The OEIS sequence [A000085](http://oeis.org/A000085) is defined by


$$ a\_n \!=\! (n-1)a\_{n-2} + a\_{n-1} \;\text{with }\; a\_0\!=\!1, a\_1\!=\!1.$$


If $n$ of the form $b^2-b+1, b \in \mathbb{N}, b > 2, \;\text{then: }\;$ $$ \left\lfloor \frac{a\_n}{a\_{n-1}} \right\rfloor > \left\lfloor \frac{a\_{n-1}}{a\_{n-2}} \right\rfloor$$


How to prove this?



Question on OEIS A000085
Your statement is (almost) proved in the paper [On solutions of $x^d=1$ in symmetric groups](https://www.cambridge.org/core/journals/canadian-journal-of-mathematics/article/on-solutions-of-xd-1-in-symmetric-groups/8F90642D9472FA7326164E54BE3BE57B). The $a\_n$ in your post corresponds to $T\_n$ in the paper, and the author defines $R\_n = T\_n/T\_{n-1}$. The author then set out to prove $n-1 \leq R\_n^2-R\_n \leq n$\*. The proof is on p.161 and can be modified to make the inequalities strict, simply by changing all $\leq$s into $<$ and all $\geq$s into $>$.


If $n=b^2-b+1$, then $b^2-b<R\_n^2-R\_n<b^2-b+1$, and $b^2-b-1<R\_{n-1}^2-R\_{n-1}<b^2-b$. Thus, $R\_n>b$ and $R\_{n-1}<b$, and your statement follows.


\*This is paraphrased from p.161 of the paper; formula 2.8 in the paper is wrong, so I circumvented it.


