A002813 a(0) = 4; for n > 0, a(n) = a(n-1)^3 - 3*a(n-1)^2 + 3. (Formerly M3561 N1443)

4, 19, 5779, 192900153619, 7177905237579946589743592924684179, 369822356418414944143680173221426891716916679027557977938929258031490127514207143830378340325399155219

Offset

0,1

Comments

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

The next term, a(7), has 305 digits. - Harvey P. Dale, Jul 19 2011

From Peter Bala, Nov 22 2012: (Start)

The present sequence is the case x = 1 of the following general remarks about the recurrence a(n+1) = a(n)^3 - 3*a(n-1)^2 + 3. Cf. A002814.

Define a sequence of polynomials P(n,x) inductively by setting P(0,x) = x^2 + 3 and P(n+1,x) = P(n,x^3 + 3*x) for n >= 0. Then P(n,x) satisfies the cubic recurrence P(n+1,x) = P(n,x)^3 - 3*P(n-1,x)^2 + 3 with the initial condition P(0,x) = x^2 + 3.

An explicit formula is P(n,x) = Q(3^(n+1),x)/Q(3^n,x), where Q(n,x) = ((x + sqrt(x^2 + 4))/2)^n + ((x - sqrt(x^2 + 4))/2)^n.

Alternatively, P(n,x) = ((x^2 + 2 + sqrt(x^4 + 4*x^2))/2)^(3^n) + ((x^2 + 2 - sqrt(x^4 + 4*x^2))/2)^(3^n) + 1.

Iterating the algebraic identity x/sqrt(x^2 + 4) = (1 - 2/(x^2 + 3))*y/sqrt(y^2 + 4), where y = x^3 + 3*x, leads to the product expansion x/sqrt(x^2 + 4) = Product_{n = 0..oo} (1 - 2/P(n,x)). See Escott and also Fine.

The sequence A(n,x) := x*Product_{k = 0..n} P(k,x) satisfies the recurrence A(n+1,x) = A(n,x)^3 + 3*A(n,x). These sequences occur in the continued cotangent expansions of Lehmer. Cases currently in the database are A006267 (x = 1), A006266 (x = 2), A006268 (x = 3), A006269 (x = 5) and A145180 through A145189 (x = 6 through x = 15).

(End)

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. 397.

E. Lucas, Nouveaux théorèmes d'arithmétique supérieure, Comptes Rend., 83 (1876), 1286-1288.

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..8

E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.

N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8, Oct. 1977.

E. Lucas, Nouveaux théorèmes d'arithmétique supérieure (annotated scanned copy)

J. Shallit Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.

Eric Weisstein's World of Mathematics, MathWorld: Lehmer Cotangent Expansion>

Formula

a(n) = L(2*3^n)+1 where L=Lucas numbers.

a(n) = L(3^(n+1))/L(3^n). - Benoit Cloitre, Sep 18 2005

a(n) = A001999(n)+1. - R. J. Mathar, Apr 22 2007

From Peter Bala, Nov 22 2012: (Start)

a(n) = ((3 + sqrt(5))/2)^(3^n) + ((3 - sqrt(5))/2)^(3^n) + 1.

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

A006267(n+1) = Product_{k = 0..n} a(k).

A002814(n+1) = a(n) - 2. (End)

Mathematica

NestList[#^3-3#^2+3&, 4, 6] (* Harvey P. Dale, Jul 19 2011 *)

Prog

(PARI) a(n)=if(n<1, 4*(n==0), a(n-1)^3-3*a(n-1)^2+3)
(PARI) a(n)=if(n<0, 0, n=2*3^n; fibonacci(n+1)+fibonacci(n-1)+1)
(Magma) [Lucas(2*3^n)+1: n in [0..5]]; // Vincenzo Librandi, Jul 20 2011

Crossrefs

Cf. A002814, A006267.

Sequence in context: A000863 A291947 A023994 * A263973 A364519 A104159

Adjacent sequences: A002810 A002811 A002812 * A002814 A002815 A002816

Keyword

nonn,easy,nice

Author

N. J. A. Sloane

Status

approved