Generating function of $\log(\log(n))$ What do we know about the following function?
$$f(x)=\sum_{n=2}^\infty \log{(\log{(n)})}x^n$$
Do we know any closed form for it? 
I have tried to work with it through different means, but, as it is difficult both to calculate sums over $\log{(\log{(n)})}$ and evaluate integrals involving that function, I have not reached any conclusion.
Thank you.

We can write
$$
\eqalign{
  & F(x) = \sum\limits_{2\, \le \,n} {\ln \left( {\ln \left( n \right)} \right)x^{\,n} }  = x^{\,2} \sum\limits_{0\, \le \,n} {\ln \left( {\ln \left( {n + 2} \right)} \right)x^{\,n} }  =   \cr 
  &  = x^{\,2} \ln \left( {\ln \left( 2 \right)} \right) + x^{\,3} \sum\limits_{0\, \le \,n} {\ln \left( {\ln \left( {n + 3} \right)} \right)x^{\,n} }  \cr} 
$$
Then we write
$$
\eqalign{
  & \ln \left( {\ln \left( {n + 3} \right)} \right) =   \cr 
  &  = \ln \left( {\ln \left( {0 + 3} \right)} \right) - \ln \left( {\ln \left( {0 + 3} \right)} \right)
 + \ln \left( {\ln \left( {1 + 3} \right)} \right) +  \cdots  - \ln \left( {\ln \left( {n - 1 + 3} \right)} \right) + \ln \left( {\ln \left( {n + 3} \right)} \right) =   \cr 
  &  = \ln \left( {\ln \left( 3 \right)} \right) + \sum\limits_{0\, \le \,k\, \le \,n - 1}
 {\ln \left( {{{\ln \left( {k + 4} \right)} \over {\ln \left( {k + 3} \right)}}} \right)}  \cr} 
$$
so that
$$
\eqalign{
  & \sum\limits_{0\, \le \,n} {\ln \left( {\ln \left( {n + 3} \right)} \right)x^{\,n} } \quad \left| {\;\left| x \right| < 1} \right.\quad  =   \cr 
  &  = \ln \left( {\ln \left( 3 \right)} \right)\sum\limits_{0\, \le \,n} {x^{\,n} }  
 + \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,k\, \le \,n - 1} {\ln \left( {{{\ln \left( {k + 4} \right)} \over {\ln \left( {k + 3} \right)}}} \right)x^{\,n} } }  =   \cr 
  &  = \ln \left( {\ln \left( 3 \right)} \right)\sum\limits_{0\, \le \,n} {x^{\,n} } 
 + \sum\limits_{1\, \le \,n} {\sum\limits_{0\, \le \,k\, \le \,n - 1} {\ln \left( {{{\ln \left( {k + 4} \right)} \over {\ln \left( {k + 3} \right)}}} \right)x^{\,n} } }  =   \cr 
  &  = {{\ln \left( {\ln \left( 3 \right)} \right)} \over {1 - x}} + x\sum\limits_{0\, \le \,n}
 {\left( {\sum\limits_{0\, \le \,k\, \le \,n} {\ln \left( {{{\ln \left( {k + 4} \right)} \over {\ln \left( {k + 3} \right)}}} \right)} } \right)x^{\,n} }  \cr} 
$$
It is known that
$$
A(z) = \sum\limits_{n\, \ge \,0} {a_n } \;z^n \quad  \Leftrightarrow \quad {{A(z)} \over {\left( {1 - z} \right)}} = \sum\limits_{0\, \le \,n} {\left( {\sum\limits_{0\, \le \,k\, \le \,n} {a_k } } \right)z^n } 
$$
therefore
$$
\eqalign{
  & \sum\limits_{0\, \le \,n} {\ln \left( {\ln \left( {n + 3} \right)} \right)x^{\,n} } \quad \left| {\;\left| x \right| < 1} \right.\quad  =   \cr 
  &  = {{\ln \left( {\ln \left( 3 \right)} \right)} \over {1 - x}} + x\sum\limits_{0\, \le \,n} 
{\left( {\sum\limits_{0\, \le \,k\, \le \,n} {\ln \left( {{{\ln \left( {k + 4} \right)} \over {\ln \left( {k + 3} \right)}}} \right)} } \right)x^{\,n} }  =   \cr 
  &  = {{\ln \left( {\ln \left( 3 \right)} \right)} \over {1 - x}} + {x \over {1 - x}}\sum\limits_{0\, \le \,n}
 {\ln \left( {{{\ln \left( {n + 4} \right)} \over {\ln \left( {n + 3} \right)}}} \right)x^{\,n} }  \cr} 
$$
which finally gives:
$$ \bbox[lightyellow] {  
\eqalign{
  & F(x) = \sum\limits_{2\, \le \,n} {\ln \left( {\ln \left( n \right)} \right)x^{\,n} } \quad \left| {\;\left| x \right| < 1} \right.\quad  =   \cr 
  &  = x^{\,2} \ln \left( {\ln \left( 2 \right)} \right) + {{\ln \left( {\ln \left( 3 \right)} \right)x^{\,3} } \over {1 - x}}
 + {{x^{\,4} } \over {1 - x}}\sum\limits_{0\, \le \,n} {\ln \left( {{{\ln \left( {n + 4} \right)} \over {\ln \left( {n + 3} \right)}}} \right)x^{\,n} }  \cr} 
}$$
which is convergent (for $|x| < 1$) and checks to be correct.
Eventually, it remains to fix some appropriate bounds on
the coefficient of the last sum, which are positive, less than 
$\ln (\ln (4) / \ln (3)) \approx 0.2326$
and decreasing to zero, and with those to sandwich $F(z)$ between two known functions ...
