Elim Qiu on Sun, 08 Oct 2017 05:44:41 +0200


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


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