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

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

• To: pari-users@pari.math.u-bordeaux.fr
• Subject: a(n+1) = log(1+a(n))
• From: Elim Qiu <elim.qiu@gmail.com>
• Date: Sat, 7 Oct 2017 21:44:33 -0600
• Delivery-date: Sun, 08 Oct 2017 05:44:41 +0200
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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