|
|
A006267
|
|
Continued cotangent for the golden ratio.
(Formerly M3699)
|
|
21
|
|
|
|
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
|
|
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, A002813, A045529.
Sequence in context: A052271 A184272 A080989 * A273952 A201984 A210519
Adjacent sequences: A006264 A006265 A006266 * A006268 A006269 A006270
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
N. J. A. Sloane
|
|
EXTENSIONS
|
The next term is too large to include.
|
|
STATUS
|
approved
|
|
|
|