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A071579
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a(n) = 2*a(n-1)*A002812(n-1), starting a(0)=1.
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8
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OFFSET
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0,2
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COMMENTS
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Also the denominators of the convergents to sqrt(3) using Newton's recursion x = (3/x+x)/2. - Cino Hilliard, Sep 28 2008
For n>1, egyptian fraction of 2-sqrt(3): 2-sqrt(3) = 1/4 + 1/56 + 1/10864 + 1/408855776 + ... - Simon Plouffe, Feb 20 2011
Apart from giving the numerators in the Engel series representation of 2 - sqrt(3), as stated above by Plouffe, this sequence is also a Pierce expansion of the real number x = 2 - sqrt(3) to the base b := 1/sqrt(12) (see A058635 for a definition of this term).
The associated series representation begins 2 - sqrt(3) = b/1 - b^2/(1*4) + b^3/(1*4*56) - b^4/(1*4*56*10864) + .... Cf. A230338.
More generally, for n >= 0, the sequence [a(n), a(n+1), a(n+2), ...] gives a Pierce expansion of (2 - sqrt(3))^(2^n) to the base b = 1/sqrt(12). Some examples are given below. (End)
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LINKS
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FORMULA
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a(n) = 1/sqrt(12)*( (2 + sqrt(3))^2^n - (2 - sqrt(3))^2^n ) = A001353(2^n).
Recurrence equations:
a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 2.
a(n) = a(n-1)*sqrt(12*a(n-1)^2 + 4) for n >= 1. - Peter Bala, Oct 30 2013
0 = 6*a(n)^2*a(n+2) - 6*a(n+1)^3 - 2*a(n+1) + a(n+2) for n>=1. - Michael Somos, Dec 05 2016
0 = a(n)^2*(2*a(n+1) + a(n+2)) - a(n+1)^3 for n>=1. - Michael Somos, Dec 05 2016
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EXAMPLE
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Let b = 1/sqrt(12) and x = 2 - sqrt(3). We have the following Pierce expansions to base b:
x = b/1 - b^2/(1*4) + b^3/(1*4*56) - b^4/(1*4*56*10864) + b^5/(1*4*56*10864*408855776) - ....
x^2 = b/4 - b^2/(4*56) + b^3/(4*56*10864) - b^4/(4*56*10864*408855776) + ....
x^4 = b/56 - b^2/(56*10864) + b^3/(56*10864*408855776) - ....
x^8 = b/10864 - b^2/(10864*408855776) + .... - Peter Bala, Oct 30 2013
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PROG
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(PARI) g(n, p) = x=1; for(j=1, p, x=(n/x+x)/2; print1(denominator(x)", "))
(Magma) I:=[1, 4]; [n le 2 select I[n] else 2*Self(n-1)*(6*Self(n-2)^2+1): n in [1..8]]; // Vincenzo Librandi, Dec 19 2011
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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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Joe Keane (jgk(AT)jgk.org), May 31 2002
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STATUS
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approved
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