Are all numbers of the type $\frac{n!}{2}+1$ deficient? Are all numbers of the type $\frac{n!}{2}+1$ deficient? Deficient numbers are such numbers $k$, that the divisor sum of $k$ is less than $2k$.
I checked all numbers of this type, with $n$ ranging from $1$ to $11$. They really are. However, I do not know, how to prove this statement in general. Probably, it has something to do with $\frac{n!}{2}+1$ being coprime with all numbers that do not exceed $n$, thus making it a number without small prime divisors. But probably, that fact is not that helpful either...
Any help will be appreciated.
 A: For a given positive integer $n$, we will write $h(n) = \frac{\sigma(n)}{n}$ where $\sigma(n)$ is the sum of the positive divisors of $n$. We will write $H(n)= \prod_{p|n} \frac{p}{p-1}$ where the product is over all primes $p$ which divide $n$. It is not too hard to show that $h(n) \leq H(n)$ with equality if and only if $n=1.$ 
Claim: Let $n$ be a perfect or abundant number with smallest prime factor $q>2$, and let $k$ be the number of distinct prime divisors of $n$. Then we must have $k \geq q$.
Proof: Assume as given. Note that $n>1$.  $$2 \leq h(n) < H(n) \leq \frac{q}{q-1}\frac{q+1}{q}\frac{q+2}{q+1} \cdots \frac{q+k-1}{q+k-2}.$$
Note that the last inequality arises from the fact that $\frac{x}{x-1}$ is a decreasing function in $x$ for $x>1$. Notice that all the terms on the right hand side cancel except the denominator of the first term and the numerator of the last term. We have then $$2 < \frac{q+k-1}{q-1}$$ and so $$2q -2 < q+k -1.$$ Simplifying, we get that $k > q-1$ and thus $k \geq q$. 
Now assume that $\frac{n!}{2}+1$ is perfect or abundant. It must have smallest prime factor at least $n+1$ and thus have at least $n+1$ distinct prime factors and so we would have $\frac{n!}{2}+1 \geq (n+1)^{n+1}$ which is clearly impossible. 
