A commutative ring $R$ is called Noetherian if every ascending chain of ideals in $R$ stabilizes, that is, $$ I_1\subseteq I_2\subseteq I_3\subseteq\cdots $$ implies the existence of $n\in\mathbb{N}$ such that $I_n=I_{n+1}=I_{n+2}=\cdots$.
My question is the following:
Suppose $R$ is commutative ring such that for every ascending chain of ideals $$ I_1\subseteq I_2\subseteq I_3\subseteq\cdots $$ there exist infinitely many $n\in\mathbb{N}$ such that $I_{n}=I_{n+1}$. Does it follow that $R$ is Noetherian?
Motivation. Let's call this condition on ascending chain "stabilizing temporarily" infinitely many times. If a counter-example to my question is found, we would obtain a nice family of non-Noetherian rings. The canonical example of non-Noetherian ring I was taught is given by $k[x_1, x_2, x_3,...]$, and the ascending chain $$(x_1)\subsetneq (x_1, x_2)\subsetneq (x_1, x_2, x_3)\subsetneq\cdots $$ which doesn't stabilize. Note that it doesn't "stabilize temporarily" (not even once).
The answer is yes, but for boring reasons.
Suppose $R$ is non-Noetherian and every ascending chain has an infinite number of successive equalities. Our hypothesis implies there is an ascending chain $I_1\subseteq I_2\subseteq\cdots$ that never stabilizes.
Then define the ascending chain $J_1\subset J_2\subset\cdots$ obtained by deleting all repeats from the original ascending chain. Since the original chain never fully stabilized, it must have an infinite number of distinct ideals, so this new chain never stabilizes and it does not have any successive equalites, let alone an infinite number of them guaranteed by our second hypothesis on $R$, a contradiction.