A019429 Continued fraction for tan(1/6).
0, 5, 1, 16, 1, 28, 1, 40, 1, 52, 1, 64, 1, 76, 1, 88, 1, 100, 1, 112, 1, 124, 1, 136, 1, 148, 1, 160, 1, 172, 1, 184, 1, 196, 1, 208, 1, 220, 1, 232, 1, 244, 1, 256, 1, 268, 1, 280, 1, 292, 1, 304, 1, 316, 1, 328, 1, 340, 1, 352, 1, 364, 1, 376, 1, 388, 1, 400, 1, 412, 1, 424, 1, 436, 1
Offset: 0
Examples
0.16822721830224246125721608... = 0 + 1/(5 + 1/(1 + 1/(16 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 13 2009
Links
- Harry J. Smith, Table of n, a(n) for n = 0..20000
- G. Xiao, Contfrac
- Index entries for linear recurrences with constant coefficients, signature (0, 2, 0, -1).
Programs
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Mathematica
Block[{$MaxExtraPrecision=1000},ContinuedFraction[Tan[1/6],100]] (* Harvey P. Dale, May 14 2014 *) -
PARI
{ allocatemem(932245000); default(realprecision, 95000); x=contfrac(tan(1/6)); for (n=0, 20000, write("b019429.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 13 2009
Formula
Conjecture: a(n) = (-1+3*(-1)^n-6*(-1+(-1)^n)*n)/2 for n>1. a(n) = 2*a(n-2)-a(n-4) for n>5. G.f.: x*(x^4-x^3+6*x^2+x+5) / ((x-1)^2*(x+1)^2). - Colin Barker, May 28 2013
From Peter Bala, Nov 18 2019: (Start)
a(2*n) = 1 and a(2*n+1) = 12*n + 4, both for n >= 1.
The above conjectures are correct. The simple continued fraction expansion for tan(1/6) may be derived by setting z = 1/6 in Lambert's continued fraction tan(z) = z/(1 - z^2/(3 - z^2/(5 - ... ))) and then, after using an equivalence transformation, making repeated use of the identity 1/(n - 1/m) = 1/((n - 1) + 1/(1 + 1/(m - 1))).
A similar approach produces the related simple continued fraction expansions
2*tan(1/6) = [0, 2, 1, 34, 1, 13, 1, 82, 1, 25, 1, 130, 1, 37, 1, 178, 1, 49, ...], with denominators c(2*n) = 1, c(4*n+1) = 12*n + 1, both for n >= 1, and c(4*n+3) = 48*n + 34 for n >= 0;
3*tan(1/6) = [0; 1, 1, 52, 1, 8, 1, 124, 1, 16, 1, 196, 1, 24, 1, 268, 1, 32, ...];
6*tan(1/6) = [1; 106, 1, 3, 1, 250, 1, 7, 1, 394, 1, 11, 1, 538, 1, 15, 1, 682,..];
(1/2)*tan(1/6) = [0, 11, 1, 7, 1, 58, 1, 19, 1, 106, 1, 31, 1, 154, 1, 43, 1, ...];
(1/3)*tan(1/6) = [0, 17, 1, 4, 1, 88, 1, 12, 1, 160, 1, 20, 1, 232, 1, 28, 1, ...];
(1/6)*tan(1/6) = [0, 35, 1, 1, 1, 178, 1, 5, 1, 322, 1, 9, 1, 466, 1, 13, 1, ...];
(End)