A001999 a(n) = a(n-1)*(a(n-1)^2 - 3). (Formerly M3055 N1239)

3, 18, 5778, 192900153618, 7177905237579946589743592924684178

Offset

0,1

Comments

The next terms in the sequence contain 102 and 305 digits. - Harvey P. Dale, Jun 09 2011

From Peter Bala, Nov 13 2012: (Start)

The present sequence is the case x = 3 of the following general remarks. For other cases see A219160 (x = 4), A219161 (x = 5) and A112845 (x = 6).

Let x > 2 and let alpha := {x + sqrt(x^2 - 4)}/2. Define a sequence a(n) (which depends on x) by setting a(n) = alpha^(3^n) + (1/alpha)^(3^n). Then it is easy to verify that the sequence a(n) satisfies the recurrence equation a(n+1) = a(n)^3 - 3*a(n) with the initial condition a(0) = x.

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. Iterating the identity produces the product expansion sqrt((x+2)/(x-2)) = Product_{n = 0..oo} (1 + 2/(a(n) - 1)), with a(0) = x and a(n+1) = a(n)^3 - 3*a(n). The rate of convergence is cubic (Fine).

For similar results to the above see A001566 and A219162. (End)

Let b(n) = a(n) - 3. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. - Peter Bala, Dec 08 2022

References

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

G. C. Greubel, Table of n, a(n) for n = 0..7

A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437; alternative link.

E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, Vol. 44, No. 10 (1937), pp. 644-646.

N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8 (Oct. 1977), pp. 629-630.

Walther Janous, Problem B-916, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 39, No. 2 (2001), p. 181; Subscript Is Power, Solution to Problem B-916 by H.-J. Seiffert, ibid., Vol. 40, No. 1 (2002), p. 86.

Hideyuki Ohtsu, Problem B-1316, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 60, No. 4 (2022), p. 365.

Eric Weisstein's World of Mathematics, Pierce Expansion.

Formula

a(n) = 2*F(2*3^n+1) - F(2*3^n) = ceiling(tau^(2*3^n)) where F(k) = A000045(k) is the k-th Fibonacci number and tau is the golden ratio. - Benoit Cloitre, Nov 29 2002

From Peter Bala, Nov 13 2012: (Start)

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

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

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

a(n) = 2*T(3^n,3/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. Cf. A219161. - Peter Bala, Feb 01 2017

From Amiram Eldar, Jan 12 2022: (Start)

a(n) = A000032(2*3^n).

a(n) = A006267(n)^2 + 2.

Product_{k=0..n} (a(k)-1) = Fibonacci(3^(n+1)) = A045529(n+1) (Janous, 2001). (End)

Sum_{n>=0} arctanh(1/a(n)) = log(5)/4 (Ohtsu, 2022). - Amiram Eldar, Dec 15 2022

Mathematica

NestList[#(#^2-3)&, 3, 6] (* Harvey P. Dale, Jun 09 2011 *)
RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 3}, a, {n,
  0, 5}] (* G. C. Greubel, Dec 30 2016 *)

Prog

(PARI) a(n)=2*fibonacci(2*3^n+1)-fibonacci(2*3^n)

Crossrefs

Cf. A000032, A006276, A001566, A002814, A045529, A112845, A219160, A219161, A219162.

Sequence in context: A214442 A057133 A321249 * A157580 A101293 A227882

Adjacent sequences: A001996 A001997 A001998 * A002000 A002001 A002002

Keyword

nonn,easy,nice

Author

N. J. A. Sloane

Status

approved