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%I #44 Feb 24 2023 03:10:46
%S 3,2,1,7,5,0,5,5,4,3,9,6,6,4,2,1,9,3,4,0,1,4,0,4,6,1,4,3,5,8,6,6,1,3,
%T 1,9,0,2,0,7,5,5,2,9,5,5,5,7,6,5,6,1,9,1,4,3,2,8,0,3,0,5,9,3,5,6,7,5,
%U 6,2,3,7,4,0,5,8,1,0,5,4,4,3,5,6,4,0,8,4,2,2,3,5,0,6,4,1,3,7,4,4,3,9,0,0,7
%N Decimal expansion of arctan 1/3.
%C arctan(1/3) + A073000 = 2*arctan(1/3) + A105533 = Pi/4.
%H Peter Bala, <a href="/A002117/a002117.pdf">New series for old functions</a>
%H Kunle Adegoke, <a href="http://arxiv.org/abs/1603.08097">Infinite arctangent sums involving Fibonacci and Lucas numbers</a>, arXiv:1603.08097 [math.NT], 2016.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Machin-LikeFormulas.html">Machin-Like Formulas</a>
%F From _Peter Bala_, Feb 04 2015: (Start)
%F arctan(1/3) = (1/3)*Sum_{k >= 0} (-1)^k/((2*k + 1)*9^k).
%F Define a pair of integer sequences A(n) = 9^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} (-1)^k/((2*k + 1)*9^k). Both sequences satisfy the same recurrence equation u(n) = (32*n + 20)*u(n-1) + 36*(2*n - 1)^2*u(n-2). From this observation we find the continued fraction expansion arctan(1/3) = (1/3)*(1 - 2/(54 + 36*3^2/(84 + 36*5^2/(116 + ... + 36*(2*n - 1)^2/((32*n + 20) + ...))))).
%F arctan(1/3) = (3/10) * Sum_{k >= 0} (2/5)^k/( (2*k + 1)*binomial(2*k,k) ).
%F Define a pair of integer sequences C(n) = 10^n*(2*n + 1)!/n! and D(n) = C(n)*Sum_{k = 0..n} (2/5)^k/( (2*k + 1)*binomial(2*k,k) ). Both sequences satisfy the same recurrence equation u(n) = (44*n + 20)*u(n-1) - 80*n*(2*n - 1)*u(n-2). From this observation we obtain the continued fraction expansion arctan(1/3) = (3/10)*( 1 + 4/(60 - 480/(108 - 1200/(152 - ... - 80*n*(2*n - 1)/((44*n + 20) - ...))))). (End)
%F arctan(1/3) = Sum_{k>=0} arctan((L(4k+2)/F(4k+2)^2) where L=A000032 and F=A000045. See also A033890 and A246453. - _Michel Marcus_, Mar 29 2016
%F From _Amiram Eldar_, Aug 09 2020: (Start)
%F Equals Sum_{k>=2} arctan(1/(2*k^2)) = Sum_{k>=2} (-1)^k arctan(2/k^2).
%F Equals Integral_{x=1..2} 1/(x^2 + 1) dx. (End)
%F Equals Sum_{n>=0} arctan(1/F(2*n+5)) = Sum_{n>=0} (-1)^n arctan(F(2*n+1)) where F=A000045. - _Gleb Koloskov_, Oct 01 2021
%e 0.3217505543966421934014046143...
%t RealDigits[ArcTan[1/3],10,120][[1]] (* _Harvey P. Dale_, Oct 28 2011 *)
%o (PARI) atan(1/3) \\ _Michel Marcus_, Mar 29 2016
%K cons,nonn
%O 0,1
%A Bryan Jacobs (bryanjj(AT)gmail.com), Apr 12 2005
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