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A019429 Continued fraction for tan(1/6).

%I A019429 #25 Mar 03 2024 10:51:25
%S A019429 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,
%T A019429 1,160,1,172,1,184,1,196,1,208,1,220,1,232,1,244,1,256,1,268,1,280,1,
%U A019429 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
%N A019429 Continued fraction for tan(1/6).
%H A019429 Harry J. Smith, <a href="/A019429/b019429.txt">Table of n, a(n) for n = 0..20000</a>
%H A019429 G. Xiao, <a href="http://wims.unice.fr/~wims/en_tool~number~contfrac.en.html">Contfrac</a>
%H A019429 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0, 2, 0, -1).
%F A019429 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
%F A019429 From _Peter Bala_, Nov 18 2019: (Start)
%F A019429 a(2*n) = 1 and a(2*n+1) = 12*n + 4, both for n >= 1.
%F A019429 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))).
%F A019429 A similar approach produces the related simple continued fraction expansions
%F A019429 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;
%F A019429 3*tan(1/6) = [0; 1, 1, 52, 1, 8, 1, 124, 1, 16, 1, 196, 1, 24, 1, 268, 1, 32, ...];
%F A019429 6*tan(1/6) = [1; 106, 1, 3, 1, 250, 1, 7, 1, 394, 1, 11, 1, 538, 1, 15, 1, 682,..];
%F A019429 (1/2)*tan(1/6) = [0, 11, 1, 7, 1, 58, 1, 19, 1, 106, 1, 31, 1, 154, 1, 43, 1, ...];
%F A019429 (1/3)*tan(1/6) = [0, 17, 1, 4, 1, 88, 1, 12, 1, 160, 1, 20, 1, 232, 1, 28, 1, ...];
%F A019429 (1/6)*tan(1/6) = [0, 35, 1, 1, 1, 178, 1, 5, 1, 322, 1, 9, 1, 466, 1, 13, 1, ...];
%F A019429 (End)
%e A019429 0.16822721830224246125721608... = 0 + 1/(5 + 1/(1 + 1/(16 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 13 2009
%t A019429 Block[{$MaxExtraPrecision=1000},ContinuedFraction[Tan[1/6],100]] (* _Harvey P. Dale_, May 14 2014 *)
%o A019429 (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
%Y A019429 Cf. A161015 (decimal expansion). Cf. A019426 through A019433.
%K A019429 nonn,cofr
%O A019429 0,2
%A A019429 _David W. Wilson_