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A091476
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Decimal expansion of Pi^2/4.
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20
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2, 4, 6, 7, 4, 0, 1, 1, 0, 0, 2, 7, 2, 3, 3, 9, 6, 5, 4, 7, 0, 8, 6, 2, 2, 7, 4, 9, 9, 6, 9, 0, 3, 7, 7, 8, 3, 8, 2, 8, 4, 2, 4, 8, 5, 1, 8, 1, 0, 1, 9, 7, 6, 5, 6, 6, 0, 3, 3, 3, 7, 3, 4, 4, 0, 5, 5, 0, 1, 1, 2, 0, 5, 6, 0, 4, 8, 0, 1, 3, 1, 0, 7, 5, 0, 4, 4, 3, 3, 5, 0, 9, 2, 9, 6, 3, 8, 0, 5, 7, 9, 5
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OFFSET
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1,1
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LINKS
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FORMULA
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Equals Integral_{x=0..Pi} x*sin(x)/(1+cos(x)^2) dx.
Equals Integral_{x=0..oo} K_0(x)^2 dx, where K_0 is a modified Bessel function (see Gradstein-Ryshik 6.576.4). - R. J. Mathar, Oct 09 2015
Equals 2*Sum_{k > 0} 1/(2*k - 1)^2. (End)
Pi^2/4 = Integral_{x = 0..oo} x/sinh(x) dx. More generally, Pi^2/4 = 2*(1 + 1/3^2 + ... + 1/(2*n-1)^2) + Integral_{x = 0..oo} exp(-2*n*x)*x/sinh(x). - Peter Bala, Nov 05 2019
Equals Integral_{x=0..oo} log(x)/(x^2 - 1) dx. - Amiram Eldar, Aug 12 2020
Equals Sum_{n >= 0} 2^(n+1)/((n+1)^2*binomial(2*n+1,n)). See my entry in A002544 dated Apr 18 2017. Cf. A253191. - Peter Bala, Jan 30 2023
Pi^2/4 = 16*Sum_{k >= 1} k^2/(4*k^2 - 1)^2 = (2*16^2)*Sum_{k >= 1} k^2/((4*k^2 - 1)*(4*k^2 - 9))^2.
The general result, which can be proved using the WZ method (see Wilf for examples of this method), is that for n >= 0 there holds
Pi^2/4 = 16^(n+1)*(2*n + 1)*(2*n)!^4/(4*n)! * Sum_{k >= 1} k^2/( (4*k^2 - 1)*(4*k^2 - 9)*...*(4*k^2 - (2*n+1)^2) )^2. (End)
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EXAMPLE
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2.46740110027233965470862274996903778...
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MAPLE
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MATHEMATICA
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First[RealDigits[Pi^2/4, 10, 100]] (* Paolo Xausa, Oct 31 2023 *)
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PROG
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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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