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%I #31 Jul 04 2023 03:05:59
%S 9,0,9,1,7,2,7,9,4,5,4,6,9,2,9,7,0,0,7,3,9,7,7,8,8,5,4,2,8,2,6,5,1,2,
%T 2,5,7,2,0,5,2,7,2,9,9,5,9,2,2,0,5,2,2,8,3,8,6,4,1,4,0,2,1,8,3,7,2,2,
%U 3,6,4,8,1,1,1,2,7,1,8,9,9,3,2,3,2,5,6,7,4,0,5,7,0,5,1,3,7,9,5,3,3,7,3
%N Decimal expansion of (sqrt(Pi)/(2^(1/4)*Gamma(5/8)*Gamma(7/8)))^2.
%C This is the value of hypergeometric([1/4,1/4],[1],-1)^2. See A278143/A241756 for the partial sums of the hypergeometric series hypergeometric([1/2/,1/2,1/2],[1,1],-1) which has this value due to Clausen's formula. See the Hardy reference, p. 106, eq. (7.4.4) where this value is written as (Gamma(9/8)/(Gamma(5/4)*Gamma(7/8)))^2.
%D G. H. Hardy, Ramanujan, AMS Chelsea Publ., Providence, RI, 2002, p. 106, eq. (7.4.4)
%H G. C. Greubel, <a href="/A278144/b278144.txt">Table of n, a(n) for n = 0..5000</a>
%F Equals hypergeometric([1/2/,1/2,1/2],[1,1],-1) = hypergeometric([1/4,1/4],[1],-1)^2 = Sum_{k>=0} (-1)^k*(risefac(k,1/2)/k!)^3, where risefac(x,m) = Product_{j =0..m-1} (x+j), and risefac(x,0) = 1.
%F Equals (Gamma(9/8)/(Gamma(5/4)*Gamma(7/8)))^2 = (sqrt(Pi)/(2^(1/4)*Gamma(5/8)*Gamma(7/8)))^2.
%F Equals Sum_{k>=0} (-1)^k * binomial(2*k,k)^3/64^k. - _Amiram Eldar_, Jul 04 2023
%F Equals Gamma(1/8)^4 * (2 - sqrt(2)) / (16 * Pi^2 * Gamma(1/4)^2). - _Vaclav Kotesovec_, Jul 04 2023
%e The value of the series 1 - (1/2)^3 + (1*3/(2*4))^3 - (1*3*5/(2*4*6) + ... is 0.909172794546929700739778854282651225720527299592205228386414021837...
%e This is also the value of the series Sum_{n>=0} c(n) with c(n) = Sum_{k=0..n} f(k)*f(n-k), where f(0)=1 and f(k) = (-1)^k*(1*5*9 *** (4*k-3)/(4*8*12 *** (4*k)))^2, k >= 1 (self-convolution of the hypergeometric([1/4,1/4],[1],-1) series).
%t RealDigits[(Pi/Sqrt[2])*(1/(Gamma[5/8]*Gamma[7/8]))^2, 10, 50][[1]] (* _G. C. Greubel_, Jan 12 2017 *)
%o (PARI) (sqrt(Pi)/(2^(1/4)*gamma(5/8)*gamma(7/8)))^2 \\ _Felix Fröhlich_, Nov 15 2016
%o (Magma) pi:=Pi(RealField(110)); (Sqrt(pi)/(2^(1/4)*Gamma(5/8)*Gamma(7/8)))^2 // _Felix Fröhlich_, Nov 15 2016
%Y Cf. A278143.
%K nonn,cons
%O 0,1
%A _Wolfdieter Lang_, Nov 14 2016
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