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A003423 a(n) = a(n-1)^2 - 2.
(Formerly M4215)
+0
5
6, 34, 1154, 1331714, 1773462177794, 3145168096065837266706434, 9892082352510403757550172975146702122837936996354 (list; graph; listen)
OFFSET

0,1

REFERENCES

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

E. Lucas, "Th\'eorie des Fonctions Num\'eriques Simplement P\'eriodiques, II", Amer. J. Math., 1 (1878), 289-321.

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.

J. O. Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.

FORMULA

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/( prod_{k=0..n} a(k) ) = 3-sqrt(8). - Paul D. Hanna (pauldhanna(AT)juno.com), Aug 11 2004

a(n)=2*A001601(n+1).

a(n-1)=Round[(1 + Sqrt[2])^(2^n)] [From Artur Jasinski (grafix(AT)csl.pl), Sep 25 2008]

MATHEMATICA

a[1] := 6; a[n_] := a[n - 1]^2 - 2; Table[a[n], {n, 1, 8}] - Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Apr 11 2006

Table[Round[(1 + Sqrt[2])^(2^n)], {n, 1, 7}] [From Artur Jasinski (grafix(AT)csl.pl), Sep 25 2008]

PROGRAM

(PARI) a(n)=if(n<1, 6*(n==0), a(n-1)^2-2)

CROSSREFS

Cf. A001566 (starting with 3), A003010 (starting with 4), A003487 (starting with 5)

Sequence in context: A062819 A092336 A161323 this_sequence A145000 A046025 A009583

Adjacent sequences: A003420 A003421 A003422 this_sequence A003424 A003425 A003426

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane (njas(AT)research.att.com).

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Last modified November 27 22:38 EST 2009. Contains 167602 sequences.


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