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

2, 7, 97, 18817, 708158977, 1002978273411373057, 2011930833870518011412817828051050497, 8095731360557835890888779535060256832479295062749579257164654370487894017

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

Comments

An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004

2^p-1 is prime iff it divides a(p-2), since a(n) = A003010(n)/2, where A003010 is the Lucas-Lehmer sequence used for Mersenne number primality testing. - M. F. Hasler, Mar 09 2007

From Cino Hilliard, Sep 28 2008: (Start)

Also numerators of the convergents to the square root of 3 using the following recursion for initial x = 1: x1=x, x=3/x, x=(x+x1)/2.

This recursion was derived by experimenting with polynomial recursions of the form x = -a(0)/(a(n-1)x^(n-1) + ... + a(1)) in an effort to find a root for the polynomial a(n)x^n + a(n-1)x^(n-1) + ... + a(0). The process was hit-and-miss until I introduced the averaging step described above. This method is equivalent to Newton's Method although derived somewhat differently. (End)

The sequence satisfies the Pell equation a(n)^2 - 3*A071579(n)^2 = 1. - Vincenzo Librandi, Dec 19 2011

From Peter Bala, Nov 2012: (Start)

The present sequence corresponds to the case x = 2 of the following general remarks. Let x > 1 and let alpha := {x + sqrt(x^2 - 1)}. Define a sequence a(n) (which depends on x) by setting a(n) = (1/2)*(alpha^(2^n) + (1/alpha)^(2^n)) for n >= 0. It is easy to verify that a(n) is a solution to the recurrence equation a(n+1) = 2*a(n)^2 - 1 with the initial condition a(0) = x.

The following algebraic identity is valid for x > 1:

sqrt(4*x^2 - 4)/(2*x + 1) = (1 - 1/(2*x))*sqrt(4*y^2 - 4)/(2*y + 1), where y = 2*x^2 - 1. Iterating the identity yields the product expansion sqrt(4*x^2 - 4)/(2*x + 1) = Product_{n >= 0} (1 - 1/(2*a(n))). A second expansion is Product_{n >= 0} (1 + 1/a(n)) = sqrt((x + 1)/(x - 1)). See Mendes-France and van der Poorten. (End)

Examples

			G.f. = 2 + 7*x + 97*x^2 + 18817*x^3 + 708158977*x^4 + ...
		

References

• L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 399.
• E. Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.
• W. Sierpiński, Sur les développements systématiques des nombres en produits infinis, in Œuvres choisies, tome 1, PWN Editions scientifiques de Pologne, 1974.
• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10
• Georg Cantor, Zwei Sätze über eine gewisse Zerlegung der Zahlen in unendliche Producte, Zeitschrift für Mathematik und Physik. Band 14, 1869, S. 152-158. (See page 155, II, error in the fourth term.)
• E. Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy)
• M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.
• Jeffrey Shallit, Rational numbers with non-terminating, non-periodic modified Engel-type expansions, Fib. Quart., 31 (1993), 37-40.
• Eric Weisstein's World of Mathematics, Newton's Iteration
• Index entries for sequences related to Engel expansions

Crossrefs

Cf. A001075, A002531, A003010, A071579.

Cf. A177879 (lpf).

Programs

GAP

a:=[2];; for n in [2..10] do a[n]:=2*a[n-1]^2-1; od; a; # Muniru A Asiru, Aug 12 2018

Magma

I:=[2]; [n le 1 select I[n] else 2*Self(n-1)^2-1: n in [1..8]]; // Vincenzo Librandi, Dec 19 2011

Maple

a:=n->((2+sqrt(3))^(2^n)+(2-sqrt(3))^(2^n))/2: seq(floor(a(n)),n=0..10); # Muniru A Asiru, Aug 12 2018

Mathematica

Table[((2 + Sqrt[3])^2^n + (2 - Sqrt[3])^2^n)/2, {n, 0, 7}] (* Bruno Berselli, Dec 20 2011 *)
NestList[2#^2-1&,2,10] (* Harvey P. Dale, May 04 2013 *)
a[ n_] := If[ n < 0, 0, ChebyshevT[2^n, 2]]; (* Michael Somos, Dec 06 2016 *)

PARI

{a(n) = if( n<1, 2 * (n==0), 2 * a(n-1)^2 - 1)}; /* Michael Somos, Mar 14 2004 */

PARI

/* Roots by recursion. Find first root of ax^2 + b^x + c */
rroot2(a,b,c,p) = { local(x=1,x1=1,j); for(j=1,p, x1=x; x=-c/(a*x+b); x=(x1+x)/2; /* Let x be the average of the last 2 values */ print1(numerator(x)","); ); } \\ Cino Hilliard, Sep 28 2008

Formula

a(n) = A001075(2^n).

a(n) = ((2+sqrt(3))^(2^n) + (2-sqrt(3))^(2^n))/2. - Bruno Berselli, Dec 20 2011

From Peter Bala, Nov 11 2012: (Start)

2*sqrt(3)/5 = Product_{n >= 0} (1 - 1/(2*a(n))).

sqrt(3) = Product_{n >= 0} (1 + 1/a(n)).

a(n) = (1/2)*A003010(n). (End)

a(n) = cos(2^n*arccos(2)). - Peter Luschny, Oct 12 2022

a(n) = A002531(2^(n+1)). - Robert FERREOL, Apr 13 2023