Take the 2-minute tour ×  I wonder whether it is impossible to
write the nth Motzkin number as a sum of a fixed number of, say,
hypergeometric terms. To illustrate what I mean: $n!+(2n)!$ is
not a hypergeometric term, but it is written as a sum of two
hypergeometric terms.  I'd also appreciate other examples,
especially if they come from counting weighted Motzkin paths.
share|improve this question Martin, I added the tag. The
simplest formula I found for your numbers is
$a_n=\dfrac1{n+1}\sum_i \dfrac{(n+1)!}{i!(i+1)!(n-2*i)!}$. But
there are so many other examples (practically everything I know!
:-) ), like Apery's numbers
$\sum_k{\dbinom{n+k}{k}}^2{\dbinom{n}{k}}^2$, or Domb numbers,
or whatever. Proving that there is no finite hypergeometric sum
is the subject of Petkovcek's algorith in $A=B$. –  Wadim
Zudilin Jun 3 '10 at 11:54 Do I understand correctly that
formula (10) at mathworld.wolfram.com/MotzkinNumber.html for the
$n$th Motzkin number as a hypergeometric $\textit{function}$ is
a sum of $n/2$ hyper