A145180 Continued cotangent recurrence a(n+1) = a(n)^3 + 3*a(n) and a(1) = 6.

6, 234, 12813606, 2103846732371087589834, 9311985549495522884757461748592522243432897275494229148348315206

Offset: 1

Author

Artur Jasinski, Oct 03 2008

Keywords

• nonn
• easy

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

Crossrefs

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)).

Cf. A002813.

Programs

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 *)

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