%I M3055 N1239 #62 Dec 15 2022 13:35:28

%S 3,18,5778,192900153618,7177905237579946589743592924684178

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

%C The next terms in the sequence contain 102 and 305 digits. - _Harvey P. Dale_, Jun 09 2011

%C From _Peter Bala_, Nov 13 2012: (Start)

%C 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).

%C 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.

%C 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).

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

%C 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

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H G. C. Greubel, <a href="/A001999/b001999.txt">Table of n, a(n) for n = 0..7</a>

%H A. V. Aho and N. J. A. Sloane, <a href="https://www.fq.math.ca/Scanned/11-4/aho-a.pdf">Some doubly exponential sequences</a>, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437; <a href="http://neilsloane.com/doc/doubly.html">alternative link</a>.

%H E. B. Escott, <a href="http://www.jstor.org/stable/2301484">Rapid method for extracting a square root</a>, Amer. Math. Monthly, Vol. 44, No. 10 (1937), pp. 644-646.

%H N. J. Fine, <a href="http://www.jstor.org/stable/2321014">Infinite products for k-th roots</a>, Amer. Math. Monthly Vol. 84, No. 8 (Oct. 1977), pp. 629-630.

%H Walther Janous, <a href="https://www.fq.math.ca/Scanned/39-2/elementary39-2.pdf">Problem B-916</a>, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 39, No. 2 (2001), p. 181; <a href="https://www.fq.math.ca/Scanned/40-1/elementary40-1.pdf">Subscript Is Power</a>, Solution to Problem B-916 by H.-J. Seiffert, ibid., Vol. 40, No. 1 (2002), p. 86.

%H Hideyuki Ohtsu, <a href="https://www.fq.math.ca/Problems/ElemProbSolnNov2022.pdf">Problem B-1316</a>, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 60, No. 4 (2022), p. 365.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PierceExpansion.html">Pierce Expansion</a>.

%F 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

%F From _Peter Bala_, Nov 13 2012: (Start)

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

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

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

%F 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

%F From _Amiram Eldar_, Jan 12 2022: (Start)

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

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

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

%F Sum_{n>=0} arctanh(1/a(n)) = log(5)/4 (Ohtsu, 2022). - _Amiram Eldar_, Dec 15 2022

%t NestList[#(#^2-3)&,3,6] (* _Harvey P. Dale_, Jun 09 2011 *)

%t RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 3}, a, {n,

%t 0, 5}] (* _G. C. Greubel_, Dec 30 2016 *)

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

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

%K nonn,easy,nice

%O 0,1

%A _N. J. A. Sloane_