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

0,1

Comments

arctan(1/3) + A073000 = 2*arctan(1/3) + A105533 = Pi/4.

Links

Table of n, a(n) for n=0..104.

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

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

Example

0.3217505543966421934014046143...

Mathematica

RealDigits[ArcTan[1/3], 10, 120][[1]] (* Harvey P. Dale, Oct 28 2011 *)

Prog

(PARI) atan(1/3) \\ Michel Marcus, Mar 29 2016

Crossrefs

Cf. A003881 (Pi/4), A073000 (arctan(1/2)), A105533 (arctan(1/7)).

Sequence in context: A067050 A001355 A364855 * A369041 A129689 A115990

Adjacent sequences: A105528 A105529 A105530 * A105532 A105533 A105534

Keyword

cons,nonn

Author

Bryan Jacobs (bryanjj(AT)gmail.com), Apr 12 2005

Status

approved