3 solved out of 20 shown (show only solved or open)
OPEN
We call $m$ practical if every integer $n<m$ is the sum of distinct divisors of $m$. If $m$ is practical then let $h(m)$ be such that $h(m)$ many divisors always suffice.

Are there infinitely many practical $m$ such that $h(m) < (\log\log m)^{O(1)}?$ Is it true that $h(n!)<n^{o(1)}$? Or perhaps even $h(n!)<(\log n)^{O(1)}$?

It is easy to see that almost all numbers are not practical. Erdős originally showed that $h(n!) <n$. Vose [Vo85] proved the existence of infinitely many practical $m$ such that $h(m)\ll (\log m)^{1/2}$.

The sequence of pratical numbers is A005153 in the OEIS.

Related to [304].

OPEN
Show that the equation $n! = a_1!a_2!\cdots a_k!,$ with $n-1>a_1\geq a_2\geq \cdots \geq a_k$, has only finitely many solutions.
This would follow if $P(n(n+1))/\log n\to \infty$, where $P(m)$ denotes the largest prime factor of $m$ (see Problem [368]). Hickerson conjectured the largest solution is $16! = 14! 5!2!.$ The condition $a_1<n-1$ is necessary to rule out the trivial solutions when $n=a_2!\cdots a_k!$.

Surányi was the first to conjecture that the only non-trivial solution to $a!b!=n!$ is $6!7!=10!$.

OPEN
Let $f(n)$ be the minimal $m$ such that $n! = a_1\cdots a_k$ with $n< a_1<\cdots <a_k=m$. Is there (and what is it) a constant $c$ such that $f(n)-2n \sim c\frac{n}{\log n}?$
Erdős, Guy, and Selfridge [EGS82] have shown that $f(n)-2n \asymp \frac{n}{\log n}.$
OPEN
Let $t(n)$ be the maximum $m$ such that $n!=a_1\cdots a_n$ with $m=a_1\leq \cdots \leq a_n$. Obtain good upper bounds for $t(n)$. In particular does there exist some constant $c>0$ such that $t(n) \leq \frac{n}{e}-c\frac{n}{\log n}$ for infinitely many $n$?
Erdős, Selfridge, and Straus have shown that $\lim \frac{t(n)}{n}=\frac{1}{e}.$ Alladi and Grinstead [AlGr77] have obtained similar results when the $a_i$ are restricted to prime powers.
OPEN
Let $A(n)$ denote the least value of $t$ such that $n!=a_1\cdots a_t$ with $a_1\leq \cdots \leq a_t\leq n^2$. Is it true that $A(n)=\frac{n}{2}-\frac{n}{2\log n}+o\left(\frac{n}{\log n}\right)?$
If we change the condition to $a_t\leq n$ it can be shown that $A(n)=n-\frac{n}{\log n}+o\left(\frac{n}{\log n}\right)$ via a greedy decomposition (use $n$ as often as possible, then $n-1$, and so on). Other questions can be asked for other restrictions on the sizes of the $a_t$.
OPEN
Let $f(n)$ denote the minimal $m$ such that $n! = a_1\cdots a_t$ with $a_1<\cdots <a_t=a_1+m$. What is the behaviour of $f(n)$?
Erdős and Graham write that they do not even know whether $f(n)=1$ infinitely often (i.e. whether a factorial is the product of two consecutive integers infinitely often).
OPEN
Are the only solutions to $n!=x^2-1$ when $n=4,5,7$?
The Brocard-Ramanujan conjecture. Erdős and Graham describe this as an old conjecture, and write it 'is almost certainly true but it is intractable at present'.

Overholt [Ov93] has shown that this has only finitely many solutions assuming a weak form of the abc conjecture.

There are no other solutions below $10^9$ (see the OEIS page).

OPEN
Is it true that there are no solutions to $n! = x^k\pm y^k$ with $x,y,n\in \mathbb{N}$, with $xy>1$ and $k>2$?
Erdős and Obláth [ErOb37] proved this is true when $(x,y)=1$ and $k\neq 4$. Pollack and Shapiro [PoSh73] proved this is true when $(x,y)=1$ and $k=4$. The known methods break down without the condition $(x,y)=1$.
OPEN
For any $k\geq 2$ let $g_k(n)$ denote the maximal value of $n-(a_1+\cdots+a_k)$ where $a_1,\ldots,a_k$ are integers such that $a_1!\cdots a_k! \mid n!$. Can one show that $\sum_{n\leq x}g_k(n) \sim c_k x\log x$ for some constant $c_k$? Is it true that there is a constant $c_k$ such that for almost all $n<x$ we have $g_k(n)=c_k\log x+o(\log x)?$
Erdős and Graham write that it is easy to show that $g_k(n) \ll_k \log n$ always, but the best possible constant is unknown.

OPEN
Is there some function $\omega(r)$ such that $\omega(r)\to \infty$ as $r\to\infty$, such that for all large $n$ there exist $a_1,a_2$ with $a_1+a_2> n+\omega(r)\log n$ such that $a_1!a_2! \mid n!2^n3^n\cdots p_r^n$?
SOLVED
Does the equation $2^m=a_1!+\cdots+a_k!$ with $a_1<a_2<\cdots <a_k$ have only finitely many solutions?
Asked by Burr and Erdős. Frankl and Lin [Li76] independently showed that the answer is yes, and the largest solution is $2^7=2!+3!+5!.$ In fact Lin showed that the largest power of $2$ which can divide a sum of distinct factorials containing $2$ is $2^{254}$, and that there are only 5 solutions to $3^m=a_1!+\cdots+a_k!$ (when $m=0,1,2,3,6$).

OPEN
Let $f(a,p)$ be the largest $k$ such that there are $a=a_1<\cdots<a_k$ such that $p^k \mid (a_1!+\cdots+a_k!).$ Is $f(a,p)$ bounded by some absolute constant? What if this constant is allowed to depend on $a$ and $p$?

Is there a prime $p$ and an infinite sequence $a_1<a_2<\cdots$ such that if $p^{m_k}$ is the highest power of $p$ dividing $\sum_{i\leq k}a_i!$ then $m_k\to \infty$?

See also [403]. Lin [Li76] has shown that $f(2,2) \leq 254$.
OPEN
Let $p$ be an odd prime. Is it true that the equation $(p-1)!+a^{p-1}=p^k$ has only finitely many solutions?
Erdős and Graham remark that it is probably true that in general $(p-1)!+a^{p-1}$ is rarely a power at all (although this can happen, for example $6!+2^6=28^2$).

Erdős and Graham ask this allowing the case $p=2$, but this is presumably an oversight, since clearly there are infinitely many solutions to this equation when $p=2$.

SOLVED
If $\tau(n)$ counts the number of divisors of $n$, then what is the set of limit points of $\frac{\tau((n+1)!)}{\tau(n!)}?$
Erdős and Graham noted that any number of the shape $1+1/k$ for $k\geq 1$ is a limit point (and thus so too is $1$), but knew of no others.

Mehtaab Sawhney has shared the following simple argument that proves that the above limit points are in fact the only ones.

If $v_p(m)$ is the largest $k$ such that $p^k\mid m$ then $\tau(m)=\prod_p (v_p(m)+1)$ and so $\frac{\tau((n+1)!)}{\tau(n!)} = \prod_{p|n+1}\left(1+\frac{v_p(n+1)}{v_p(n!)+1}\right).$ Note that $v_p(n!)\geq n/p$, and furthermore $n+1$ has $<\log n$ prime divisors, each of which satisfy $v_p(n+1)<\log n$. It follows that the contribution from $p\leq n^{2/3}$ is at most $\left(1+\frac{\log n}{n^{1/3}}\right)^{\log n}\leq 1+o(1).$

There is at most one $p\mid n+1$ with $p\geq n^{2/3}$ which (if present) contributes exactly $\left(1+\frac{1}{\frac{n+1}{p}}\right).$ We have proved the claim, since these two facts combined show that the ratio in question is either $1+o(1)$ or $1+1/k+o(1)$, the latter occurring if $n+1=pk$ for some $p>n^{2/3}$.

After receiving Sawhney's argument I found that this had already been proved, with essentially the same argument, by Erdős, Graham, Ivić, and Pomerance [EGIP].

Additional thanks to: Zachary Chase, Mehtaab Sawhney
OPEN
Let $p$ be a prime and $A_p = \{ k! \pmod{p} : 1\leq k<p\}.$ Is it true that $\lvert A_p\rvert \sim (1-\tfrac{1}{e})p?$
SOLVED
Let $p_1,\ldots,p_k$ be distinct primes. Are there infinitely many $n$ such that $n!$ is divisible by an even power of each of the $p_i$?
The answer is yes, proved by Berend [Be97], who further proved that the sequence of such $n$ has bounded gaps (where the bound depends on the initial set of primes).
Additional thanks to: Euro Vidal Sampaio
OPEN
Let $k\geq 2$. Does $(n+k)!^2 \mid (2n)!$ for infinitely many $n$?
A conjecture of Erdős, Graham, Ruzsa, and Straus [EGRS75]. It is open even for $k=2$.

Balakran [Ba29] proved this holds for $k=1$ - that is, $(n+1)^2\mid \binom{2n}{n}$ for infinitely many $n$. It is a classical fact that $(n+1)\mid \binom{2n}{n}$ for all $n$ (see Catalan numbers).

Erdős, Graham, Ruzsa, and Straus observe that the method of Balakran can be further used to prove that there are infinitely many $n$ such that $(n+k)!(n+1)! \mid (2n)!$ (in fact this holds whenever $k<c \log n$ for some small constant $c>0$).

Erdős [Er68c] proved that if $a!b!\mid n!$ then $a+b\leq n+O(\log n)$.

OPEN
Let $\epsilon,C>0$. Are there integers $a,b,n$ such that $a>\epsilon n$, $b>\epsilon n$, $a! b! \mid n!(a+b-n)!$ and $a+b>n+C\log n$?
A question of Erdős, Graham, Ruzsa, and Straus [EGRS75]. Erdős [Er68c] proved that if $a!b!\mid n!$ then $a+b\leq n+O(\log n)$.

By Legendre's formula $a! b! \mid n!(a+b-n)!$ is true if and only if for all primes $p$ $s_p(n)+s_p(a+b-n) \leq s_p(a)+s_p(b),$ where $s_p(n)$ is the sum of the base $p$ digits of $n$.

Let $C>0$ be a constant. Are there infinitely many integers $a,b,n$ with $a+b> n+C\log n$ such that the denominator of $\frac{n!}{a!b!}$ contains only primes $\ll_C 1$?
Erdős [Er68c] proved that if $a!b!\mid n!$ then $a+b\leq n+O(\log n)$. The proof is easy, and can be done with powers of $2$ alone: Legendre's formula implies that if $2^k$ is the highest power of $2$ dividing $n!$ then $k=n+O(\log n)$, and hence if $a!b!\mid n!$ then $a+b\leq n+O(\log n)$.
This problem is asking if $a!b!\mid n!$ 'ignoring what happens on small primes' still implies $a+b+\leq n+O(\log n)$.
Find some reasonable function $f(n)$ such that, for almost all integers $n$, the least integer $m$ such that $m\nmid \binom{2n}{n}$ satisfies $m\sim f(n).$
A problem of Erdős, Graham, Ruzsa, and Straus [EGRS75], who say it is 'not hard to show that', for almost all $n$, the minimal such $m$ satisfies $m=\exp((\log n)^{1/2+o(1)}).$