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A002814 For n > 1: a(n) = a(n-1)^3 + 3a(n-1)^2 - 3; a(0) = 1, a(1) = 2.
(Formerly M2105 N0833)
14
1, 2, 17, 5777, 192900153617, 7177905237579946589743592924684177 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

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

Next terms have 102 and 305 digits. - Harvey P. Dale, Jun 06 2011

From Peter Bala, Nov 15 2012: (Start)

The present sequence is the case x = 3 of the following general remarks. The recurrence equation a(n+1) = a(n)^3 + 3*a(n)^2 - 3 with the initial condition a(1) = x - 1 > 1 has the explicit solution a(n+1) = alpha^(3^n) + (1/alpha)^(3^n) - 1 for n >= 0, where alpha := {x + sqrt(x^2 - 4)}/2.

Two other recurrences satisfied by the sequence are a(n+1) = (a(1) + 3)*(Product_{k = 1..n} a(k)^2) - 3 and a(n+1) = 1 + (a(1) - 1)*Product_{k = 1..n} (a(k) + 2)^2, both with a(1) = x - 1.

The associated sequence b(n) := a(n) + 1 satisfies the recurrence equation b(n+1) = b(n)^3 - 3*b(n) with the initial condition b(1) = x. See A001999 for the case x = 3. The sequence c(n) := a(n) + 2 satisfies the recurrence equation c(n+1) = c(n)^3 - 3*c(n)^2 + 3 with the initial condition c(1) = x + 1.

The sequences a(n) and b(n) have been considered by Fine and Escott in connection with a product expansion for quadratic irrationals. We have the following identity, valid for x > 2: sqrt((x + 2)/(x - 2)) = (1 + 2/(x-1))*sqrt((y + 2)/(y - 2)), where y = x^3 - 3*x or, equivalently, y - 1 = (x - 1)^3 + 3*(x - 1)^2 - 3. Iterating the identity produces the product expansion sqrt((x+2)/(x-2)) = Product_{n >= 1} (1 + 2/a(n)), with a(1) = x - 1 and a(n+1) = a(n)^3 + 3*a(n)^2 - 3.

For similar results to the above see A145502. See also A219162.

(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

Reinhard Zumkeller, 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)

M. Mendes France and A. J. van der Poorten, From geometry to Euler identities, Theoret. Comput. Sci., 65 (1989), 213-220.

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

FORMULA

a(n) = Fibonacci(3^n)/Fibonacci(3^(n-1)). - Henry Bottomley, Jul 10 2001

a(n+1) = 5*(f(n))^2 - 3, where f(n) = Fibonacci(3^n) = product of first n entries. - Lekraj Beedassy, Jun 16 2003

From Artur Jasinski, Oct 05 2008: (Start)

a(n+2) = (G^(3^(n + 1)) - (1 - G)^(3^(n + 1)))/((G^(3^n)) - (1 - G)^(3^n)) where G = (1 + sqrt(5))/2;

a(n+2) = A045529(n+1)/A045529(n). (End)

From Peter Bala, Nov 15 2012: (Start)

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

The sequence b(n):= a(n) + 2 is a solution to the recurrence b(n+1) = b(n)^3 - 3*b(n)^2 + 3 with b(1) = 4.

Other recurrence equations:

a(n+1) = -3 + 5*(Product_{k = 1..n} a(k)^2) with a(1) = 2.

a(n+1) = 1 + Product_{k = 1..n} (a(k) + 2)^2 with a(1) = 2.

Thus Y := Product_{k = 1..n} a(k) and X := Product_{k = 1..n} (a(k) + 2) give a solution to the Diophantine equation X^2 - 5*Y^2 = -4.

sqrt(5) = Product_{n >= 1} (1 + 2/a(n)). The rate of convergence is cubic. Fine remarks that twelve factors of the product would give well over 300,000 correct decimal digits for sqrt(5).

5 - {Product_{n = 1..N} (1 + 2/a(n))}^2 = 20/(a(N+1) + 3).

(End)

MATHEMATICA

Join[{1}, NestList[#^3+3#^2-3&, 2, 5]] (* Harvey P. Dale, Apr 01 2011 *)

PROG

(PARI) a(n)=if(n<2, max(0, n+1), a(n-1)^3+3*a(n-1)^2-3)

(Maxima) a[0]:1$ a[1]:2$ a[n]:=a[n-1]^3 + 3*a[n-1]^2-3$ A002814(n):=a[n]$

makelist(A002814(n), n, 0, 6); /* Martin Ettl, Nov 12 2012 */

(Haskell)

a002814 n = a002814_list !! n

a002814_list = 1 : zipWith div (tail xs) xs

   where xs = map a000045 a000244_list

-- Reinhard Zumkeller, Nov 24 2012

CROSSREFS

Cf. A000045, A001566.

Cf. A045529. A001999, A145502, A219162.

Cf. A000244.

Sequence in context: A274015 A279884 A060353 * A122207 A174305 A003819

Adjacent sequences:  A002811 A002812 A002813 * A002815 A002816 A002817

KEYWORD

nonn,easy,nice

AUTHOR

N. J. A. Sloane

EXTENSIONS

Definition improved by Reinhard Zumkeller, Feb 29 2012

STATUS

approved

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Last modified November 14 22:45 EST 2019. Contains 329135 sequences. (Running on oeis4.)