This post is **almost** an answer to the question. It provides a method to directly calculate $F(z;t)$ and bypasses the obstacles related to computing the expectation values $\langle E_k \rangle_n$. 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 $\Bbb{f}_t(n) := {1 \over {n!}} \langle H_t \rangle_n$ \begin{equation} n^2 \Bbb{f}_t(n) \ = \ \big(n^2 - t \big)\Bbb{f}_t(n-1) \ + \ \big(1 - t \big)\big(n^2 - t \big) \Bbb{f}_t(n-2) \end{equation} Of course $F(z;t) = \sum_{n \geq 0} \, \Bbb{f}_t(n) z^n$ 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 for general values of the parameter $t$. Nevertheless, by construction, $F(z;0)$ must be the generating function for the Fibonacci numbers, i.e. \begin{equation} \begin{array}{l} \displaystyle F(z;0) &\displaystyle = \ 1 + z + 2z^2 + 3z^3 + 5z^4 + \cdots \\ &\displaystyle = \ {1 \over {(1 - z -z^2)}} \end{array} \end{equation} and one can check directly that $(1 - z -z^2)^{-1}$ is indeed a solution to the ODE when $t=0$. When $t=1$ the ODE's general solution is $c_1 + c_2 \int z^{-1} \, (z-1)^{-2} \, dz$ where $c_1, c_2$ are constants. However, any fibonacci word $u$ with $|u| > 0$ must contain a box $\Box \in u$ with $\mathrm{h}(\Box)=1$ and so the statistic $H_1(u)$ must vanish. Combinatorial realities thus force us to select the constant solution $F(z;1) = 1$. The recurrence above is clearly implemented by tridiagonal determinants. Specifically $\Bbb{f}_t(n)$ equals the following $n \times n$ tridiagonal determinant: \begin{equation} \det \begin{pmatrix} 1-t & 1 - {1 \over 4} t & 0 & 0 & \\ t-1 & 1 - {1 \over 4} t & 1 - {1 \over 9} t & 0 & \\ 0 & t-1 & 1 - {1 \over 9} t & 1 - {1 \over 16}t & \\ 0 & 0 & t-1 & 1 - {1 \over 16}t & & \\ & & & & \ddots & \end{pmatrix} \end{equation} So the problem of finding $F(z;t)$ can be viewed as a special case of the general problem of evaluating generating functions of the sort \begin{equation} \sum_{n \geq 0} \, \det (T_n) \, z^n \end{equation} where $T_n$ is the $n \times n$ leading, principal submatrix of a infinite $\Bbb{N} \times \Bbb{N}$ tridiagonal matrix $T$. As of yet, I don't know enough of the theory (presumably results in the study of orthogonal functions and/or continued fractions) to be able to answer. Any help would be appreciated. ines.