a(n+1) = log(1+a(n))

[Elim Qiu <elim.qiu@gmail.com>]

I'm study the behavior of   n(n a(n) -2) / log(n)

where a(1) > 0, a(n+1) = log(1+a(n))

Using Pari:

f(n) =

{ my(v = 2);

  for(k=1,n, v = log(1+v));

  return(n*(n*v -2) / (log(n)));

}

It turns out the program runs very slowly. The same logic in python runs 100 time faster but not have the accuracy I need.

Any ideas?

Thanks