Number of coefficients equal to $k$ in certain "Fibonacci polynomials" Let $F_i$ denote the $i$th Fibonacci number (with $F_1=F_2=1$). Define
$$ P_n(x) = \prod_{i=1}^n (1+x^{F_{i+1}}). $$
Let $\nu_k(n)$ denote the number of coefficients of the polynomial $P_n(x)$
that are equal to the positive integer $k$. Evidence suggests that for
sufficiently large $n$ (depending on $k$), $\nu_k(n)$ is a linear
polynomial in $n$. These polynomials for $1\leq k\leq 12$ are
empirically given by $2n$, $4n-8$, $8n-32$, $12n-68$, $16n-112$,
$24n-192$, $24n-224$, $36n-352$, $40n-432$, $48n-544$, $40n-512$, and
$88n-1056$. How can one prove these observations and generalize to any
$k$? Similar conjectures can be made for a wide class of more
general polynomials. See my papers https://arxiv.org/abs/1901.04647
and https://arxiv.org/abs/2101.02131 for some related results. The
linear algebraic techniques in these papers might be applicable to
the present problem.
Addendum. The case $k=1$ follows easily from properties of the Fibonacci triangle poset discussed in Section 3 of https://arxiv.org/abs/2101.02131.
 A: This is just a `formulas-free' version of the proof that the numbers $\nu_k(n)$ are indeed linear in $n$, for a fixed $k$ and large enough $n\geq n_0(k)$. The following lemma is the same as Max Alekseyev's starting point. As in his answer, we assume that the Zeckendorf representation is a $01$-string, where the leftmost digit is the least impotrant one.
Lemma. Each representation of a positive integer $M$ as a sum of distinct Fibonacci numbers is obtained from the Zeckendorf representation of $M$ by unrollings of the form $001\mapsto 110$.
Proof. Clearly, all unrollings produce representations of $M$.
Conversely, we show that each representation (also viewed as a $01$-string) can be rolled to the Zeckendorf representation by the operations $110\mapsto 001$. Take an arbitrary representation. Find the rightmost occurrence of $11$; it is followed by $0$ (or by the end of line, in which case we just augment the string by $0$ on the right). Replace this $110$ with $001$ and repeat.
This process stops, as the sum of elements in the string decreases. At the end, we get a string without occurrences of $11$, i.e., the Zeckendorf representation of $M$.  $\Box$
Now let the Zeckendorf representation of $M$ have the form $0^{i_p}10^{i_{p-1}}1\dots 0^{i_1}1$, where $i_p\geq 0$ and $i_t>0$ for $1\leq t<p$. Assume that $M$ has exactly $k$ representations.
1. We have $i_t\leq 2k-1$ for all $t$. Indeed, if $i:=i_t\geq 2k$, then we can make the sequence of unrollings
$$
  0^i1\mapsto 0^{i-2}110\mapsto 0^{i-4}11010\mapsto\dots\mapsto
  0^{i-2k}1(10)^k,
$$
obtaining $k+1$ distinct representations, which cannot appear.
2. We have $i_t\leq1$ for all $t\geq k$ (thus $i_t=1$ for $p<t\leq k$ and $i_p\in\{0,1\}$ if $p\geq k$). Indeed, if $i_t>1$, then we can successively unroll the $t$th rightmost $1$, then the $(t-1)$th rightmost $1$, and so on, again obtaining at least $k+1$ distinct representations.
The two properties above show that the Zeckendorf representations of all numbers $M$  under consideration look like either $0101\dots01 $ or $1010\dots 1$ followed by a tail with bounded number of $1$s and bounded number of $0$s between consecutive $1$s. Moreover, the number of representations of such tail is still $n$, as the unrollings cannot touch the starting periodical part. There are finitely many such tails.
In other words, $\nu_k(n)$ can be computed  as follows. There are finitely many possible tails corresponding to numbers having $k$ representations and having the form $10^{i_p}1\dots 0^{i_1}1$, where $i_k\geq 2$. Each of those can be augmented by a periodic string of the form $\dots 01010$. in order to produce a string of length at most $n$. The number of such strings is $\nu_k(n)$.
Now it is clear that this number is linear in $n$ for all sufficiently large $n\geq n_0$ (where $n_0$ is the largest length of the tail).
