Fwd: Re: a(n+1) = log(1+a(n))

[John Cremona <john.cremona@gmail.com>]

---------- Forwarded message ----------
From: "John Cremona" <john.cremona@gmail.com>
Date: 8 Oct 2017 08:18
Subject: Re: a(n+1) = log(1+a(n))
To: "Elim Qiu" <elim.qiu@gmail.com>
Cc:

Try starting with v=2.0

On 8 Oct 2017 04:44, "Elim Qiu" <elim.qiu@gmail.com> wrote:

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