Set and bounded gap Let $S$ be the set of positive integers $n$ such that $n!$ cannot be expressed of form $a^3+b^3+c^3+d^3$. Does the set $S$ have bounded gap?
 A: The set $S$ is very likely finite.
It is unclear if you intend for $a$, $b$, $c$ and $d$ to be positive.  If you don't assume that $a$, $b$, $c$ and $d$ are positive, then $n!$ has such a representation for $n \geq 3$. This is due to the identity $$6x =(x-1)^3 + (-x)^3 + (-x)^3 +(x+1)^3.$$
If one assumes that $a$, $b$, $c$ and $d$ must be positive, then there is strong reason to believe that $7373170279850$ is the largest number which is not the sum of four positive cubes. See the paper by that title  by Deshouillers, Hennecart,  Landreau, and  I. Gusti Putu Purnaba, in Mathematics of Computation from 2000. We also have good reason to think that the set of numbers representable as the sum of three cubes has positive density (which I still find very hard to wrap my head around). See "Sums of powers: an arithmetic refinement to the probabilistic model of Erdős and Rényi." by  Deshouillers, Hennecart,  Landreau, in Acta Arithmetica, 1998, as well followup papers by the same three authors.
There is also work by Trevor Wooley showing that the set of numbers which are the sums of three non-negative cubes and are at most $x$ grows at least like $x^C$ for a constant $C$ which is a little greater than $9/10$. A reasonably "generic" set T with that high a  density at least is already enough such that the minimum number of ways of writing any positive integer $n$ as an element in $T$ plus a cube should go to infinity as $n$ goes to infinity.
In the very unlikely event that your set $S$ is infinite, it seems highly unlikely that the elements will be at all common. So I don't see an immediate proof, but it seems really unlikely that $S$ has bounded gaps between elements.
