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Recurrence a(n) = a(n-1)^3 - 3*a(n-1) with a(0) = 6.
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%I #23 Oct 04 2017 09:14:32

%S 6,198,7761798,467613464999866416198,

%T 102249460387306384473056172738577521087843948916391508591105798

%N Recurrence a(n) = a(n-1)^3 - 3*a(n-1) with a(0) = 6.

%C Identical to A006243 apart from the initial term. For some general remarks on this recurrence see A001999. - _Peter Bala_, Nov 13 2012

%H G. C. Greubel, <a href="/A112845/b112845.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.

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

%F a(n) = -2*cos(3^n*arccos(-3)).

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

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

%F Product {n = 0..inf} (1 + 2/(a(n) - 1)) = sqrt(2).

%F (End)

%t RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 6}, a, {n,

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

%Y Cf. A006275, A006276.

%Y Cf. A006243. - _R. J. Mathar_, Aug 15 2008

%Y Cf. A001999, A219160, A219161.

%K nonn,easy

%O 0,1

%A _Eric W. Weisstein_, Sep 21 2005