A006267 Continued cotangent for the golden ratio. (Formerly M3699)

1, 4, 76, 439204, 84722519070079276, 608130213374088941214747405817720942127490792974404

Offset

0,2

References

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

Links

Harry J. Smith, Table of n, a(n) for n = 0..7

Mohammad K. Azarian, Problem 123, Missouri Journal of Mathematical Sciences, Vol. 10, No. 3 (Fall 1998), p. 176; Solution, ibid., Vol. 12, No. 1 (Winter 2000), pp. 61-62.

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

Zalman Usiskin, Problem B-266, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 11, No. 3 (1973), p. 334; Lucas Numbers for Powers of 3, Solution to Problem B-266 by David Zeitlin, ibid., Vol. 12, No. 3 (1974), p. 315-316.

Eric Weisstein's World of Mathematics, Lehmer Cotangent Expansion.

Formula

(1+sqrt(5))/2 = cot(Sum_{n>=0} (-1)^n*acot(a(n))); let b(0) = (1+sqrt(5))/2, b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1)) then a(n) = floor(b(n)). - Benoit Cloitre, Apr 10 2003

a(n) = A000204(3^n). - Benoit Cloitre, Sep 18 2005

a(n) = Round(c^(3^n)) where c = GoldenRatio = 1.6180339887498948482... = (sqrt(5)+1)/2 (A001622). - Artur Jasinski, Sep 22 2008

Recurrence a(n+1) = a(n)^3 + 3*a(n), a(0) = 4. - Artur Jasinski, Sep 24 2008

a(n+1) = Product_{k = 0..n} A002813(k). Thus a(n) divides a(n+1). - Peter Bala, Nov 22 2012

Sum_{n>=0} a(n)^2/A045529(n+1) = 1. - Amiram Eldar, Jan 12 2022

a(n) = Product_{k=0..n-1} (Lucas(2*3^k) + 1) (Usiskin, 1973). - Amiram Eldar, Jan 29 2022

From Peter Bala, Nov 15 2022: (Start)

a(n) = Lucas(3^n) for n >= 1.

a(n) == 1 (mod 3) for n >= 1.

a(n+1) == a(n) (mod 3^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).

The smallest positive residue of a(n) mod 3^n = A268924(n).

In the ring of 3-adic integers the limit_{n -> oo} a(n) exists and is equal to A271223. Cf. A006266. (End)

Maple

a := proc(n) option remember; if n = 1 then 4 else a(n-1)^3 + 3*a(n-1) end if; end: seq(a(n), n = 1..5); # Peter Bala, Nov 15 2022

Mathematica

c = N[GoldenRatio, 1000]; Table[Round[c^(3^n)], {n, 1, 8}] (* Artur Jasinski, Sep 22 2008 *)
a = {}; x = 4; Do[AppendTo[a, x]; x = x^3 + 3 x, {n, 1, 10}]; a (* Artur Jasinski, Sep 24 2008 *)

Prog

(PARI) bn=vector(100); b(n)=if(n<0, 0, bn[n]); bn[1]=(1+sqrt(5))/2; for(n=2, 10, bn[n]=(b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1)))) a(n)=floor(b(n+1))
(PARI) { default(realprecision, 10000); bn=vector(8); bn[1]=(1+sqrt(5))/2; for(n=2, 8, bn[n]=(bn[n-1]*floor(bn[n-1]) + 1)/(bn[n-1] - floor(bn[n-1]))); for (n=1, 8, write("b006267.txt", n-1, " ", floor(bn[n]))); } \\ Harry J. Smith, May 04 2009

Crossrefs

Cf. A000032, A000204, A001622, A001999, A002666, A002667, A002668, A006266, A002813, A045529, A271223, A268924.

Sequence in context: A052271 A184272 A080989 * A273952 A370953 A201984

Adjacent sequences: A006264 A006265 A006266 * A006268 A006269 A006270

Keyword

nonn,easy

Author

N. J. A. Sloane

Extensions

The next term is too large to include.

Status

approved