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%I #64 Apr 03 2023 10:36:10
%S 3,1,0,0,6,2,7,6,6,8,0,2,9,9,8,2,0,1,7,5,4,7,6,3,1,5,0,6,7,1,0,1,3,9,
%T 5,2,0,2,2,2,5,2,8,8,5,6,5,8,8,5,1,0,7,6,9,4,1,4,4,5,3,8,1,0,3,8,0,6,
%U 3,9,4,9,1,7,4,6,5,7,0,6,0,3,7,5,6,6,7,0,1,0,3,2,6,0,2,8,8,6,1,9
%N Decimal expansion of Pi^3.
%C Surface area of the 6-dimensional unit sphere. - _Stanislav Sykora_, Nov 08 2013
%C Surface area of a sphere of diameter Pi equals the volume of the circumscribed cube. - _Omar E. Pol_, Dec 25 2013
%C Area of a circle of radius Pi. - _Omar E. Pol_, Jan 31 2016
%H Harry J. Smith, <a href="/A091925/b091925.txt">Table of n, a(n) for n = 2..20000</a>
%H G. L. Honaker, Jr. and Chris Caldwell, <a href="https://t5k.org/curios/cpage/827.html">Prime Curios! 31</a>
%H Jean-Christophe Pain, <a href="https://arxiv.org/abs/2206.15281">Series representations for Pi^3 involving the golden ratio</a>, arXiv:2206.15281 [math.NT], 2022.
%H Khodabakhsh Hessami Pilehrood, Tatiana Hessami Pilehrood, <a href="https://dmtcs.episciences.org/504">Series acceleration formulas for beta values</a>, Discrete Mathematics & Theoretical Computer Science 12:2 (2010), pp. 223-236.
%H Stanislav Sykora, <a href="http://dx.doi.org/10.3247/SL1Math05.002">Surface Integrals over n-Dimensional Spheres</a>.
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%F Sum_{k >= 0} binomial(2*k,k)/((2*k + 1)^3*16^k) = 7*Pi^3/216. (Kh. Hessami Pilehrood and T. Hessami Pilehrood).
%F From _Peter Bala_, Feb 05 2015: (Start)
%F The integer sequences A(n) := 2^n*(2*n + 1)!^3/n!^2 and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)^3*(1/16)^k ) both satisfy the second order recurrence equation u(n) = (160*n^4 + 128*n^3 + 144*n^2 + 2)*u(n-1) - 32*(n - 1)*(2*n - 1)^7*u(n-2). From this observation we can obtain the continued fraction expansion 7/216*Pi^3 = 1 + 2/(432 - 32*3^7/(4162 - 32*2*5^7/(17714 - ... - 32*(n - 1)*(2*n - 1)^7/((160*n^4 + 128*n^3 + 144*n^2 + 2) - ... )))). Cf. A002388, A019670 and A093954. (End)
%F From _Peter Bala_, Oct 31 2019: (Start)
%F Pi^3 = (1/7) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/6)^3 + 1/(n + 5/6)^3 ).
%F Pi^3 = (1/31) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/10)^3 - 1/(n + 3/10)^3 - 1/(n + 7/10)^3 + 1/(n + 9/10)^3 ). Cf. A019692, A092731 and A092735. (End)
%F Equals Integral_{x=-oo..oo} x^2/(exp(x/2) + exp(-x/2)) dx. - _Amiram Eldar_, May 21 2021
%e 31.00627668029982017547631506710139520222528856588510769414453810380639...
%t First@ RealDigits@ N[Pi^3, 120] (* _Michael De Vlieger_, Jan 31 2016 *)
%o (PARI) default(realprecision, 20080); x=Pi^3/10; for (n=2, 20000, d=floor(x); x=(x-d)*10; write("b091925.txt", n, " ", d)); \\ _Harry J. Smith_, Jun 22 2009
%o (Magma) R:= RealField(100); (Pi(R))^3; // _G. C. Greubel_, Mar 09 2018
%Y Cf. A000796, A002388, A058285 (continued fraction), A019670, A093954, A092731, A092735.
%K easy,nonn,cons
%O 2,1
%A _Mohammad K. Azarian_, Mar 16 2004
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