$(q,t)$-Fibonacci polynomials: area & bounce statistics This is related to my earlier (unanswered) MO post. Preserve notations from there.
We take advantage of the one-to-one correspondence between the $(s,s+1)$-core partitions and $(s,s+1)$-Dyck paths. Let $\mathcal{F}_s$ denote the set of $(s,s+1)$-Dyck paths whose associated core partitions have distinct parts.
Also, we recall the two notions area and bounce statistics defined on Dyck paths. We write $[n]_t=\frac{1-t^n}{1-t}$ and the $t$-binomials $\binom{n}k_t=\frac{[n]_t}{[k]_t[n-k]_t}$.
Experimental observations lead to the following

QUESTION 1. Is this true?
$$\sum_{\lambda\in\mathcal{F}_s}q^{\text{area}(\lambda)}\,t^{\text{bounce}(\lambda)}
=\sum_{j\geq0}\binom{s-j}j_tq^j\,t^{\binom{s}2-j(s-j)}.$$


QUESTION 2. Alternatively, if $H_s(q,t):=\sum_{\lambda\in\mathcal{F}_s}q^{\text{area}(\lambda)}\,t^{\text{bounce}(\lambda)}$ then does this hold?
$$H_{s+1}(q,t)=t^{s+1}\,H_s(q,t)+q\,t^s\,H_{s-1}(q,t)$$
with initial conditions $H_1(q,t)=1$ and $H_2(q,t)=q+t$.

 A: I mentioned in my previous answer that the order ideals corresponding to $(s,s+1)$-cores with distinct parts are precisely the subsets of $\{1,2,\dots,s-1\}$ that contain no consecutive elements. This means that the Dyck paths under investigation are all of height 2, with peaks at each element of the order ideal. Since there are no consecutive elements, the paths will always be "bouncing", in the sense that the associated bounce path is the same as the Dyck path itself.
Let's restrict our attention to the Dyck paths with exactly $j$ peaks. There are exactly $\binom{s-j}{j}$ of these. They will all have area $j$, so it remains to prove that the generating function for their bounce statistic is given by
$$t^{\binom{s}{2}-j(s-j)}\binom{s-j}{j}_t=t^{\binom{j}{2}+\binom{s-j}{2}}\binom{s-j}{j}_t.$$
To each subset $\{a_1,a_2,\dots,a_j\}$ of $\{1,2,\dots,s-1\}$ we can associate the subset $\{b_1,b_2,\dots,b_j\}$ of $\{0,1,\dots,s-j-1\}$ given by $b_i=a_i-i$. We can check that the bounce statistic of this subset is equal to $\binom{s-j}{2}+b_1+b_2+\cdots+b_j$, therefore the generating function we want is the coefficient of $x^j$ in the expression
$$t^{\binom{s-j}{2}}\prod_{i=0}^{s-j-1}(1+t^ix).$$
From here the desired result follows from the q-binomial theorem.
