Question on OEIS A000085 The OEIS sequence 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?
 A: $\newcommand{\fl}[1]{\lfloor #1 \rfloor}\newcommand\N{\mathbb N}$User LeechLattice gave a complete answer to the original post.
This post is to complement that answer by confirming the empirical observation, made in my previous comment, that the inequality
$$ \left\lfloor \frac{a_n}{a_{n-1}} \right\rfloor > \left\lfloor \frac{a_{n-1}}{a_{n-2}} \right\rfloor \tag{1}$$
does not hold if $n\ge3$ and $n\notin\{b^2−b+1\colon b\ge3,b\in\N\}$. (Everywhere here, $n\in\{0,1,\dots\}$.)
The proof of this observation is based on the inequality
$$n-1<r_n^2-r_n<n \tag{2}$$
for $n\ge4$, used in the LeechLattice's answer, where
$$r_n:=\frac{a_n}{a_{n-1}}.$$
It is enough to show that
$$\fl{r_n}=\fl{r_{n-1}} \tag{3}$$
if $n\ge3$ and $n\notin\{b^2−b+1\colon b\ge3,b\in\N\}$.
Note that $r_1=1$ and
$$r_n=1+\frac{n-1}{r_{n-1}} \tag{4}$$
for $n\ge2$.
It is straightforward to check that (3) holds for $n=3,4,5,6$, whereas $7=b^2−b+1$ for $b=3$. So, it remains to show that (3) holds if $b\ge3$, $b\in\N$, and $b^2−b+1<n<(b+1)^2−(b+1)+1$, that is, if $b\in\{3,4,\dots\}$ and
$$b^2−b+2\le n\le(b+1)^2−(b+1).$$
For such $n$ and $b$, by (2),
$$b^2−b<n-1<r_n^2-r_n<n\le(b+1)^2−(b+1) \tag{5}$$
and
$$b^2−b\le n-2<r_{n-1}^2-r_{n-1}<n-1<(b+1)^2−(b+1). \tag{6}$$
By (4), $r_n\ge1$ for all $n$. Also, $r^2-r$ is strictly increasing in $r\ge1$. So, (5) and (6) imply $b<r_n<b+1$ and $b<r_{n-1}<b+1$, whence $\fl{r_n}=b=\fl{r_{n-1}}$, so that (3) follows.

The proof of the crucial inequality (2) (given in the paper cited by LeechLattice) is very simple but perhaps not easy to find. Indeed, (2) can be rewritten as
$$c_{n-1}<r_n<c_n, \tag{7}$$
where $c_n:=(1+\sqrt{4n+1})/2$, the positive root of the equation $c^2-c-n=0$, and, in turn, the bracketing (7) of $r_n$ is easily verified by induction on $n$ using the recurrence (4).
A: Your statement is (almost) proved in the paper On solutions of $x^d=1$ in symmetric groups. 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.
