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Continued fraction for tan(1/9).
4

%I #34 Apr 28 2024 22:27:20

%S 0,8,1,25,1,43,1,61,1,79,1,97,1,115,1,133,1,151,1,169,1,187,1,205,1,

%T 223,1,241,1,259,1,277,1,295,1,313,1,331,1,349,1,367,1,385,1,403,1,

%U 421,1,439,1,457,1,475,1,493,1,511,1,529,1,547,1,565,1,583,1,601,1,619,1,637,1,655

%N Continued fraction for tan(1/9).

%C The odd-indexed terms from and after a(3) are equal to 18n+7. - _Harvey P. Dale_, Sep 26 2021

%H Harry J. Smith, <a href="/A019432/b019432.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-9*(-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+9*x^2+x+8) / ((x-1)^2*(x+1)^2). (End)

%e 0.11157062783380058372650480... = 0 + 1/(8 + 1/(1 + 1/(25 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 14 2009

%t Block[{$MaxExtraPrecision=1000}, ContinuedFraction[Tan[1/9],100]] (* or *) LinearRecurrence[{0,2,0,-1},{0,8,1,25,1,43},80] (* _Harvey P. Dale_, Sep 26 2021 *)

%o (PARI) { allocatemem(932245000); default(realprecision, 98000); x=contfrac(tan(1/9)); for (n=0, 20000, write("b019432.txt", n, " ", x[n+1])); } \\ _Harry J. Smith_, Jun 14 2009

%o (PARI) Vec(x*(x^4-x^3+9*x^2+x+8)/((x-1)^2*(x+1)^2) + O(x^100)) \\ _Colin Barker_, Sep 08 2013

%Y Cf. A161018 (decimal expansion), A019425 through A019433.

%K nonn,cofr,easy

%O 0,2

%A _David W. Wilson_