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%I #42 Mar 03 2024 10:49:30
%S 0,1,1,4,1,8,1,12,1,16,1,20,1,24,1,28,1,32,1,36,1,40,1,44,1,48,1,52,1,
%T 56,1,60,1,64,1,68,1,72,1,76,1,80,1,84,1,88,1,92,1,96,1,100,1,104,1,
%U 108,1,112,1,116,1,120,1,124,1,128,1,132,1,136,1,140,1,144,1,148,1,152,1,156,1
%N Continued fraction for tan(1/2).
%C From _Peter Bala_, Nov 17 2019: (Start)
%C The simple continued fraction expansion for tan(1/2) may be derived by setting z = 1/2 in Lambert's continued fraction tan(z) = z/(1 - z^2/(3 - z^2/(5 - ... ))) and, after using an equivalence transformation, making repeated use of the identity 1/(n - 1/m) = 1/((n - 1) + 1/(1 + 1/(m - 1))).
%C The same approach produces the simple continued fraction expansions for the numbers tan(1/n), n*tan(1/n) and 1/n*tan(1/n) for n = 1,2,3,.... [added Oct 03 2023: and, even more generally, the simple continued fraction expansions for the numbers d*tan(1/n) and 1/d*tan(1/n), where d divides n. See A019429 for an example]. (End)
%H Harry J. Smith, <a href="/A019425/b019425.txt">Table of n, a(n) for n = 0..20000</a>
%H Dan Romik, <a href="https://doi.org/10.1090/S0002-9947-08-04467-X">The dynamics of Pythagorean Triples</a>, Trans. Amer. Math. Soc. 360 (2008), 6045-6064.
%H G. Xiao, <a href="http://wims.unice.fr/~wims/en_tool~number~contfrac.en.html">Contfrac</a>
%H <a href="/index/Con#confC">Index entries for continued fractions for constants</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0, 2, 0, -1).
%F a(n) = n - 1/2 - (n-3/2)*(-1)^n + binomial(1,n) - 2*binomial(0,n). - _Paul Barry_, Oct 25 2007
%F From _Philippe Deléham_, Feb 10 2009: (Start)
%F a(n) = 2*a(n-2) - a(n-4), n>=6.
%F G.f.: (x + x^2 + 2*x^3 - x^4 + x^5)/(1-x^2)^2. (End)
%F From _Peter Bala_, Nov 17 2019; (Start)
%F Related simple continued fraction expansions:
%F 2*tan(1/2) = [1, 10, 1, 3, 1, 26, 1, 7, 1, 42, 1, 11, 1, 58, 1, 15, 1, 74, 1, 19, 1, 90, ...]
%F (1/2)*tan(1/2) = [0; 3, 1, 1, 1, 18, 1, 5, 1, 34, 1, 9, 1, 50, 1, 13, 1, 66, 1, 17, 1, 82, ...]. (End)
%e 0.546302489843790513255179465... = 0 + 1/(1 + 1/(1 + 1/(4 + 1/(1 + ...)))). - _Harry J. Smith_, Jun 13 2009
%p a := n -> if n < 2 then n else ifelse(irem(n, 2) = 0, 1, 2*n - 2) fi:
%p seq(a(n), n = 0..80); # _Peter Luschny_, Oct 03 2023
%t Join[{0, 1}, LinearRecurrence[{0, 2, 0, -1}, {1, 4, 1, 8}, 100]] (* _Vincenzo Librandi_, Jan 03 2016 *)
%o (PARI) { allocatemem(932245000); default(realprecision, 85000); x=contfrac(tan(1/2)); for (n=0, 20000, write("b019425.txt", n, " ", x[n+1])); } \\ _Harry J. Smith_, Jun 13 2009
%o (Magma) [0,1] cat [n-1/2-(n-3/2)*(-1)^n+Binomial(1,n)- 2*Binomial(0,n): n in [2..80]]; // _Vincenzo Librandi_, Jan 03 2016
%Y Cf. A161011 (decimal expansion). Cf. A019426 through A019433.
%K nonn,cofr
%O 0,4
%A _David W. Wilson_
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