I'm very sorry if this is a duplicate in any way. There's a lot of material out there on connections between these sequences so it's a possibility . . .
Let $P_n$ be the number of partitions of $n$ and $F_n$ be the $n$th Fibonacci number.
I've been learning how to use GAP. Naturally, the accompanying tutorial gives a few examples of its use and a couple of very basic ones are functions to determine $P_n$ and $F_n$ for a given $n\in\mathbb{N}$. So I played around with them and soon I noticed $(P_{10} \operatorname{mod} F_{10}) = P_{10}$ and asked the following question.
Question 1 For which $n$ does $$(P_n \operatorname{mod} F_n) = P_n$$ (i.e., $0 \le P_n < F_n$)?
Based on what I've seen, I think it's for all $n>8$ (because it looks like $P_n<F_n$ in that case). Thus I have two more questions:
Question 2: What's so special about $9$ in this context?
and
Question 3: Is it true that $P_n<F_n$ for all $n>8$?
I'm not sure how to approach these.
NB: I hope it's clear that, a priori, Question 2 and Question 3 could be considered separate from one another (so this question won't be so stupid if the answer to Question 3 is well-known & trivial).]
The partition function is asymptotically $\frac{1}{4n\sqrt{3}}e^{\pi\sqrt{2n/3}}$ and the Fibonacci numbers are asymptotically $\frac{1}{\sqrt{5}}\varphi^n$, where $\varphi$ is the golden ratio. The quotient of these two is less than $1$ if $n$ is large enough: $$\frac{\sqrt{5}e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}\varphi^n} = \frac{\sqrt{5}}{4\sqrt{3}}\cdot\frac{\left(e^{\pi\sqrt{2/3}}\right)^{\sqrt{n}}}{n\varphi^n}.$$ So $P_n < F_n$ for large enough $n$.