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%I #64 Feb 16 2022 23:42:15
%S 6,3,6,2,2,6,5,1,3,1,5,6,7,3,2,8,3,9,3,6,9,1,2,4,5,4,4,0,5,8,6,8,0,4,
%T 4,1,0,6,9,9,7,1,4,9,8,5,1,3,8,9,8,9,6,8,6,5,8,2,0,4,1,6,1,7,0,4,5,9,
%U 9,8,5,8,7,3,3,1,7,8,4,8,5,4,1,3,4,5,5,0,8,7,7,1,3
%N Decimal expansion of 1/sin(arctan(1/t)) or t/sin(arctan(t)) where t = 2*Pi: hypotenuse for a right triangle of equal area to a disk.
%C "The great mathematician Archimedes used the tools of Euclidean geometry to show that the area inside a circle is equal to that of a right triangle whose base has the length of the circle's circumference and whose height equals the circle's radius in his book Measurement of a Circle." (Quote from Wikipedia link)
%H G. C. Greubel, <a href="/A233700/b233700.txt">Table of n, a(n) for n = 1..10000</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Area_of_a_disk#Triangle_method">Area of a disk: Triangle method</a>
%F Equals sqrt(1+(2*Pi)^2) = sqrt(1 + (A019692)^2) = sqrt(1 + A212002) = 1/sin(A233527) = A019692/sin(A233528) = 1/cos(A233528) = A019692/cos(A233527).
%e 6.362265131567328393691245440586804410699714985138989686582041617045998587331...
%t RealDigits[(2*Pi)/Sin[ArcTan[2*Pi]],10,120][[1]] (* _Harvey P. Dale_, Jul 12 2014 *)
%t RealDigits[ Sqrt[1 + 4*Pi^2], 10, 111][[1]] (* _Robert G. Wilson v_, Mar 12 2015 *)
%o (PARI) sqrt(1+(2*Pi)^2)
%o (Magma) C<i> := ComplexField(); Sqrt(1 + 4*Pi(C)^2) // _G. C. Greubel_, Jan 08 2018
%o (Magma) R:=RealField(110); SetDefaultRealField(R); n:=Sqrt(1+4*Pi(R)^2); Reverse(Intseq(Floor(10^108*n))); // _Bruno Berselli_, Mar 13 2018
%o (Julia)
%o using Nemo
%o RR = RealField(310)
%o t = const_pi(RR) + const_pi(RR)
%o t/sin(atan(t)) |> println # _Peter Luschny_, Mar 13 2018
%Y Cf. A233527, A019692, A233528.
%K nonn,cons,nice
%O 1,1
%A _John W. Nicholson_, Dec 16 2013
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