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A105531
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Decimal expansion of arctan 1/3.
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6
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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
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OFFSET
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0,1
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COMMENTS
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LINKS
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FORMULA
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arctan(1/3) = (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) + ...))))).
arctan(1/3) = (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>=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
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EXAMPLE
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0.3217505543966421934014046143...
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MATHEMATICA
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RealDigits[ArcTan[1/3], 10, 120][[1]] (* Harvey P. Dale, Oct 28 2011 *)
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PROG
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CROSSREFS
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KEYWORD
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AUTHOR
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Bryan Jacobs (bryanjj(AT)gmail.com), Apr 12 2005
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STATUS
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approved
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