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%I #31 Oct 03 2023 08:59:06
%S 0,4,1,13,1,23,1,33,1,43,1,53,1,63,1,73,1,83,1,93,1,103,1,113,1,123,1,
%T 133,1,143,1,153,1,163,1,173,1,183,1,193,1,203,1,213,1,223,1,233,1,
%U 243,1,253,1,263,1,273,1,283,1,293,1,303,1,313,1,323,1,333,1,343,1,353,1,363,1
%N Continued fraction for tan(1/5).
%C The simple continued fraction expansion of 5*tan(1/5) begins [1; 73, 1, 3, 1, 173, 1, 7, 1, 273, 1, 11, 1, 373, 1, 15, 1, 473, 1, 19, 1, 573, ...], while the simple continued fraction expansion of (1/5)*tan(1/5) begins [0; 24, 1, 1, 1, 123, 1, 5, 1, 223, 1, 9, 1, 323, 1, 13, 1, 423, 1, 17, 1, 523, ...]. See my comment in A019425. - _Peter Bala_, Sep 30 2023
%H Harry J. Smith, <a href="/A019428/b019428.txt">Table of n, a(n) for n = 0..20000</a>
%H G. Xiao, <a href="http://wims.unice.fr/~wims/en_tool~number~contfrac.en.html">Contfrac</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,0,-1).
%F From _Colin Barker_, Sep 08 2013: (Start)
%F a(n) = (-1 + 3*(-1)^n - 5*(-1 + (-1)^n)*n)/2 for n > 1.
%F a(n) = 2*a(n-2) - a(n-4) for n > 5.
%F G.f.: x*(x^4-x^3+5*x^2+x+4) / ((x-1)^2*(x+1)^2). (End)
%e 0.20271003550867248332135827... = 0 + 1/(4 + 1/(1 + 1/(13 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 13 2009
%t Join[{0, 4}, LinearRecurrence[{0, 2, 0, -1}, {1, 13, 1, 23}, 100]] (* _Vincenzo Librandi_, Jan 03 2016 *)
%o (PARI) { allocatemem(932245000); default(realprecision, 93000); x=contfrac(tan(1/5)); for (n=0, 20000, write("b019428.txt", n, " ", x[n+1])); } \\ _Harry J. Smith_, Jun 13 2009
%o (PARI) Vec(x*(x^4-x^3+5*x^2+x+4)/((x-1)^2*(x+1)^2) + O(x^100)) \\ _Colin Barker_, Sep 08 2013
%o (Magma) [0,4] cat [(-1+3*(-1)^n-5*(-1+(-1)^n)*n)/2: n in [2..80]]; // _Vincenzo Librandi_, Jan 03 2016
%Y Cf. A161014 (decimal expansion), A019425 through A019433.
%K nonn,cofr,easy
%O 0,2
%A _David W. Wilson_
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