So I'm trying to compute the Galois group of family of polynomials (indexed by their degree) using the technique of the Newton polygon. In order to apply this technique I need to find some good prime number $p$.

So this is the motivation behind my question that might seem a little bit unmotivated:

Let $N$ be a large integer. Then it is not too difficult to show the following statement:

**Theorem**: For every prime $p$ such that $N/2< p< 2N/3$ one has that $p$ divides the following sum
$$
S_N:=\sum_{k=0}^N \binom{N+k}{k}2^{N-k}(-1)^k
$$

After many numerical examples, it always happens that most of the primes in the interval $N/2 < p < 2N/3$ divide $S_N$ with multiplicity one. So here is my question:

Q: How would you show that there exists at least one prime $p$ in the interval $N/2 < p < 2N/3$ that divides exactly $S_N$, i.e., $p|S_N$ but $p^2\nmid S_N$ ,?

Note that the square of the product of all primes in the interval $(\frac{N}{2},\frac{2N}{3})$ is less that $\binom{2N}{N}$, so a naive counting argument does not seem to work here.

If you think that this problem is intractable then let me know, I'll try a different strategy.

allthe primes up to $67$ is about $6.2 \cdot 10^{49}$. – Aaron Meyerowitz Mar 28 '11 at 0:12