What is the monotonicity of the sequence $a_{n+1} = -a_{n}^{2}+8a_n-10$ given that $a_1=0$? I am trying to find the  monotonicity of the sequence $a_{n+1} = -a_n^{2}+8a_n-10$ where $a_1=0$

Taking the first terms we notice that $a_1 = 0, a_2 = -10, a_3 = - 190$.
Therefore, I made the hypothesis that the sequence is strictly decreasing and proved it by deduction.

*

*For $n=1: a_1  > a_2 \iff 0 > -10$

*For $n=k$ let $a_k > a_{k+1}$

*For $n=k+1:$ An algebraic mess occurs...


Another approach is to prove that $a_n - a_{n+1} > 0 \iff$
$$a_{n} - a_n^{2}-8a_n+10 = ...$$ but that's a dead end.

Any other ideas on how to study the monotonicity of this sequence? There must be an easier way.
 A: If $a_n ≤ 0$, then $a_{n+1} - a_n = a_n(7 - a_n) - 10 < 0$, and so also $a_{n+1} < a_n ≤ 0$.
A: Write $a_{n+1} = -a_{n}^{2}+8a_n-10$ as $a_{n+1} = -(a_{n}-4)^{2}+6$
That shows for $a_n \le 0$, $a_{n+1} \lt a_{n}$.
A: You got $$a_{n} - (-a_n^{2}+8a_n-10) = a_n^2-7a_n+10 = (a_n-5)(a_n-2)$$
Since $a_n\leq 0$ this product is negative, so the sequance is decreasing.
A: What you consider to be a dead end is viable to show
$$a_{n+1} - a_n \:=\: -\,(a_n^2 -7a_n +10) \:=\: (a_n-2)(5-a_n) \;
\begin{cases}\geqslant 0 & \text{if }\:2\leqslant a_n\leqslant 5 \\
< 0 & \text{else} \end{cases}$$
the desired monotonicity, as in Aqua's answer.
When the iteration starts with $a_1=0$, then always the second line in the case distinction applies, hence the sequence $\,a_1,a_2,a_3,\ldots\,$ is strictly decreasing.

Another idea how to study the monotonicity properties, not necessarily an easier way, yields more quantitative insight via a closed-form expression for $\,a_n\,$ from which the monotonicity is deduced.
Similar to the answer by Math Lover one has
$$4-a_{n+1} \:=\: (4-a_n)^2 -2\,,$$
thus the linear term on recursion's RHS drops out. Let $\,b_n =4-a_n$,
and go for the ansatz
$$b_n \:=\: \beta^{2^{n-1}} + \big(\beta^{-1}\big)^{2^{n-1}} \\[1.5ex]
\implies b_n^2 \:=\: \beta^{2^n} +2 +\big(\beta^{-1}\big)^{2^n} \:=\: b_{n+1}+2$$
so that the recursion is satisfied.
$\beta\,$ is determined by the start value. From $\,4=b_1=\beta+\beta^{-1}$ we have that $\,\beta=2\pm\sqrt3\iff\beta^{-1}=2\mp\sqrt3$, hence
$$b_n \:=\: \big(2+\sqrt3\big)^{2^{n-1}} + \big(2-\sqrt3\big)^{2^{n-1}}$$
which is super-exponentially (and also strictly) growing.
Notice that it is a particular property of the given quadratic recursion that the iterates can be written in closed-form. In general this is unavailable.
