I ran across a seemingly relatively simple combinatorics problem that appears open. For an alphabet of size $n$, let $A(n)$ be the number of strings over the alphabet that have no substring of length $>1$ that appears more than once in the string. For example, with $n=2$ and alphabet $\{0,1\}$, the string $10011$ would be counted, but $1010$ would not because substring $10$ appears twice, and similarly $000$ would not be counted because substring $00$ appears twice (note that occuring substrings can overlap). The empty string counts as a valid string too. I computed $A(n)$ for $n=1,2,3,4,5$ and got $3,25,1885,3636297,327094648711$. However, when I entered this into OEIS, I got no match (and similarly when I subtracted $1$ and entered $A(n)-1$ for $n=2,3,4$). So I added the sequence A246536 to the OEIS database.

Anyway, I was wondering if anyone can help either come up with a combinatorial formula (either closed form or in terms of somewhat tractable summation etc.), or similarly come up with an asymptotic formula $G(n)$ such that $\lim_{n\to\infty} A(n)/G(n)=1$. The only progress I've been able to make is noting that a circular De Bruijn sequence where every substring of length 2 occurs exactly once can be cut in any of the $n^2$ possible cut points, and then the full string obtained or the string with the last character removed can be counted, and this gives a somewhat good lower bound of $A(n)>G(n)=2n^2B(n,2)$ where $B(n,2)$ is the De Bruijn number for alphabet of size $n$ and substrings of length $2$ (which has an exact known formula). However this does not give an asymptotic lower bound $G(n)$ that satisfies $\lim_{n\to\infty}A(n)/G(n)=1$, because the number of left out shorter strings not counted by $G(n)$ that are counted by $A(n)$ is non-trivial. Any help or pointers to the literature are greatly appreciated.