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A318741
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Decimal expansion of Pi^3/48 + Pi*log(2)^2/4.
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0
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1, 0, 2, 3, 3, 1, 1, 0, 1, 2, 2, 3, 6, 3, 7, 0, 3, 2, 3, 0, 8, 4, 8, 2, 0, 5, 0, 4, 0, 8, 8, 4, 8, 6, 7, 3, 8, 3, 1, 8, 7, 2, 0, 9, 7, 6, 7, 4, 7, 3, 2, 8, 1, 3, 0, 3, 0, 5, 1, 3, 4, 2, 7, 6, 3, 6, 2, 9, 5, 3, 3, 4, 3, 9, 7, 5, 6, 0, 8, 6, 6, 8, 2, 9, 2, 3, 4
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OFFSET
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1,3
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COMMENTS
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The first part of Ramanujan's question 308 in the Journal of the Indian Mathematical Society (III, 168) asked "Show that Integral_{t=0..Pi/2} t * cotan(t) * log(sin(t)) dt = -Pi^3/48 - Pi*log(2)^2/4".
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LINKS
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Pedro Ribeiro, Problem 12051, The American Mathematical Monthly, Vol. 125, No. 6 (2018), p. 563; A Series Involving Central Binominal [sic] Coefficients, Solution to Problem 12051 by Hongwei Chen, ibid., Vol. 127, No. 2 (2020), pp. 182-183.
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FORMULA
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Equals Sum_{k>=0} binomial(2*k,k)/(4^k*(2*k+1)^3) (Ribeiro, 2018). - Amiram Eldar, Oct 04 2021
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EXAMPLE
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1.0233110122363703230848205040884867383187209767473281303051342763...
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MATHEMATICA
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RealDigits[Pi^3/48 + Pi*Log[2]^2/4, 10, 100][[1]] (* Amiram Eldar, Oct 04 2021 *)
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PROG
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(PARI) Pi^3/48+Pi*log(2)^2/4
(PARI) -intnum(x=0, Pi/2, x*cotan(x)*log(sin(x)))
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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