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A003423
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a(n) = a(n-1)^2 - 2.
(Formerly M4215)
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9
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OFFSET
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0,1
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COMMENTS
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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
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REFERENCES
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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.
E. Lucas, "Théorie des Fonctions Numériques Simplement Périodiques, II", Amer. J. Math., 1 (1878), 289-321.
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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FORMULA
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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
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*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
4*sqrt(2)/7 = Product_{n >= 0} (1 - 1/a(n))
sqrt(2) = Product_{n >= 0} (1 + 2/a(n)).
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)
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MAPLE
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a:= n-> simplify(2*ChebyshevT(2^n, 3), 'ChebyshevT'):
seq(a(n), n=0..7);
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MATHEMATICA
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Table[Round[(1 + Sqrt[2])^(2^n)], {n, 1, 7}] (* Artur Jasinski, Sep 25 2008 *)
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PROG
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(PARI) a(n)=if(n<1, 6*(n==0), a(n-1)^2-2)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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