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Search: id:A003010
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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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+0 21
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| 4, 14, 194, 37634, 1416317954, 2005956546822746114, 4023861667741036022825635656102100994, 16191462721115671781777559070120513664958590125499158514329308740975788034
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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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 (qntmpkt(AT)yahoo.com), Jun 07 2003
These are the terms in A003500 whose indices are powers of 2. - John Blythe Dobson (j.dobson(AT)uwinnipeg.ca), Oct 28 2007
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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.
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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.
Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
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 (benoit7848c(AT)orange.fr), 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-1))+(2-sqrt(3))^(2^(n-1)) - 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 (pauldhanna(AT)juno.com), 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 (Maximilian.Hasler(AT)gmail.com), Mar 09 2007
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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 (Maximilian.Hasler(AT)gmail.com), Mar 09 2007
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PROGRAM
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(PARI:) a(n)=if(n, a(n-1)^2-2, 4) vector(10, i, a(i-1)) - M. F. Hasler (Maximilian.Hasler(AT)gmail.com), 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.
Sequence in context: A129224 A129225 A129226 this_sequence A118770 A112514 A001140
Adjacent sequences: A003007 A003008 A003009 this_sequence 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 (njas(AT)research.att.com).
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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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