OFFSET
1,1
COMMENTS
General formula for continued cotangent recurrences type:
a(n+1) = a(n)3 + 3*a(n) and a(1)=k is following:
a(n) = Floor[((k+Sqrt[k^2+4])/2)^(3^(n-1))].
The next term (a(6)) has 192 digits. - Harvey P. Dale, Mar 09 2013
LINKS
J. Shallit, Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.
Eric W. Weisstein, MathWorld: Lehmer Cotangent Expansion
FORMULA
a(n+1)=a(n)^3 + 3*a(n) and a(1)=6
a(n)=Floor[((6+Sqrt[6^2+4])/2)^(3^(n-1))]
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
MATHEMATICA
a = {}; k = 6; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a
or
Table[Floor[((6 + Sqrt[40])/2)^(3^(n - 1))], {n, 1, 5}] (* Artur Jasinski *)
NestList[#^3+3#&, 6, 5] (* Harvey P. Dale, Mar 09 2013 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Artur Jasinski, Oct 03 2008
STATUS
approved