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A051009
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Reduced denominators of Newton's iteration for sqrt(2).
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10
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OFFSET
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1,2
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COMMENTS
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For n>1, Egyptian fraction expansion of 2-sqrt(2), i.e., 2-sqrt(2) = 1/2 + 1/12 + 1/408 + 1/470832 + ... - Simon Plouffe, Feb 22 2011
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LINKS
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FORMULA
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a(n) = 2*a(n-1)*A001601(n-1). - Joe Keane (jgk(AT)jgk.org), May 31 2002
For n>1, a(n) = 2*a(n-1)*sqrt(2*a(n-1)^2+1). - Mario Catalani (mario.catalani(AT)unito.it), May 27 2003
For n>0: a(n) = Sum_{r=0..2^(n-1)-1} binomial(2^n, 2*r+1)*2^r. - Mario Catalani (mario.catalani(AT)unito.it), May 30 2003
a(n+1) = (1/(2*sqrt(2)))*((1 + sqrt(2))^(2^n) - (1 - sqrt(2))^(2^n)). - Artur Jasinski, Oct 10 2008
a(1)=1, a(2)=2, a(n) = 2 * a(n-1) * cos(2^(n-3) * arccos(3)). - Daniel Suteu, Dec 01 2016
0 = a(n)^2*(2*a(n+1) + a(n+2)) - a(n+1)^3 if n>0. - Michael Somos, Dec 01 2016
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EXAMPLE
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G.f. = x + 2*x^2 + 12*x^3 + 408*x^4 + 470832*x^5 + ...
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MATHEMATICA
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Table[Simplify[Expand[(1/(2 Sqrt[2])) ((1 + Sqrt[2])^(2^n) - (1 - Sqrt[2])^(2^n))]], {n, 0, 7}] (* Artur Jasinski, Oct 10 2008 *)
Do[Print[Fibonacci[2^n, 2]], {n, 0, 10}] (* Sergio Falcon, Dec 04 2008 *)
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CROSSREFS
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KEYWORD
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nonn,frac
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AUTHOR
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STATUS
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approved
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