0..inf} (1 + 2/a(n)). a(n) - 1 = A145503(n+1). a(n) = 2*A002812(n). - Peter Bala, Nov 11 2012
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MAPLE
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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);
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MATHEMATICA
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a[0] = 4; a[n_] := a[n - 1]^2 - 2; Array[a, 8, 0] (* Robert G. Wilson v, Jun 28 2012 *)
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PROG
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(PARI:) a(n)=if(n, a(n-1)^2-2, 4) vector(10, i, a(i-1)) - M. F. Hasler, Mar 09 2007
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CROSSREFS
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Cf. A001566 (starting with 3), A003423 (starting with 6), A003487 (starting with 5).
Cf. A002812. A145503 .
Sequence in context: A129224 A129225 A129226 * A118770 A112514 A001140
Adjacent sequences: A003007 A003008 A003009 * A003011 A003012 A003013
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane.
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EXTENSIONS
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One more term from Thomas A. Rockwell (LlewkcoRAT(AT)aol.com), Jan 18 2005
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STATUS
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approved
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