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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 \ = \ \left\{ \begin{array}{ç} \displaystyle \ \ \ \, {1 \over n} \big( n^2 - t \big) \langle H_t \rangle_{n-1} \ \ + \\ \displaystyle {n-1 \over n} \big( 1 - t\big) \big(n^2 - t\big) \langle H_t \rangle_{n-2} \end{array} \right. \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) \ = \ \left\{ \begin{array}{c} \displaystyle \big(n^2 - t \big)\Bbb{f}_t(n-1) \\ + \\ \displaystyle \big(1 - t \big)\big(n^2 - t \big) \Bbb{f}_t(n-2) \end{array} \right. \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 $F(z;1) = 1$.

We can eliminate the first order term in the $(*)$-ODE by introducing an appropriate phase in the solution $F(z;t) := e^{\varphi(x;t)} \, K(z;t)$ where $\varphi(z;t):= - \log \big[\sqrt{z} \, \big( (1-t)z^2 + z -1\big) \big]$ the $(*)$-ODE becomes

\begin{equation} \begin{array}{c} \displaystyle (**) \ \ A(z) {d^2 \over {dz^2}} K(z;t) \ + \ D(z) \, K(z;t) \ = \ 0 \\ \text{where} \\ \begin{array}{l} \displaystyle A(z) \ = \ (1-t)z^3 + z^2 - z \\ \displaystyle D(z) \ = \ {(1-t)(4-t)z^2 + (1-4t)z - 1 \over {4z}} \\ \end{array} \end{array} \end{equation}

I'm not sure if this regauging will necessarily help; nevertheless I offer it as a potential avenue to crack the nut, so to speak.

The recurrence above is clearly implemented by tridiagonal determinants. Specifically $\Bbb{f}_t(n)$ equals the initial $n \times n$ principal minor of the following semi-infinite tridiagonal matrix:

\begin{equation} \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 $T$. If it were possible to show that the associated semi-infinite Hankel matrix

\begin{equation} H := \ \begin{pmatrix} \Bbb{f}_t(0) & \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \\ \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \\ \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \\ \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \Bbb{f}_t(6) & \\ & & & & \ddots \end{pmatrix} \end{equation}

were positive semi-definite (i.e. all finite, principal minors of $H$ are non-negative) then a result of Alan Sokal (see theorem 2 of https://arxiv.org/pdf/1804.04498.pdf) would imply that $F(z;t)$ has an expansion as $J$-type continued fraction

\begin{equation} {\alpha_0 \over {1 - z \gamma_0 \ - \ {\displaystyle z^2 \beta_1 \over {\displaystyle 1 - z \gamma_1 \ - \ {z^2 \beta_2 \over {\displaystyle 1 - z \gamma_2 \ - \ {z^2 \beta_3 \over {\ddots}}}}}}}} \end{equation}

where $\alpha_0 \geq 0$ is a real number and

\begin{equation} \begin{array}{ll} \displaystyle \underline{\beta} &\displaystyle = \ \big(\beta_1, \ \beta_2, \ \beta_3, \ \dots \big) \ \ \text{with $\beta_k \geq 0$ for all $k \geq 1$} \\ \displaystyle \underline{\gamma} &\displaystyle = \ \big(\gamma_0, \ \gamma_1, \ \gamma_2, \ \, \dots \big) \ \ \text{with $\gamma_k \in \Bbb{R}$ for all $k \geq 0$} \end{array} \end{equation}

When $t=0$ the Hankel matrix $H$ will consists of Fibonacci numbers

\begin{equation} H := \ \begin{pmatrix} 1 & 1 & 2 & 3 & \\ 1 & 2 & 3 & 5 & \\ 2 & 3 & 5 & 8 & \\ 3 & 5 & 8 & 13 & \\ & & & & \ddots \end{pmatrix} \end{equation}

which is clearly positive semi-definite. As WimC pointed out to me here (https://math.stackexchange.com/questions/4055848/what-is-the-j-type-continued-fraction-of-the-generating-function-of-the-fibonacc) the $J$-type continued fraction of $F(z;0) = \big(1 - z -z^2 \big)^{-1}$ is itself, i.e.

\begin{equation} {1 \over {1 - 1 \cdot z \ - \ {\displaystyle 1 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\ddots}}}}}}}} \end{equation}

Initial calculations using Mathematica endorse the claim that $H$ is positive semi-definite for generic values of $t$, but the initial parameters $\beta_1, \beta_2, \beta_3$ and $\gamma_0, \gamma_1, \gamma_2$ in the $J$-fraction expansion
do not have a particularly enlightening shape from which to discern a pattern.

Any help in these matters would be greatly appreciated.

ines.

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 \ = \ \left\{ \begin{array}{ç} \displaystyle \ \ \ \, {1 \over n} \big( n^2 - t \big) \langle H_t \rangle_{n-1} \ \ + \\ \displaystyle {n-1 \over n} \big( 1 - t\big) \big(n^2 - t\big) \langle H_t \rangle_{n-2} \end{array} \right. \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) \ = \ \left\{ \begin{array}{c} \displaystyle \big(n^2 - t \big)\Bbb{f}_t(n-1) \\ + \\ \displaystyle \big(1 - t \big)\big(n^2 - t \big) \Bbb{f}_t(n-2) \end{array} \right. \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} (\dagger \! \dagger \! \dagger) \ \ \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 $F(z;1) = 1$.

We can eliminate the first order term in the $(*)$-ODE by introducing an appropriate phase in the solution $F(z;t) := e^{\varphi(x;t)} \, K(z;t)$ where $\varphi(z;t):= - \log \big[\sqrt{z} \, \big( (1-t)z^2 + z -1\big) \big]$ the $(\dagger \! \dagger \! \dagger)$-ODE becomes

\begin{equation} \begin{array}{c} \displaystyle (**) \ \ A(z) {d^2 \over {dz^2}} K(z;t) \ + \ D(z) \, K(z;t) \ = \ 0 \\ \text{where} \\ \begin{array}{l} \displaystyle A(z) \ = \ (1-t)z^3 + z^2 - z \\ \displaystyle D(z) \ = \ {(1-t)(4-t)z^2 + (1-4t)z - 1 \over {4z}} \\ \end{array} \end{array} \end{equation}

I'm not sure if this regauging will necessarily help; nevertheless I offer it as a potential avenue to crack the nut, so to speak.

The recurrence above is clearly implemented by tridiagonal determinants. Specifically $\Bbb{f}_t(n)$ equals the initial $n \times n$ principal minor of the following semi-infinite tridiagonal matrix:

\begin{equation} \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 $T$. If it were possible to show that the associated semi-infinite Hankel matrix

\begin{equation} H := \ \begin{pmatrix} \Bbb{f}_t(0) & \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \\ \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \\ \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \\ \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \Bbb{f}_t(6) & \\ & & & & \ddots \end{pmatrix} \end{equation}

were positive semi-definite (i.e. all finite, principal minors of $H$ are non-negative) then a result of Alan Sokal (see theorem 2 of https://arxiv.org/pdf/1804.04498.pdf) would imply that $F(z;t)$ has an expansion as $J$-type continued fraction

\begin{equation} {\alpha_0 \over {1 - z \gamma_0 \ - \ {\displaystyle z^2 \beta_1 \over {\displaystyle 1 - z \gamma_1 \ - \ {z^2 \beta_2 \over {\displaystyle 1 - z \gamma_2 \ - \ {z^2 \beta_3 \over {\ddots}}}}}}}} \end{equation}

where $\alpha_0 \geq 0$ is a real number and

\begin{equation} \begin{array}{ll} \displaystyle \underline{\beta} &\displaystyle = \ \big(\beta_1, \ \beta_2, \ \beta_3, \ \dots \big) \ \ \text{with $\beta_k \geq 0$ for all $k \geq 1$} \\ \displaystyle \underline{\gamma} &\displaystyle = \ \big(\gamma_0, \ \gamma_1, \ \gamma_2, \ \, \dots \big) \ \ \text{with $\gamma_k \in \Bbb{R}$ for all $k \geq 0$} \end{array} \end{equation}

When $t=0$ the Hankel matrix $H$ will consists of Fibonacci numbers

\begin{equation} H := \ \begin{pmatrix} 1 & 1 & 2 & 3 & \\ 1 & 2 & 3 & 5 & \\ 2 & 3 & 5 & 8 & \\ 3 & 5 & 8 & 13 & \\ & & & & \ddots \end{pmatrix} \end{equation}

which is clearly positive semi-definite. As WimC pointed out to me here (https://math.stackexchange.com/questions/4055848/what-is-the-j-type-continued-fraction-of-the-generating-function-of-the-fibonacc) the $J$-type continued fraction of $F(z;0) = \big(1 - z -z^2 \big)^{-1}$ is itself, i.e.

\begin{equation} {1 \over {1 - 1 \cdot z \ - \ {\displaystyle 1 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\ddots}}}}}}}} \end{equation}

Initial calculations using Mathematica endorse the claim that $H$ is positive semi-definite for generic values of $t$, but the initial parameters $\beta_1, \beta_2, \beta_3$ and $\gamma_0, \gamma_1, \gamma_2$ in the $J$-fraction expansion
do not have a particularly enlightening shape from which to discern a pattern.

Any help in these matters would be greatly appreciated.

ines.

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 \ = \ \left\{ \begin{array}{ç} \displaystyle \ \ \ \, {1 \over n} \big( n^2 - t \big) \langle H_t \rangle_{n-1} \ \ + \\ \displaystyle {n-1 \over n} \big( 1 - t\big) \big(n^2 - t\big) \langle H_t \rangle_{n-2} \end{array} \right. \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) \ = \ \left\{ \begin{array}{c} \displaystyle \big(n^2 - t \big)\Bbb{f}_t(n-1) \\ + \\ \displaystyle \big(1 - t \big)\big(n^2 - t \big) \Bbb{f}_t(n-2) \end{array} \right. \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 $F(z;1) = 1$.

We can eliminate the first order term in the $(*)$-ODE by introducing an appropriate phase in the solution $F(z;t) := e^{\varphi(x;t)} \, K(z;t)$ where $\varphi(z;t):= z^{-{1 \over 2}} \big( (1-t)z^2 + z -1\big)^{-1}$ the $(*)$-ODE becomes

\begin{equation} \begin{array}{c} \displaystyle (**) \ \ A(z) {d^2 \over {dz^2}} K(z;t) \ + \ D(z) \, K(z;t) \ = \ 0 \\ \text{where} \\ \begin{array}{l} \displaystyle A(z) \ = \ (1-t)z^3 + z^2 - z \\ \displaystyle D(z) \ = \ {(1-t)(4-t)z^2 + (1-4t)z - 1 \over {4z}} \\ \end{array} \end{array} \end{equation}

I'm not sure if this regauging will necessarily help; nevertheless I offer it as a potential avenue to crack the nut, so to speak.

The recurrence above is clearly implemented by tridiagonal determinants. Specifically $\Bbb{f}_t(n)$ equals the initial $n \times n$ principal minor of the following semi-infinite tridiagonal matrix:

\begin{equation} \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 $T$. If it were possible to show that the associated semi-infinite Hankel matrix

\begin{equation} H := \ \begin{pmatrix} \Bbb{f}_t(0) & \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \\ \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \\ \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \\ \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \Bbb{f}_t(6) & \\ & & & & \ddots \end{pmatrix} \end{equation}

were positive semi-definite (i.e. all finite, principal minors of $H$ are non-negative) then a result of Alan Sokal (see theorem 2 of https://arxiv.org/pdf/1804.04498.pdf) would imply that $F(z;t)$ has an expansion as $J$-type continued fraction

\begin{equation} {\alpha_0 \over {1 - z \gamma_0 \ - \ {\displaystyle z^2 \beta_1 \over {\displaystyle 1 - z \gamma_1 \ - \ {z^2 \beta_2 \over {\displaystyle 1 - z \gamma_2 \ - \ {z^2 \beta_3 \over {\ddots}}}}}}}} \end{equation}

where $\alpha_0 \geq 0$ is a real number and

\begin{equation} \begin{array}{ll} \displaystyle \underline{\beta} &\displaystyle = \ \big(\beta_1, \ \beta_2, \ \beta_3, \ \dots \big) \ \ \text{with $\beta_k \geq 0$ for all $k \geq 1$} \\ \displaystyle \underline{\gamma} &\displaystyle = \ \big(\gamma_0, \ \gamma_1, \ \gamma_2, \ \, \dots \big) \ \ \text{with $\gamma_k \in \Bbb{R}$ for all $k \geq 0$} \end{array} \end{equation}

When $t=0$ the Hankel matrix $H$ will consists of Fibonacci numbers

\begin{equation} H := \ \begin{pmatrix} 1 & 1 & 2 & 3 & \\ 1 & 2 & 3 & 5 & \\ 2 & 3 & 5 & 8 & \\ 3 & 5 & 8 & 13 & \\ & & & & \ddots \end{pmatrix} \end{equation}

which is clearly positive semi-definite. As WimC pointed out to me here (https://math.stackexchange.com/questions/4055848/what-is-the-j-type-continued-fraction-of-the-generating-function-of-the-fibonacc) the $J$-type continued fraction of $F(z;0) = \big(1 - z -z^2 \big)^{-1}$ is itself, i.e.

\begin{equation} {1 \over {1 - 1 \cdot z \ - \ {\displaystyle 1 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\ddots}}}}}}}} \end{equation}

Initial calculations using Mathematica endorse the claim that $H$ is positive semi-definite for generic values of $t$, but the initial parameters $\beta_1, \beta_2, \beta_3$ and $\gamma_0, \gamma_1, \gamma_2$ in the $J$-fraction expansion
do not have a particularly enlightening shape from which to discern a pattern.

Any help in these matters would be greatly appreciated.

ines.

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 \ = \ \left\{ \begin{array}{ç} \displaystyle \ \ \ \, {1 \over n} \big( n^2 - t \big) \langle H_t \rangle_{n-1} \ \ + \\ \displaystyle {n-1 \over n} \big( 1 - t\big) \big(n^2 - t\big) \langle H_t \rangle_{n-2} \end{array} \right. \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) \ = \ \left\{ \begin{array}{c} \displaystyle \big(n^2 - t \big)\Bbb{f}_t(n-1) \\ + \\ \displaystyle \big(1 - t \big)\big(n^2 - t \big) \Bbb{f}_t(n-2) \end{array} \right. \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 $F(z;1) = 1$.

We can eliminate the first order term in the $(*)$-ODE by introducing an appropriate phase in the solution $F(z;t) := e^{\varphi(x;t)} \, K(z;t)$ where $\varphi(z;t):= - \log \big[\sqrt{z} \, \big( (1-t)z^2 + z -1\big) \big]$ the $(*)$-ODE becomes

\begin{equation} \begin{array}{c} \displaystyle (**) \ \ A(z) {d^2 \over {dz^2}} K(z;t) \ + \ D(z) \, K(z;t) \ = \ 0 \\ \text{where} \\ \begin{array}{l} \displaystyle A(z) \ = \ (1-t)z^3 + z^2 - z \\ \displaystyle D(z) \ = \ {(1-t)(4-t)z^2 + (1-4t)z - 1 \over {4z}} \\ \end{array} \end{array} \end{equation}

I'm not sure if this regauging will necessarily help; nevertheless I offer it as a potential avenue to crack the nut, so to speak.

The recurrence above is clearly implemented by tridiagonal determinants. Specifically $\Bbb{f}_t(n)$ equals the initial $n \times n$ principal minor of the following semi-infinite tridiagonal matrix:

\begin{equation} \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 $T$. If it were possible to show that the associated semi-infinite Hankel matrix

\begin{equation} H := \ \begin{pmatrix} \Bbb{f}_t(0) & \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \\ \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \\ \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \\ \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \Bbb{f}_t(6) & \\ & & & & \ddots \end{pmatrix} \end{equation}

were positive semi-definite (i.e. all finite, principal minors of $H$ are non-negative) then a result of Alan Sokal (see theorem 2 of https://arxiv.org/pdf/1804.04498.pdf) would imply that $F(z;t)$ has an expansion as $J$-type continued fraction

\begin{equation} {\alpha_0 \over {1 - z \gamma_0 \ - \ {\displaystyle z^2 \beta_1 \over {\displaystyle 1 - z \gamma_1 \ - \ {z^2 \beta_2 \over {\displaystyle 1 - z \gamma_2 \ - \ {z^2 \beta_3 \over {\ddots}}}}}}}} \end{equation}

where $\alpha_0 \geq 0$ is a real number and

\begin{equation} \begin{array}{ll} \displaystyle \underline{\beta} &\displaystyle = \ \big(\beta_1, \ \beta_2, \ \beta_3, \ \dots \big) \ \ \text{with $\beta_k \geq 0$ for all $k \geq 1$} \\ \displaystyle \underline{\gamma} &\displaystyle = \ \big(\gamma_0, \ \gamma_1, \ \gamma_2, \ \, \dots \big) \ \ \text{with $\gamma_k \in \Bbb{R}$ for all $k \geq 0$} \end{array} \end{equation}

When $t=0$ the Hankel matrix $H$ will consists of Fibonacci numbers

\begin{equation} H := \ \begin{pmatrix} 1 & 1 & 2 & 3 & \\ 1 & 2 & 3 & 5 & \\ 2 & 3 & 5 & 8 & \\ 3 & 5 & 8 & 13 & \\ & & & & \ddots \end{pmatrix} \end{equation}

which is clearly positive semi-definite. As WimC pointed out to me here (https://math.stackexchange.com/questions/4055848/what-is-the-j-type-continued-fraction-of-the-generating-function-of-the-fibonacc) the $J$-type continued fraction of $F(z;0) = \big(1 - z -z^2 \big)^{-1}$ is itself, i.e.

\begin{equation} {1 \over {1 - 1 \cdot z \ - \ {\displaystyle 1 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\ddots}}}}}}}} \end{equation}

Initial calculations using Mathematica endorse the claim that $H$ is positive semi-definite for generic values of $t$, but the initial parameters $\beta_1, \beta_2, \beta_3$ and $\gamma_0, \gamma_1, \gamma_2$ in the $J$-fraction expansion
do not have a particularly enlightening shape from which to discern a pattern.

Any help in these matters would be greatly appreciated.

ines.

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 \ = \ \left\{ \begin{array}{ç} \displaystyle \ \ \ \, {1 \over n} \big( n^2 - t \big) \langle H_t \rangle_{n-1} \ \ + \\ \displaystyle {n-1 \over n} \big( 1 - t\big) \big(n^2 - t\big) \langle H_t \rangle_{n-2} \end{array} \right. \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) \ = \ \left\{ \begin{array}{c} \displaystyle \big(n^2 - t \big)\Bbb{f}_t(n-1) \\ + \\ \displaystyle \big(1 - t \big)\big(n^2 - t \big) \Bbb{f}_t(n-2) \end{array} \right. \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 $F(z;1) = 1$.

We can eliminate the first order term in the $(*)$-ODE by introducing an appropriate phase in the solution $F(z;t) := e^{\varphi(x;t)} \, K(z;t)$. For $\varphi(z;t):= z^{-{1 \over 2}} \big( (1-t)z^2 + z -1\big)^{-1}$ the $(*)$-ODE becomes

\begin{equation} \begin{array}{c} \displaystyle (**) \ \ A(z) {d^2 \over {dz^2}} K(z;t) \ + \ D(z) \, K(z;t) \ = \ 0 \\ \text{where} \\ \begin{array}{l} \displaystyle A(z) \ = \ (1-t)z^3 + z^2 - z \\ \displaystyle D(z) \ = \ {(1-t)(4-t)z^2 + (1-4t)z - 1 \over {4z}} \\ \end{array} \end{array} \end{equation}

I'm not sure if this regauging will necessarily help; nevertheless I offer it as a potential avenue to crack the nut, so to speak.

The recurrence above is clearly implemented by tridiagonal determinants. Specifically $\Bbb{f}_t(n)$ equals the initial $n \times n$ principal minor of the following semi-infinite tridiagonal matrix:

\begin{equation} \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 $T$. If it were possible to show that the associated semi-infinite Hankel matrix

\begin{equation} H := \ \begin{pmatrix} \Bbb{f}_t(0) & \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \\ \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \\ \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \\ \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \Bbb{f}_t(6) & \\ & & & & \ddots \end{pmatrix} \end{equation}

were positive semi-definite (i.e. all finite, principal minors of $H$ are non-negative) then a result of Alan Sokal (see theorem 2 of https://arxiv.org/pdf/1804.04498.pdf) would imply that $F(z;t)$ has an expansion as $J$-type continued fraction

\begin{equation} {\alpha_0 \over {1 - z \gamma_0 \ - \ {\displaystyle z^2 \beta_1 \over {\displaystyle 1 - z \gamma_1 \ - \ {z^2 \beta_2 \over {\displaystyle 1 - z \gamma_2 \ - \ {z^2 \beta_3 \over {\ddots}}}}}}}} \end{equation}

where $\alpha_0 \geq 0$ is a real number and

\begin{equation} \begin{array}{ll} \displaystyle \underline{\beta} &\displaystyle = \ \big(\beta_1, \ \beta_2, \ \beta_3, \ \dots \big) \ \ \text{with $\beta_k \geq 0$ for all $k \geq 1$} \\ \displaystyle \underline{\gamma} &\displaystyle = \ \big(\gamma_0, \ \gamma_1, \ \gamma_2, \ \, \dots \big) \ \ \text{with $\gamma_k \in \Bbb{R}$ for all $k \geq 0$} \end{array} \end{equation}

When $t=0$ the Hankel matrix $H$ will consists of Fibonacci numbers

\begin{equation} H := \ \begin{pmatrix} 1 & 1 & 2 & 3 & \\ 1 & 2 & 3 & 5 & \\ 2 & 3 & 5 & 8 & \\ 3 & 5 & 8 & 13 & \\ & & & & \ddots \end{pmatrix} \end{equation}

which is clearly positive semi-definite. As WimC pointed out to me here (https://math.stackexchange.com/questions/4055848/what-is-the-j-type-continued-fraction-of-the-generating-function-of-the-fibonacc) the $J$-type continued fraction of $F(z;0) = \big(1 - z -z^2 \big)^{-1}$ is itself, i.e.

\begin{equation} {1 \over {1 - 1 \cdot z \ - \ {\displaystyle 1 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\ddots}}}}}}}} \end{equation}

Initial calculations using Mathematica endorse the claim that $H$ is positive semi-definite for generic values of $t$, but the initial parameters $\beta_1, \beta_2, \beta_3$ and $\gamma_0, \gamma_1, \gamma_2$ in the $J$-fraction expansion
do not have a particularly enlightening shape from which to discern a pattern.

Any help in these matters would be greatly appreciated.

ines.

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 \ = \ \left\{ \begin{array}{ç} \displaystyle \ \ \ \, {1 \over n} \big( n^2 - t \big) \langle H_t \rangle_{n-1} \ \ + \\ \displaystyle {n-1 \over n} \big( 1 - t\big) \big(n^2 - t\big) \langle H_t \rangle_{n-2} \end{array} \right. \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) \ = \ \left\{ \begin{array}{c} \displaystyle \big(n^2 - t \big)\Bbb{f}_t(n-1) \\ + \\ \displaystyle \big(1 - t \big)\big(n^2 - t \big) \Bbb{f}_t(n-2) \end{array} \right. \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 $F(z;1) = 1$.

We can eliminate the first order term in the $(*)$-ODE by introducing an appropriate phase in the solution $F(z;t) := e^{\varphi(x;t)} \, K(z;t)$ where $\varphi(z;t):= z^{-{1 \over 2}} \big( (1-t)z^2 + z -1\big)^{-1}$ the $(*)$-ODE becomes

\begin{equation} \begin{array}{c} \displaystyle (**) \ \ A(z) {d^2 \over {dz^2}} K(z;t) \ + \ D(z) \, K(z;t) \ = \ 0 \\ \text{where} \\ \begin{array}{l} \displaystyle A(z) \ = \ (1-t)z^3 + z^2 - z \\ \displaystyle D(z) \ = \ {(1-t)(4-t)z^2 + (1-4t)z - 1 \over {4z}} \\ \end{array} \end{array} \end{equation}

I'm not sure if this regauging will necessarily help; nevertheless I offer it as a potential avenue to crack the nut, so to speak.

The recurrence above is clearly implemented by tridiagonal determinants. Specifically $\Bbb{f}_t(n)$ equals the initial $n \times n$ principal minor of the following semi-infinite tridiagonal matrix:

\begin{equation} \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 $T$. If it were possible to show that the associated semi-infinite Hankel matrix

\begin{equation} H := \ \begin{pmatrix} \Bbb{f}_t(0) & \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \\ \Bbb{f}_t(1) & \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \\ \Bbb{f}_t(2) & \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \\ \Bbb{f}_t(3) & \Bbb{f}_t(4) & \Bbb{f}_t(5) & \Bbb{f}_t(6) & \\ & & & & \ddots \end{pmatrix} \end{equation}

were positive semi-definite (i.e. all finite, principal minors of $H$ are non-negative) then a result of Alan Sokal (see theorem 2 of https://arxiv.org/pdf/1804.04498.pdf) would imply that $F(z;t)$ has an expansion as $J$-type continued fraction

\begin{equation} {\alpha_0 \over {1 - z \gamma_0 \ - \ {\displaystyle z^2 \beta_1 \over {\displaystyle 1 - z \gamma_1 \ - \ {z^2 \beta_2 \over {\displaystyle 1 - z \gamma_2 \ - \ {z^2 \beta_3 \over {\ddots}}}}}}}} \end{equation}

where $\alpha_0 \geq 0$ is a real number and

\begin{equation} \begin{array}{ll} \displaystyle \underline{\beta} &\displaystyle = \ \big(\beta_1, \ \beta_2, \ \beta_3, \ \dots \big) \ \ \text{with $\beta_k \geq 0$ for all $k \geq 1$} \\ \displaystyle \underline{\gamma} &\displaystyle = \ \big(\gamma_0, \ \gamma_1, \ \gamma_2, \ \, \dots \big) \ \ \text{with $\gamma_k \in \Bbb{R}$ for all $k \geq 0$} \end{array} \end{equation}

When $t=0$ the Hankel matrix $H$ will consists of Fibonacci numbers

\begin{equation} H := \ \begin{pmatrix} 1 & 1 & 2 & 3 & \\ 1 & 2 & 3 & 5 & \\ 2 & 3 & 5 & 8 & \\ 3 & 5 & 8 & 13 & \\ & & & & \ddots \end{pmatrix} \end{equation}

which is clearly positive semi-definite. As WimC pointed out to me here (https://math.stackexchange.com/questions/4055848/what-is-the-j-type-continued-fraction-of-the-generating-function-of-the-fibonacc) the $J$-type continued fraction of $F(z;0) = \big(1 - z -z^2 \big)^{-1}$ is itself, i.e.

\begin{equation} {1 \over {1 - 1 \cdot z \ - \ {\displaystyle 1 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\ddots}}}}}}}} \end{equation}

Initial calculations using Mathematica endorse the claim that $H$ is positive semi-definite for generic values of $t$, but the initial parameters $\beta_1, \beta_2, \beta_3$ and $\gamma_0, \gamma_1, \gamma_2$ in the $J$-fraction expansion
do not have a particularly enlightening shape from which to discern a pattern.

Any help in these matters would be greatly appreciated.

ines.