<s>[INST] Is it a coincidence that the 12th Fibonacci number is 12 squared? The first $12$ Fibonacci numbers are:
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
n  & 0& 1& 2& 3& 4& 5& 6& 7& 8 & 9&  10& 11& 12\\
\hline
f_{n} & 0& 1& 1& 2& 3& 5& 8& 13& 21& 34& 55& 89& 144\\
\hline
\end{array}$$
When you get to $f_{12}$ you find it is equal to $144$.
What strikes me here is the following:
\begin{align*}
f_{12} &= 144\\
12^2 &=144.
\end{align*}
Not only is $f_{12}$ equal to $144$, but so is $12^2$. Here, $n^2= f_n$.
This is the only example as  $n^2 > f_n$ is true for all $1<n<12$ (works for $0$ and $1$) and $n^2< f_n$ is true for all $n > 12$.
So is this just a coincidence?
 [/INST] This is a very nice manifestation of the Strong Law of Small Numbers formulated by Richard K. Guy: 
"There aren't enough small numbers to meet the many demands made of them."
A more striking manifestation of the same "law" of small numbers is this:
$$
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
n  & 0& 1& 2& 3& 4& 5& 6& 7& 8 & 9&  10& 11& 12\\
\hline
\mbox{Fibonacci }
f_{n} & 0& 1& 1& 2& 3& 5& 8& 13& 21& 34& 55& 89& 144\\
\hline
\hline
\mbox{# squares in }[e^{n-1},e^n)
      & 0& 1& 1& 2& 3& 5& 8& 13& 21& 36& 58& 96& 159\\
\hline
\end{array}
$$
The last line is OEIS A306486: the number of squares in the interval $[e^{n-1},e^n)$, which happens to be the same as the Fibonacci number $f_n$ for $0\le n\le 8$. Note that both $f_n$ and A306486($n$) grow about as fast as certain geometric progressions; the common ratio is $\phi\approx1.618\ldots$ for the former and $\sqrt{e}\approx1.6487\ldots$ for the latter.
</s>