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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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2
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1, 4, 56, 10864, 408855776, 579069776145402304, 1161588808526051807570761628582646656, 4674072680304961790168962360144614650442718636276775741658113370728376064
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| Also the denominators of the convergents to sqrt(3) using Newton's recursion x = (3/x+x)/2. - Cino Hilliard (hillcino368(AT)hotmail.com), 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.
The sequence satisfies the Pell equation A002812(n)^2-3*a(n)^2 = 1. - Vincenzo Librandi, Dec 19 2011
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LINKS
| Vincenzo Librandi, Table of n, a(n) for n = 0..10
Eric Weisstein's World of Mathematics, Newton's Iteration
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FORMULA
| a(n) = A001353(2^n).
a(n) = 2*a(n-1)*(6*a(n-2)^2+1). - Max Alekseyev (maxale(AT)gmail.com), Apr 19 2006
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PROG
| (PARI) g(n, p) = x=1; for(j=1, p, x=(n/x+x)/2; print1(denominator(x)", "))
g(3, 8) \\ Cino Hilliard (hillcino368(AT)hotmail.com), Sep 28 2008
(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
| Cf. A002812, A001353.
Sequence in context: A056075 A000315 A080984 * A060497 A092273 A193745
Adjacent sequences: A071576 A071577 A071578 * A071580 A071581 A071582
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KEYWORD
| nonn,easy
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AUTHOR
| Joe Keane (jgk(AT)jgk.org), May 31 2002
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