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%I #11 Mar 09 2013 16:39:30
%S 6,234,12813606,2103846732371087589834,
%T 9311985549495522884757461748592522243432897275494229148348315206
%N Continued cotangent recurrence a(n+1) = a(n)^3 + 3*a(n) and a(1) = 6.
%C General formula for continued cotangent recurrences type:
%C a(n+1) = a(n)3 + 3*a(n) and a(1)=k is following:
%C a(n) = Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))].
%C The next term (a(6)) has 192 digits. - _Harvey P. Dale_, Mar 09 2013
%H J. Shallit, <a href="http://archive.org/details/jresv80Bn2p285">Predictable regular continued cotangent expansions</a>, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.
%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/LehmerCotangentExpansion.html">MathWorld: Lehmer Cotangent Expansion</a>
%F a(n+1)=a(n)^3 + 3*a(n) and a(1)=6
%F a(n)=Floor[((6+Sqrt[6^2+4])/2)^(3^(n-1))]
%F a(n) divides a(n+1) and b(n) = a(n+1)/a(n) satisfies the recurrence b(n+1) = b(n)^3 - 3*b(n-1)^2 + 3. See A002813. - _Peter Bala_, Nov 23 2012
%t a = {}; k = 6; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
%t or
%t Table[Floor[((6 + Sqrt[40])/2)^(3^(n - 1))], {n, 1, 5}] (* Artur Jasinski *)
%t NestList[#^3+3#&,6,5] (* _Harvey P. Dale_, Mar 09 2013 *)
%Y Cf. A006267, A006266, A006268, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189 (k = 1 to 15 with k=4 being A006267(n+1)).
%Y Cf. A002813.
%K nonn,easy
%O 1,1
%A _Artur Jasinski_, Oct 03 2008
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