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A001566
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a(0) = 3; thereafter, a(n) = a(n-1)^2 - 2.
(Formerly M2705 N1084)
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31
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OFFSET
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0,1
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COMMENTS
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Expansion of 1/phi: 1/phi = (1-1/3)*(1-1/((3-1)*7))*(1-1/(((3-1)*7-1)*47))*(1-1/((((3-1)*7-1)*47-1)*2207))... (phi being the golden ration (1+sqrt(5))/2) - Thomas Baruchel (baruchel(AT)users.sourceforge.net), Nov 06 2003
An infinite coprime sequence defined by recursion. - Michael Somos Mar 14 2004
Starting with 7, the terms end with 7,47,07,47,07,..., of the form 8a+7 where a = 0,1,55,121771,... Conjecture: Every a is squarefree, every other a is divisible by 55, the a's are a subset of A046194, the heptagonal triangular numbers (the first,2nd,3rd,6th,11th,?... terms) . - Gerald McGarvey, Aug 08 2004
Also the reduced numerator of the convergents to sqrt(5) using Newton's recursion x = (5/x+x)/2. [Cino Hilliard (hillcino368(AT)hotmail.com), Sep 28 2008]
The subsequence of primes begins a(n) for n = 0, 1, 2, 3. [Jonathan Vos Post, Feb 26, 2011].
We have sum{n=0,..,N} a(n)^2 = 2*(N+1) + sum{n=1,..,N+1} a(n), sum{n=0,..,N} a(n)^4 = 5*(sum{n=1,..,N+1} a(n)) + a(N+1)^2 + 6*N -3, etc. which is very interesting with respect the fact that a(n) = Lucas(2^n) - see W. Webb's problem in Witula-Slota's paper. - Roman Witula, Nov 02 2012
From Peter Bala, Nov 11 2012: (Start)
The present sequence corresponds to the case x = 3 of the following general remarks.
The recurrence a(n+1) = a(n)^2 - 2 with initial condition a(0) = x > 2 has the solution a(n) = ((x + sqrt(x^2 - 4))/2)^(2^n) + ((x - sqrt(x^2 - 4))/2)^(2^n).
We have the product expansion sqrt(x + 2)/sqrt(x - 2) = product {n = 0..inf} (1 + 2/a(n)) (essentially due to Euler - see Mendes-France and van der Poorten). Another expansion is sqrt(x^2 - 4)/(x + 1) = product {n = 0..inf} (1 - 1/a(n)), which follows by iterating the identity sqrt(x^2 - 4)/(x + 1) = (1 - 1/x)*sqrt(y^2 - 4)/(y + 1), where y = x^2 - 2.
The sequence b(n) := a(n) - 1 satisfies b(n+1) = b(n)^2 + 2*b(n) - 2. Cases currently in the database are A145502 through A145510. The sequence c(n) := a(n)/2 satisfies c(n+1) = 2*c(n)^2 - 1. Cases currently in the database are A002812, A001601, A005828, A084764 and A084765.
(End)
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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. 397.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 223.
E. Lucas, Nouveaux theoremes d'arithmetique superieure, Comptes Rend., 83 (1876), 1286-1288.
M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. Witula, D. Slota, delta-Fibonacci Numbers, Appl. Anal. Discrete Math., 3 (2009), 310-329.
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 0..12
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.
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) = Fibonacci(2^(n+2))/Fibonacci(2^(n+1)) = A058635(n+2)/A058635(n+1). - Len Smiley, May 08 2000, and Artur Jasinski, Oct 05 2008
a(n) = ceiling(c^(2^n)) where c = (3+sqrt(5))/2 = tau^2 is the largest root of x^2-3*x+1=0. - Benoit Cloitre, Dec 03, 2002
a(n) = round(G^(2^n)) where G is the golden ratio (A001622). - Artur Jasinski, Sep 22 2008
a(n) = (G^(2^(n+1))-(1-G)^(2^(n+1)))/((G^(2^n))-(1-G)^(2^n)) = G^(2^n)+(1-G)^(2^n) = G^(2^n)+(-G)^(-2^n) where G is the golden ratio. - Artur Jasinski, Oct 05 2008
a(n) = 2*cosh(2^n*arccosh(sqrt(5)/2). - Artur Jasinski, Oct 09 2008
a(n) = Fibonacci(2^(n+1)-1)+Fibonacci(2^(n+1)+1). (3-sqrt(5))/2 = 1/3 + 1/(3*7) + 1/(3*7*47) + 1/(3*7*47*2207) + ... (E. Lucas) - Philippe DELEHAM, Apr 21 2009
a(n)*(a(n+1)-1)/2 = A023039(2^n). - M. F. Hasler, Sep 27 2009
For n>=1, a(n) = 2+prod{i=0..n-1}(a(i)+2). - Vladimir Shevelev, Nov 28 2010]
a(n) = 2*T(2^n,3/2) where T(n,x) is the Chebyshev polynomial of the first kind. - Leonid Bedratyuk, Mar 17 2011
Engel expansion of 1/2*(3 - sqrt(5)). Thus 1/2*(3 - sqrt(5)) = 1/3 + 1/(3*7) + 1/(3*7*47) + ... as noted above by Deleham. See Liardet and Stambul. - Peter Bala, Oct 31 2012
sqrt(5)/4 = product {n = 0..inf} (1 - 1/a(n)).
sqrt(5) = product {n = 0..inf} (1 + 2/a(n)).
a(n) - 1 = A145502(n+1). - Peter Bala, Nov 11 2012
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EXAMPLE
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Contribution from Cino Hilliard (hillcino368(AT)hotmail.com), Sep 28 2008: (Start)
Init x=1.
x = (5/1+1)/2 = 3/1
x = (5/3+3)/2 = 7/3
x = (5/7/3+7/3)/2 = 47/21
x = (5/47/21+47/21)/2 = 2207/987
(2207/987)^2 = 5.000004106... (End)
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MAPLE
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a:= n-> simplify(2*ChebyshevT(2^n, 3/2), 'ChebyshevT'):
seq (a(n), n=0..8);
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MATHEMATICA
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c = N[GoldenRatio, 1000]; Table[Round[c^(2^n)], {n, 1, 10}] (* Artur Jasinski, Sep 22 2008 *)
c = (1 + Sqrt[5])/2; Table[Expand[c^(2^n) + (-c + 1)^(2^n)], {n, 1, 8}] (* Artur Jasinski, Oct 05 2008 *)
G = (1 + Sqrt[5])/2; Table[Expand[(G^(2^(n + 1)) - (1 - G)^(2^(n + 1)))/Sqrt[5]]/Expand[((G^(2^n)) - (1 - G)^(2^n))/Sqrt[5]], {n, 1, 10}] (* Artur Jasinski, Oct 05 2008 *)
Table[2*Cosh[2^n*ArcCosh[Sqrt[5]/2], {n, 1, 30}] (* Artur Jasinski, Oct 09 2008 *)
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PROG
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(PARI) a(n)=if(n<1, 3*(n==0), a(n-1)^2-2)
Contribution from Cino Hilliard (hillcino368(AT)hotmail.com), Sep 28 2008: (Start)
(PARI) g(n, p) = x=1; for(j=1, p, x=(n/x+x)/2; print1(numerator(x)", "))
g(5, 8) (End)
(Maxima)
a[0]:3$
a[n]:=a[n-1]^2-2$
A001566(n):=a[n]$
makelist(A001566(n), n, 0, 7); /* Martin Ettl, Nov 12 2012 */
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CROSSREFS
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Lucas numbers (A000032) with subscripts that are powers of 2 greater than 1 (Herb Wilf). Cf. A000045.
Cf. A003010 (starting with 4), A003423 (starting with 6), A003487 (starting with 5).
Cf. A058635. - Artur Jasinski, Oct 05 2008
Cf. A002812, A145502, A145274, A050614, A088334, A181393, A181419, A186750-A186751.
Sequence in context: A052381 A219877 A031440 * A173771 A019039 A077559
Adjacent sequences: A001563 A001564 A001565 * A001567 A001568 A001569
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KEYWORD
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easy,nonn,nice
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AUTHOR
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N. J. A. Sloane.
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STATUS
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approved
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