A071579 - OEIS

%I #52 Jul 30 2024 06:28:31

%S 1,4,56,10864,408855776,579069776145402304,

%T 1161588808526051807570761628582646656,

%U 4674072680304961790168962360144614650442718636276775741658113370728376064

%N a(n) = 2*a(n-1)*A002812(n-1), starting a(0)=1.

%C Also the denominators of the convergents to sqrt(3) using Newton's recursion x = (3/x+x)/2. - _Cino Hilliard_, Sep 28 2008

%C 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

%C The sequence satisfies the Pell equation A002812(n)^2-3*a(n)^2 = 1. - _Vincenzo Librandi_, Dec 19 2011

%C From _Peter Bala_, Oct 30 2013: (Start)

%C 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).

%C 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.

%C 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)

%H Vincenzo Librandi, <a href="/A071579/b071579.txt">Table of n, a(n) for n = 0..10</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/NewtonsIteration.html">Newton's Iteration</a>.

%H Doron Zeilberger, <a href="http://www.math.rutgers.edu/~zeilberg/tokhniot/oNesSomos1">Another Book of Somos-Like Miracles</a>, Prop. Number, 2; <a href="/A071579/a071579.txt">Local copy</a>.

%F a(n) = 1/sqrt(12)*( (2 + sqrt(3))^2^n - (2 - sqrt(3))^2^n ) = A001353(2^n).

%F a(n) = 2*a(n-1)*(6*a(n-2)^2+1). - _Max Alekseyev_, Apr 19 2006

%F Recurrence equations:

%F a(n)/a(n-1) = (a(n-1)/a(n-2))^2 - 2 for n >= 2.

%F a(n) = a(n-1)*sqrt(12*a(n-1)^2 + 4) for n >= 1. - _Peter Bala_, Oct 30 2013

%F 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

%F 0 = a(n)^2*(2*a(n+1) + a(n+2)) - a(n+1)^3 for n>=1. - _Michael Somos_, Dec 05 2016

%F a(n) = A001353(2^n). - _Michael Somos_, Jul 29 2024

%e Let b = 1/sqrt(12) and x = 2 - sqrt(3). We have the following Pierce expansions to base b:

%e 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) - ....

%e x^2 = b/4 - b^2/(4*56) + b^3/(4*56*10864) - b^4/(4*56*10864*408855776) + ....

%e x^4 = b/56 - b^2/(56*10864) + b^3/(56*10864*408855776) - ....

%e x^8 = b/10864 - b^2/(10864*408855776) + .... - _Peter Bala_, Oct 30 2013

%t a[ n_] := If[n<0, 0, Coefficient[PolynomialMod[x^2^n, x^2 - 4*x + 1], x]]; (* _Michael Somos_, Jul 29 2024 *)

%t a[ n_] := If[n<1, Boole[n==0], a[n] = a[n-1]*Sqrt[12*a[n-1]^2 + 4] ]; (* _Michael Somos_, Jul 29 2024 *)

%o (PARI) g(n,p) = x=1;for(j=1,p,x=(n/x+x)/2;print1(denominator(x)","))

%o (PARI) {a(n) = if(n<0, 0, imag((2 + quadgen(12))^2^n))}; /* _Michael Somos_, Jul 29 2024 */

%o (PARI) {a(n) = if(n<0, 0, polcoef(lift(Mod(x^2^n, x^2 - 4*x + 1)), 1))}; /* _Michael Somos_, Jul 29 2024 */

%o g(3,8) \\ _Cino Hilliard_, Sep 28 2008

%o (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

%Y Cf. A002812, A001353. A058635, A230338.

%K nonn,easy

%O 0,2

%A Joe Keane (jgk(AT)jgk.org), May 31 2002