A071579 a(n) = 2*a(n-1)*A002812(n-1), starting a(0)=1.

1, 4, 56, 10864, 408855776, 579069776145402304, 1161588808526051807570761628582646656, 4674072680304961790168962360144614650442718636276775741658113370728376064

Offset: 0

Author

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

Keywords

• nonn
• easy

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)

Examples

			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
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10
• Eric Weisstein's World of Mathematics, Newton's Iteration.
• Doron Zeilberger, Another Book of Somos-Like Miracles, Prop. Number, 2; Local copy.

Crossrefs

Cf. A002812, A001353. A058635, A230338.

Programs

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

Mathematica

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

PARI

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

PARI

{a(n) = if(n<0, 0, imag((2 + quadgen(12))^2^n))}; /* Michael Somos, Jul 29 2024 */

PARI

{a(n) = if(n<0, 0, polcoef(lift(Mod(x^2^n, x^2 - 4*x + 1)), 1))}; /* Michael Somos, Jul 29 2024 */
g(3,8) \\ Cino Hilliard, Sep 28 2008

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

a(n) = A001353(2^n). - Michael Somos, Jul 29 2024