A star number is a number of the form $6n(n-1)+1$. They are equal to the number of dots in a star, like this:
According to Wikipedia:
Unique among the star numbers is 35113, since its prime factors (i.e., 13, 37 and 73) are also consecutive star numbers.
This statement does not give a source. It is not clear in what way 35113 is 'unique'. After some experimenting, I arrived at the following conjecture:
The only star number that can be written as the product of at least 2 consecutive star numbers larger than 1, is 35113.
For products of $2,3,4$ successive star numbers, I made a Python algorithm. It checked whether there are among the first $10$ million star numbers, $i$ succesive star succesive whose product is another star number. The only result that came up was $13\times37\times73=35113$.
Now using some heuristics. Let $S_n$ be the $n$th star number. The 'probability' that a number $x$ is a star number is proportional to $\frac{1}{\sqrt x}$, so the 'probability' that $S_mS_{m+1}$ is a star number is on the order of $\frac{1}{\sqrt{n^4}}=n^{-2}$. Therefore, the 'expected' number of values of $m\geq n$ such that $S_nS_{n+1}$ is a star number is something like $\sum_{m=n}^{\infty}m^{-2}\approx n^{-1}$. So I would expect there to be only finitely many numbers $n$ such that $S_nS_{n+1}$ is a star number. This reasoning works similarly for $S_nS_{n+1}S_{n+2}$ and $S_nS_{n+1}S_{n+2}S_{n+3}$. However, it is still heuristic so it doesn't prove anything. Maybe one can use analytic number theory? Or is there a more simple method?
A simple case to start with, would be products of 2 consecutive star numbers. For example, 13 and 37 are consecutive star numbers, but 13x37=481 is not a star number. In formulas this would boil down to the equation
$$(6n(n-1)+1)(6(n+1)n+1)=6m(m-1)+1$$
for integers $n,m$ with $n\geq2$. This is a quartic equation in two variables. It can be rewritten to
$$6n^4-4n^2=m^2-m.$$
This looks similar to a Pell equation. Is there a way to prove that this equation has no solutions at all? (Here we have $n>1$)