A105531 Decimal expansion of arctan(1/3).
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, 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, 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
Offset: 0
Examples
0.3217505543966421934014046143...
References
- John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 242.
Links
- Peter Bala, New series for old functions
- Kunle Adegoke, Infinite arctangent sums involving Fibonacci and Lucas numbers, arXiv:1603.08097 [math.NT], 2016.
- Eric Weisstein's World of Mathematics, Machin-Like Formulas
- Index entries for transcendental numbers
Programs
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Mathematica
RealDigits[ArcTan[1/3],10,120][[1]] (* Harvey P. Dale, Oct 28 2011 *)
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PARI
atan(1/3) \\ Michel Marcus, Mar 29 2016
Formula
From Peter Bala, Feb 04 2015: (Start)
Equals (1/3)*Sum_{k >= 0} (-1)^k/((2*k + 1)*9^k).
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) + ...))))).
Equals (3/10) * Sum_{k >= 0} (2/5)^k/( (2*k + 1)*binomial(2*k,k) ).
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)
Equals Sum_{k>=1} arctan(L(4*k+2)/F(4*k+2)^2) where L=A000032 and F=A000045. See also A033890 and A246453. - Michel Marcus, Mar 29 2016 [corrected by Jason Yuen, Jan 18 2025]
From Amiram Eldar, Aug 09 2020: (Start)
Equals Sum_{k>=2} arctan(1/(2*k^2)) = Sum_{k>=2} (-1)^k arctan(2/k^2).
Equals Integral_{x=1..2} 1/(x^2 + 1) dx. (End)
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
Comments