%I #12 Apr 24 2015 18:29:04
%S 9,1,0,0,7,6,2,7,2,8,9,6,6,4,4,9,9,4,5,9,3,5,6,4,3,3,4,7,1,4,6,3,0,2,
%T 0,7,5,4,2,2,9,2,7,9,7,5,4,1,4,8,8,0,8,1,3,6,5,2,5,9,0,4,5,9,6,5,8,1,
%U 4,1,1,1,3,2,3,7,4,6,6,2,8,2,4,3,5,9,8,0,0,8,5,1,7,9,5,2,2,1,2,8,1,6,3,7,1
%N Decimal expansion of I3(u,v) = A248897/AG3(u,v) for u=2, v=1.
%C For positive u and v, AG3(u,v) is defined as the common limit of u_k, v_k such that u_0=u, v_0=v, u_(k+1)=(u_k+2*v_k)/3, v_(k+1)=(v_k*(u_k*u_k+u_k*v_k+v_k*v_k)/3)^(1/3). Since the iterative algorithm is similar to that for AGM, AG3 is sometimes referred to as "cubic AGM".
%C An alternative definition of I3(u,v) is by means of the definite integral I3(u,v) = Integral[x=0,inf](x/((u^3+x^3)*(v^3+x^3)^2)^(1/3)).
%H Stanislav Sykora, <a href="/A257097/b257097.txt">Table of n, a(n) for n = 0..2000</a>
%H J. M. Borwein, P. B. Borwein, <a href="http://dx.doi.org/10.1090/S0002-9947-1991-1010408-0">A cubic counterpart of Jacobi's identity and the AGM</a>, Transactions of the AMS, 323 (1991), 691-701.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Arithmetic-GeometricMean.html">Arithmetic-Geometric Mean</a>, Equations 26-32.
%F Equals Integral[x=0,inf](x/((8+x^3)*(1+x^3)^2)^(1/3)).
%e 0.9100762728966449945935643347146302075422927975414880813652590...
%o (PARI) I3(u,v)={my(an=u+0.0,bn=v+0.0,anext=0.0,ncyc=0,
%o eps=2*10^(-default(realprecision)));
%o while(1, anext=(an+2*bn)/3;
%o bn=(bn*(an*an+an*bn+bn*bn)/3)^(1/3); an=anext;
%o ncyc++; if((ncyc>3)&&(abs(an-bn)<eps),break));
%o return((2*Pi/(3*sqrt(3)))/an);}
%o a = I3(2,1)
%Y Cf. A248897, A257096.
%K nonn,cons
%O 0,1
%A _Stanislav Sykora_, Apr 16 2015
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