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A257096
Decimal expansion of I3(u,v) = A248897/AG3(u,v) for u=1, v=2.
3
7, 2, 4, 2, 3, 5, 6, 3, 3, 8, 0, 0, 9, 7, 1, 4, 2, 9, 5, 3, 8, 9, 2, 3, 3, 3, 1, 1, 1, 1, 5, 0, 1, 8, 3, 8, 3, 3, 0, 9, 7, 6, 3, 4, 4, 6, 8, 3, 2, 9, 5, 5, 3, 0, 4, 9, 8, 9, 2, 4, 7, 6, 0, 7, 2, 5, 1, 1, 4, 3, 5, 6, 4, 7, 3, 6, 3, 5, 5, 8, 5, 5, 2, 3, 5, 8, 4, 6, 2, 2, 3, 9, 6, 1, 3, 9, 4, 0, 3, 8, 9, 3, 8, 5, 4
OFFSET
0,1
COMMENTS
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".
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)).
LINKS
J. M. Borwein, P. B. Borwein, A cubic counterpart of Jacobi's identity and the AGM, Transactions of the AMS, 323 (1991), 691-701.
Eric Weisstein's World of Mathematics, Arithmetic-Geometric Mean, Equations 26-32.
FORMULA
Equals Integral[x=0,inf](x/((1+x^3)*(8+x^3)^2)^(1/3)).
EXAMPLE
0.724235633800971429538923331111501838330976344683295530...
MATHEMATICA
RealDigits[ NIntegrate[(x/((1 + x^3) (8 + x^3)^2)^(1/3)), {x, 0, Infinity}, AccuracyGoal -> 111, WorkingPrecision -> 111]][[1]] (* Robert G. Wilson v, Apr 16 2015 *)
PROG
(PARI) I3(u, v)={my(an=u+0.0, bn=v+0.0, anext=0.0, ncyc=0,
eps=2*10^(-default(realprecision)));
while(1, anext=(an+2*bn)/3;
bn=(bn*(an*an+an*bn+bn*bn)/3)^(1/3); an=anext;
ncyc++; if((ncyc>3)&&(abs(an-bn)<eps), break));
return((2*Pi/(3*sqrt(3)))/an); }
a = I3(1, 2)
CROSSREFS
Sequence in context: A021584 A021062 A176436 * A121562 A167902 A356032
KEYWORD
nonn,cons
AUTHOR
Stanislav Sykora, Apr 16 2015
STATUS
approved