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%I #13 Jul 12 2018 20:28:07
%S 4,52,140452,2770663499604052,
%T 21269209556953516583554114034636483645584976452
%N Recurrence equation a(n+1) = a(n)^3 - 3*a(n) with a(0) = 4.
%C For some general remarks on this recurrence see A001999.
%H G. C. Greubel, <a href="/A219160/b219160.txt">Table of n, a(n) for n = 0..6</a>
%H E. B. Escott, <a href="https://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="https://www.jstor.org/stable/2321014">Infinite products for k-th roots</a>, Amer. Math. Monthly Vol. 84, No. 8, Oct. 1977.
%F a(n) = (2 + sqrt(3))^(3^n) + (2 - sqrt(3))^(3^n).
%F Product {n = 0..inf} (1 + 2/(a(n) - 1)) = sqrt(3). The rate of convergence is cubic. Fine remarks that taking the first twelve factors of the product would give well over 300,000 correct decimals for sqrt(3).
%t RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 4}, a, {n,
%t 0, 5}] (* _G. C. Greubel_, Dec 30 2016 *)
%Y Cf. A001999, A112845, A219161.
%K nonn,easy
%O 0,1
%A _Peter Bala_, Nov 13 2012
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