While this post is **not** an answer the question, it does provide an observation which may help calculate $F(z;t)$ directly. It's not hard to see that \begin{equation} \begin{array}{lll} \displaystyle H_t(1v) &\displaystyle = \, \big(n^2 - t \big) H_t(v) &\text{if $|v|=n-1$} \\ \displaystyle H_t(2v) &\displaystyle = \, \big( 1 - t \big) \big( n^2 - t \big) H_t(v) &\text{if $|v|=n-2$} \end{array} \end{equation} and consequently \begin{equation} \langle H_t \rangle_n \ = \ {1 \over n} \big( n^2 - t \big) \langle H_t \rangle_{n-1} \ + \ {n-1 \over n} \big( 1 - t\big) \big(n^2 - t\big) \langle H_t \rangle_{n-2} \end{equation} or, after setting $\omega(n) := {1 \over {n!}} \langle H_t \rangle_n$ that \begin{equation} n^2 \omega(n) \ = \ \big(n^2 - t \big)\omega(n-1) \ + \ \big(1 - t \big)\big(n^2 - t \big) \omega(n-2) \end{equation} Of course $F(z;t) = \sum_{n \geq 0} \, \omega(n) z^n$ and it will satisfy the following second order homogeneous ODE in light of the linear recurrence: \begin{equation} \begin{array}{c} \displaystyle A(z) {d^2 \over {dz^2}} F(z;t) \ + \ B(z) {d \over {dz}} F(z;t) \ + \ C(z) F(z;t) \ = \ 0 \\ \text{where} \\ \begin{array}{l} A(z) \ = \ (1-t)z^3 + z^2 - z \\ B(z) \ = \ 5(1-t)z^2 + 3z - 1 \\ C(z) \ = \ (1-t)(4-t)z + (1-t) \end{array} \end{array} \end{equation} I don't know how to solve this ODE but when $s=0$ one solution is $F(z;0) = (1 - z -z^2)$ which is the denominator of the partition function for the Fibonacci numbers. When $s=1$ one solution is $F(z;1) = \int z^{-1} \, (z-1)^{-2} \, dz$ which can be computed by partial fractions. ines.