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.

  • $\begingroup$ Here are some thoughts. It is well-known that you can confirm if a number $P$ is composite by checking all factors less than $\sqrt P$. Now let $\{a_1, a_2, \cdots ,a_n\}$ be an increasing sequence of divisors in $[1, \sqrt P)$; $\{b_1, b_2, \cdots ,b_n\}$ be an decreasing sequence of divisors $[\sqrt P, P]$. Then the divisor sum is $S=\sum (a_k+b_k)\le n(a_n+b_1)=n(\sqrt P + P)\le\sqrt P (\sqrt P + P)$. $\endgroup$ Commented Aug 14, 2018 at 14:41

1 Answer 1


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.


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