The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A071579 a(n) = 2*a(n-1)*A002812(n-1), starting a(0)=1. 8
 1, 4, 56, 10864, 408855776, 579069776145402304, 1161588808526051807570761628582646656, 4674072680304961790168962360144614650442718636276775741658113370728376064 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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 The sequence satisfies the Pell equation A002812(n)^2-3*a(n)^2 = 1. - Vincenzo Librandi, Dec 19 2011 From Peter Bala, Oct 30 2013: (Start) 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) LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..10 Eric Weisstein's World of Mathematics, Newton's Iteration Eric Weisstein's World of Mathematics, Newton's Iteration Doron Zeilberger, Another Book of Somos-Like Miracles, Prop. Number, 2; Local copy FORMULA a(n) = 1/sqrt(12)*( (2 + sqrt(3))^2^n - (2 - sqrt(3))^2^n ) = A001353(2^n). a(n) = 2*a(n-1)*(6*a(n-2)^2+1). - Max Alekseyev, Apr 19 2006 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 EXAMPLE 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 PROG (PARI) g(n, p) = x=1; for(j=1, p, x=(n/x+x)/2; print1(denominator(x)", ")) g(3, 8) \\ Cino Hilliard, 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 CROSSREFS Cf. A002812, A001353. A058635, A230338. Sequence in context: A056075 A000315 A080984 * A060497 A092273 A193745 Adjacent sequences:  A071576 A071577 A071578 * A071580 A071581 A071582 KEYWORD nonn,easy AUTHOR Joe Keane (jgk(AT)jgk.org), May 31 2002 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified July 31 02:16 EDT 2021. Contains 346367 sequences. (Running on oeis4.)