

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
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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

Stanislav Sykora, Table of n, a(n) for n = 0..2000
J. M. Borwein, P. B. Borwein, A cubic counterpart of Jacobi's identity and the AGM, Transactions of the AMS, 323 (1991), 691701.
Eric Weisstein's World of Mathematics, ArithmeticGeometric Mean, Equations 2632.


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(anbn)<eps), break));
return((2*Pi/(3*sqrt(3)))/an); }
a = I3(1, 2)


CROSSREFS

Cf. A248897, A257097.
Sequence in context: A021584 A021062 A176436 * A121562 A167902 A154174
Adjacent sequences: A257093 A257094 A257095 * A257097 A257098 A257099


KEYWORD

nonn,cons


AUTHOR

Stanislav Sykora, Apr 16 2015


STATUS

approved



