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

4, 19, 5779, 192900153619, 7177905237579946589743592924684179, 369822356418414944143680173221426891716916679027557977938929258031490127514207143830378340325399155219

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

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, Lehmer Cotangent Expansion

Crossrefs

Cf. A002814, A006267.

Programs

Magma

[Lucas(2*3^n)+1: n in [0..5]]; // Vincenzo Librandi, Jul 20 2011

Mathematica

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

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)

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)