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%I #28 Oct 06 2023 10:28:47
%S 0,9,1,28,1,48,1,68,1,88,1,108,1,128,1,148,1,168,1,188,1,208,1,228,1,
%T 248,1,268,1,288,1,308,1,328,1,348,1,368,1,388,1,408,1,428,1,448,1,
%U 468,1,488,1,508,1,528,1,548,1,568,1,588,1,608,1,628,1,648,1,668,1,688,1,708,1,728
%N Continued fraction for tan(1/10).
%C From _Peter Bala_, Oct 04 2023: (Start)
%C Related simple continued fractions expansions (see my comments in A019425):
%C tan(1/(10*k)) = [0; 10*k - 1, 1, 30*k - 2, 1, 50*k - 2, 1, 70*k - 2, 1, 90*k - 2, 1, ...] for k >= 1.
%C If d is a divisor of 10 with d*d' = 10 then the simple continued fraction expansion of d*tan(1/10) begins [0; d' - 1, 1, 30*d - 2, 1, 5*d' - 2, 1, 70*d - 2, 1, 9*d' - 2, 1, 110*d - 2, 1, 13*d' - 2, ...], while the simple continued fraction expansion of (1/d)*tan(1/10) begins [ 0; 10*d - 1, 1, 3*d'- 2, 1, 50*d - 2, 1, 7*d' - 2, 1, 90*d - 2, 1, 11*d' - 2, 1, 130*d - 2, ...]. (End)
%H Harry J. Smith, <a href="/A019433/b019433.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) = -1/2+(3*(-1)^n)/2+5*n-5*(-1)^n*n for n>1.
%F a(n) = 2*a(n-2)-a(n-4) for n>5.
%F G.f.: x*(x^4-x^3+10*x^2+x+9) / ((x-1)^2*(x+1)^2). (End)
%e 0.10033467208545054505808004... = 0 + 1/(9 + 1/(1 + 1/(28 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 14 2009
%t LinearRecurrence[{0,2,0,-1},{0,9,1,28,1,48},80] (* or *) Join[{0,9},Riffle[NestList[20+#&,28,40],1,{1,-1,2}]] (* _Harvey P. Dale_, Jul 23 2023 *)
%o (PARI) { allocatemem(932245000); default(realprecision, 99000); x=contfrac(tan(1/10)); for (n=0, 20000, write("b019433.txt", n, " ", x[n+1])); } \\ _Harry J. Smith_, Jun 14 2009
%o (PARI) Vec(x*(x^4-x^3+10*x^2+x+9)/((x-1)^2*(x+1)^2) + O(x^100)) \\ _Colin Barker_, Sep 08 2013
%Y Cf. A161019 (decimal expansion), A019425 through A019432.
%K nonn,cofr,easy
%O 0,2
%A _David W. Wilson_
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