A006268 A continued cotangent.

3, 36, 46764, 102266868132036, 1069559300034650646049671039050649693658764

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

References

• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Jeffrey Shallit, Predictable regular continued cotangent expansions, J. Res. Nat. Bur. Standards Sect. B 80B (1976), no. 2, 285-290.

Crossrefs

Continued cotangents: A006267, A006266, A006269, A145180, A145181, A145182, A145183, A145184, A145185, A145186, A145187, A145188, A145189 (k = 1 to 15 with k=4 being A006267(n+1)).

Cf. A002813, A006273.

Programs

Mathematica

a = {}; k = 3; Do[AppendTo[a, k]; k = k^3 + 3 k, {n, 1, 6}]; a (* Artur Jasinski, Oct 03 2008 *)
Table[Round[N[(3/2 + Sqrt[13]/2)^(3^(n - 1)), 1000]], {n, 1, 8}] (* Artur Jasinski, Oct 03 2008 *)

PARI

a(n) = if (n==0, 3, a(n-1)^3 + 3*a(n-1)); \\ Michel Marcus, Aug 28 2020

Formula

From Artur Jasinski, Oct 03 2008: (Start)

a(n+1) = a(n)^3 + 3*a(n) and a(0)=3.

a(n) = round((3/2 + sqrt(13)/2)^(3^(n - 1))). (End)

From Peter Bala, Jan 19 2022: (Start)

a(n) = (3/2 + sqrt(13)/2)^(3^(n-1)) + (3/2 - sqrt(13)/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. For remarks about this recurrence see A002813.

1 + a(n)^2 = A006273(n+1). (End)