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

3, 18, 5778, 192900153618, 7177905237579946589743592924684178

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

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.
• A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
• 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.

Crossrefs

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

Programs

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 *)

PARI

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

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