A144184 - OEIS

%I #12 Nov 21 2013 13:11:53

%S 5,5,0,7,9,8,5,6,5,2,7,7,3,1,7,8,2,5,7,5,8,9,0,2,6,2,9,8,0,5,2,1,3,8,

%T 7,3,0,0,1,6,0,2,4,6,6,3,3,0,4,1,1,8,2,2,9,8,8,3,0,2,8,6,8,5,1,9,3,3,

%U 6,8,2,3,8,2,0,3,9,0,2,5,8,1,7,5,5,8,0,6,6,4,8,9,4,9,7,9,6,3,9,4

%N Decimal expansion of the convergent to the recurrence x = 1/(x^(1/x)-1/x-1) for all starting values of x >= 3.

%C 1/(x^(1/x)-1/x-1) ~ pi(x), the number of prime numbers <= x. This is comparable to the well known approximation Pi(x) ~ x/(log(x)-1). As x -> infinity, pi(x) - 1/(x^(1/x)-1/x-1) -> 1/2 as x-> infinity. This was derived from my original n-th root formula 1/(x^(1/x)-1) ~ pi(x). The convergent of the recurrence x = 1/(x^(1/x)-1) = 2.293166287... is expanded in A085846 and is referred to as Foias constant. The convergents 5.507985652... and 2.293166287... are both roots of 1/(x^(1/x)-1/x-1)-x = 0. 2.293166287... is also a root of 1/(x^(1/x)-1) - x = 0.

%C We have here examples of functions, f(x), for which we can solve for a root by recursion of the variable x. Another simple example is the recursion x = 1/(x+1).

%H Eric Weisstein, <a href="http://mathworld.wolfram.com/FoiasConstant.html">Foias Constant</a>

%F The convergent used to generate this sequence, 5.50798565277317825758902..., is computed with the recurrence x = 1/(x^(1/x)-1/x-1) and can also be found by solving for the roots of 1/(x^(1/x)-1/x-1)-x = 0.

%t RealDigits[ x /. FindRoot[ 1/(x^(1/x) - 1/x - 1) - x == 0, {x, 5}, WorkingPrecision -> 100]][[1]] (* _Jean-François Alcover_, Dec 20 2011 *)

%o (PARI) g(x) = 1/(x^(1/x)-1/x-1) g2(n) = a=n;for(j=1,100,a=g(a));b=eval(Vec(Str(floor(a*10^99))));

%o for(j=1,100,print1(b[j]","))

%Y Cf. A085846.

%K nonn,cons

%O 1,1

%A _Cino Hilliard_, Sep 13 2008, Sep 15 2008