login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

a(n) = a(n-1)*(a(n-1)^2 - 3).
(Formerly M3055 N1239)
16

%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_