0 solved out of 11 shown

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.

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 $a_t$.

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).

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.

See also [401].

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

See also [404].

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 $m_k\to \infty$?