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%I #48 Mar 21 2024 06:38:01
%S 4,6,3,6,4,7,6,0,9,0,0,0,8,0,6,1,1,6,2,1,4,2,5,6,2,3,1,4,6,1,2,1,4,4,
%T 0,2,0,2,8,5,3,7,0,5,4,2,8,6,1,2,0,2,6,3,8,1,0,9,3,3,0,8,8,7,2,0,1,9,
%U 7,8,6,4,1,6,5,7,4,1,7,0,5,3,0,0,6,0,0,2,8,3,9,8,4,8,8,7,8,9,2,5,5,6,5,2,9
%N Decimal expansion of arctangent of 1/2.
%C The angle at which you must shoot a cue ball on a standard pool table so that it will strike all four sides and return to its origin. [Barrow] - _Robert G. Wilson v_, Nov 29 2015
%D John D. Barrow, One Hundred Essential Things You Didn't Know You Didn't Know, W. W. Norton & Co., NY & London, 2008.
%H G. C. Greubel, <a href="/A073000/b073000.txt">Table of n, a(n) for n = 0..5000</a>
%H Peter Bala, <a href="/A002117/a002117.pdf">New series for old functions</a>.
%H R. J. Mathar, <a href="http://arxiv.org/abs/1309.3705">Hierarchical Subdivision of the Simple Cubic Lattice</a>, arXiv preprint arXiv:1309.3705 [math.MG], 2013.
%H Simon Plouffe, <a href="http://www.worldwideschool.org/library/books/sci/math/MiscellaneousMathematicalConstants/chap7.html">arctan(1/2)</a>.
%F Equals Pi/2 - A105199 = A019669 - A105199. - _R. J. Mathar_, Aug 21 2013
%F From _Peter Bala_, Feb 04 2015: (Start)
%F Arctan(1/2) = 1/2*Sum_{k >= 0} (-1)^k/((2*k + 1)*4^k).
%F Define a pair of integer sequences A(n) = 4^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} (-1)^k/((2*k + 1)*4^k). Both sequences satisfy the same second order recurrence equation u(n) = (12*n + 10)*u(n-1) + 16*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion 2*arctan(1/2) = 1 - 2/(24 + 16*3^2/(34 + 16*5^2/(46 + ... + 16*(2*n - 1)^2/((12*n + 10) + ...)))). See A002391, A105531 and A002162 for similar expansions.
%F Arctan(1/2) = 2/5 * Sum_{k >= 0} (4/5)^k/((2*k + 1)*binomial(2*k,k)).
%F Define a pair of integer sequences C(n) = 5^n*(2*n + 1)!/n! and D(n) = C(n)*Sum_{k = 0..n} (4/5)^k/((2*k + 1)*binomial(2*k,k)). Both sequences satisfy the same second order recurrence equation u(n) = (24*n + 10)*u(n-1) - 40*n*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion 5/2*arctan(1/2) = 1 + 4/(30 - 240/(58 - 600/(82 - ... - 40*n*(2*n - 1)/((24*n + 10) - ... )))).
%F Arctan(1/2) = 2/25 * Sum_{k >= 0} (24*k + 17)*(4/5)^(2*k)/( (4*k + 1)*(4*k + 3)*binomial(4*k,2*k) ).
%F Arctan(1/2) = 2/125 * Sum_{k >= 0} (1116*k^2 + 1446*k + 433)*(4/5)^(3*k)/( (6*k + 1)*(6*k + 3)*(6*k + 5)*binomial(6*k,3*k) ). (End)
%F Equals Integral_{x = 0..inf} exp(-2*x)*sin(x)/x dx. - _Peter Bala_, Nov 05 2019
%F Equals 2 * arccot(phi^3), where phi is the golden ratio (A001622). - _Amiram Eldar_, Jul 06 2023
%F Equals Sum_{n >= 1} i/(n*P(n, 2*i)*P(n-1, 2*i)) = (1/2)*Sum_{n >= 1} (-1)^(n+1)*4^n/(n*A098443(n)*A098443(n-1)), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The n-th summand of the series is O( 1/(3 + 2*sqrt(2))^n ). - _Peter Bala_, Mar 16 2024
%e Arctan(1/2)
%e =0.463647609000806116214256231461214402028537054286120263810933088720197864165... radians
%e =26°.56505117707798935157219372045329467120421429964522102798601631528806582148474...
%e =26°33'.9030706246793610943316232271976802722528579787132616791609789172839492890...
%e =26°33'54".184237480761665659897393631860816335171478722795700749658735037036957...
%e complement = 63°.43494882292201064842780627954670532879578570035477897201398368471...
%e supplement = 153°.4349488229220106484278062795467053287957857003547789720139836847...
%p evalf(arctan(0.5)) ; # _R. J. Mathar_, Aug 22 2013
%t RealDigits[ ArcTan[1/2], 10, 110] [[1]]
%o (PARI) default(realprecision,2000);atan(1/2) \\ _Anders Hellström_, Nov 30 2015
%Y Cf. A001622, A002162, A002391, A105531, A254619.
%K cons,nonn
%O 0,1
%A _Robert G. Wilson v_, Aug 03 2002
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