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 practical numbers is A005153 in the OEIS.

See also [304] and [825].

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]). Erdős [Er76d] proved that this problem would also follow from showing that $P(n(n-1))>4\log n$.

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 $6!7!=10!$.

Erdős, Guy, and Selfridge [EGS82] have shown that \[f(n)-2n \asymp \frac{n}{\log n}.\]

It is easy to see that \[\lim \frac{t(n)}{n}\leq \frac{1}{e}.\] Erdős [Er96b] wrote he, Selfridge, and Straus had proved a corresponding lower bound, so that $\lim \frac{t(n)}{n}=\frac{1}{e}$, and 'believed that Straus had written up our proof. Unfortunately Straus suddenly died and no trace was ever found of his notes. Furthermore, we never could reconstruct our proof, so our assertion now can be called only a conjecture.'

Alladi and Grinstead [AlGr77] have obtained similar results when the $a_i$ are restricted to prime powers.

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 $a_t$.

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).

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).

Erdős and Obláth [ErOb37] proved this is true when $(x,y)=1$ and $k\neq 4$. Pollack and Shapiro [PoSh73] proved there are no solutions to $n!=x^4-1$. The known methods break down without the condition $(x,y)=1$.

Jonas Barfield has found the solution \[10! = 48^4 - 36^4=12^4\cdot 175.\]

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.

See also [401].

See also [400].

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 $m=0,1,2,3,6$).

See also [404].

See also [403]. Lin [Li76] has shown that $f(2,2) \leq 254$.

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 $p=2$.

Brindza and Erdős [BrEr91] proved that are finitely many such solutions. Yu and Liu [YuLi96] showed that the only solutions are \[2!+1^2=3\] \[2!+5^2=3^3\] and \[4!+1^4=5^2.\]

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].

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).

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 $a+b\leq n+O(\log n)$.

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 $n$.

See also [729].

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 $a+b+\leq n+O(\log n)$.

See also [728].

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 \[m=\exp((\log n)^{1/2+o(1)}).\]

A problem of Erdős and Selfridge, who proved (see [Er82c]) \[h(n) \asymp \left(\frac{n}{\log n}\right)^{1/2}.\]