%I #17 Apr 23 2019 15:16:32
%S 5,110,1330670,2356194280407770990,
%T 13080769480548649962914459850235688797656360638877986030
%N Recurrence equation a(n+1) = a(n)^3 - 3*a(n) with a(0) = 5.
%C For some general remarks on this recurrence see A001999.
%C The next term (a(5)) has 166 digits. - _Harvey P. Dale_, Apr 23 2019
%H G. C. Greubel, <a href="/A219161/b219161.txt">Table of n, a(n) for n = 0..6</a>
%H E. B. Escott, <a href="http://www.jstor.org/stable/2301484">Rapid method for extracting a square root</a>, Amer. Math. Monthly, 44 (1937), 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, 629-630.
%F a(n) = (1/2*(5 + sqrt(21)))^(3^n) + (1/2*(5 - sqrt(21)))^(3^n).
%F Product_{n = 0..inf} (1 + 2/(a(n) - 1)) = sqrt(7/3).
%F a(n) = 2*T(3^n,5/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. Cf. A001999. - Peter Bala, Feb 01 2017
%t RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 5}, a, {n,
%t 0, 5}] (* _G. C. Greubel_, Dec 30 2016 *)
%t NestList[#^3-3#&,5,5] (* _Harvey P. Dale_, Apr 23 2019 *)
%Y Cf. A001999, A112845, A219160.
%K nonn,easy
%O 0,1
%A _Peter Bala_, Nov 13 2012