# Young-Fibonacci version of Nekrasov-Okounkov

This question addresses a hierarchy of linear recurrences which arise from an attempt to generalize the Nekrasov-Okounkov formula to the Young-Fibonacci setting. A related posting

extensions of the Nekrasov-Okounkov formula

asks how one might try to extend the Nekrasov-Okounkov formula by replacing the Plancherel measure on the Young lattice $$\Bbb{Y}$$ with another ergodic, central measure. In this discussion, I want to instead replace the Young lattice $$\Bbb{Y}$$ by the Young-Fibonacci lattice $$\Bbb{YF}$$ which comes equipped with its own Plancherel measure in virtue of being a $$1$$-differential poset. Allow me to briefly review some basics of the Young-Fibonacci lattice before I state the putative $$\Bbb{YF}$$-version of the Nekrasov-Okounkov partition function.

Young-Fibonacci Preliminaries: Recall that a fibonacci word $$u$$ is a word formed out of the alphabet $$\{1,2\}$$. As a set $$\Bbb{YF}$$ is the collection of a (finite) fibonacci words and $$\Bbb{YF}_n$$ will denote the set of fibonacci words $$u \in \Bbb{YF}$$ of length $$|u|=n$$ where $$|u|:= a_1 + \cdots + a_k$$ and where $$u=a_k \cdots a_1$$ is the parsing of $$u$$ into its digits $$a_1, \dots, a_k \in \{1,2 \}$$. The adjective Fibonacci reflects the fact that the cardinality of $$\Bbb{YF}_n$$ is the $$n$$-th Fibonacci number. I will skip defining the poset structure on $$\Bbb{YF}$$ and instead I point the readers to the Wikipedia page https://en.wikipedia.org/wiki/Young–Fibonacci_lattice. Suffice it to say that when endowed with an appropriate partial order $$\unlhd$$ the set $$\Bbb{YF}$$ becomes a ranked, modular (but not distributive), $$1$$-differential lattice. R. Stanley's $$1$$-differential property (see https://en.wikipedia.org/wiki/Differential_poset) is key here because it implies that the function $$\mu_\mathrm{P}: \Bbb{YF} \longrightarrow \Bbb{R}_{>0}$$ defined by

$$$$\begin{array}{ll} \mu_\mathrm{P}(u) &\displaystyle := \ { \dim^2(u) \over {|u|!}} \quad \text{where} \\ \dim(u) &\displaystyle := \ \# \left\{ \begin{array}{l} \text{all saturated chains (u_0 \lhd \cdots \lhd u_n) in \Bbb{YF}} \\ \text{starting with u_0 = \emptyset and ending at u_n =u} \end{array} \right\} \end{array}$$$$

restricts to a positive probability distribution $$\mu^{(n)}_\mathrm{P}$$ on $$\Bbb{YF}_n$$ for each $$n \geq 0$$. In fact $$\mu_\mathrm{P}$$ satisfies a stronger property known as coherence: The ratios

$$$$\tilde{\mu}_\mathrm{P}(u \lhd v) \ := \ {\mu_\mathrm{P}(v) \over {\mu_\mathrm{P}(u)}}$$$$

restrict to a probability distribution $$\tilde{\mu}_{\mathrm{P},u}$$ on the set of covering relations $$u \lhd v$$ (i.e. edges in the Hasse diagram of $$\Bbb{YF}$$) for any fixed $$u \in \Bbb{YF}_n$$. We refer to $$\mu^{(n)}_\mathrm{P}$$ as the Plancherel measure for $$\Bbb{YF}_n$$. If $$S:\Bbb{YF} \longrightarrow \Bbb{R}_{\geq 0}$$ is some statistic let $$\langle S \rangle_n$$ denote its expectation value with respect to the Plancherel measure, i.e.

$$$$\langle S \rangle_n \ := \ \sum_{|u|=n} \, {\dim^2(u) \over {n!}} \, S(u)$$$$

We may visualize a fibonacci word $$u \in \Bbb{YF}$$ using a profile of boxes akin to the way one depicts a partition by its Young diagram. The following example with $$u = 12112211$$ should illustrate the concept of a Young-Fibonacci diagram clearly. For emphasis each digit of the fibonacci word $$u$$ is written directly underneath the corresponding column of boxes:

$$$$\begin{array}{cccccccc} & \Box & & & \Box & \Box & & \\ \Box & \Box & \Box & \Box & \Box & \Box & \Box & \Box \\ 1 & 2 & 1 & 1 & 2 & 2 & 1 & 1 \end{array}$$$$

A Fibonacci word $$u$$ will be synonymous with its Young-Fibonacci diagram and $$\Box \in u$$ will indicate membership of a box. The hook length $$\mathrm{h}(\Box)$$ of a box $$\Box \in u$$ is defined to be $$1$$ whenever it is in the top row; otherwise $$\mathrm{h}(\Box)$$ equals $$1$$ plus the total number of boxes directly above it and to its right. For example the hook lengths of the boxes of $$u = 12112211$$ are indicated in the tableaux below:

$$$$\begin{array}{cccccccc} & \boxed{1 \ \ } & & & \boxed{1 \ \ } & \boxed{1 \ \ } & & \\ \boxed{11} & \boxed{10} & \boxed{8 \ \ } & \boxed{7 \ \ } & \boxed{6 \ \ } & \boxed{4 \ \ } & \boxed{2 \ \ } & \boxed{1 \ \ } \end{array}$$$$

These graphical conventions allows us to reformulate the value of $$\mu_\mathrm{P}(u)$$ in terms of (the squares of) the hook-lenghts of $$u \in \Bbb{Y}$$, i.e.

$$$$\mu_\mathrm{P}(u) \ = \ \prod_{\Box \, \in \, u} \, {|u|! \over {\mathrm{h}^2(\Box)} }$$$$

This is a non-trivial observation made by R. Stanley in the course of his work examining differential posets.

The $$\Bbb{YF}$$-version of the Nekrasov-Okounkov partition function: For a fibonacci words $$u \in \Bbb{YF}$$ define a $$t$$-statistic $$H_t(u) := \prod_{\Box \, \in \, u} \, \big(\mathrm{h}^2(\Box) - t \big)$$ and the $$\Bbb{YF}$$-Nekrasov-Okounkov partition function as

$$$$\begin{array}{ll} F(z;t) &\displaystyle = \ \sum_{n \geq 0} {z^n \over {n!}} \, \langle H_t \rangle_n \\ &\displaystyle = \ \sum_{n \geq 0} {z^n \over {n!}} \, \sum_{|u|=n} \, {\dim^2(u) \over {n!}} \, H_t(u) \end{array}$$$$

Given a fibonacci word $$u$$ let $$E_k(u)$$ be the elementary symmetric polynomial in the square hook lengths $$\mathrm{h}^2(\Box)$$ for $$\Box \in u$$ with the conventions that $$E_k(u) = 0$$ whenever $$k > |u|$$ and that $$E_0(u) = 1$$ for all $$u \in \Bbb{YF}$$. Following a hint from Stanley's notes "Partition Statistics with Respect to Plancherel Measure" (http://www-math.mit.edu/~rstan/transparencies/plancherel.ps) we will try to compute $$F(z;t)$$ by working out a recursion for the expectation values $$\langle E_k \rangle_n$$. It will be convenient to make a change of variable $$z \mapsto -z$$ and consider $$F^\vee(z;t) := F(-z;t)$$ instead; the effect of this sign-change is to replace the statistic $$H_t(u)$$ by $$H^\vee_t(u) := \prod_{\Box \, \in \, u} \, \big(t -\mathrm{h}^2(\Box) \big)$$ in the definition of the partition function. After expanding into elementary symmetric polynomials $$E_k$$ we get

$$$$\begin{array}{ll} \displaystyle H^\vee_t(u) &\displaystyle = \ \sum_{k=0}^{|u|} \, (-1)^k \, E_{k}(u) \, t^{|u|-k} \\ \\ &\text{--- and so ---} \\ \\ \displaystyle F^\vee(z;t) &\displaystyle = \ \sum_{n \geq 0} \, {z^n \over {n!}} \, \langle H^\vee_t \rangle_n \\ &\displaystyle = \ \sum_{n \geq 0} \, {z^n \over {n!}} \, \sum_{k=0}^n \, (-1)^k \, \langle E_k \rangle_n \, t^{n-k} \\ &\displaystyle = \ \sum_{k \geq 0} \, (-t)^{-k} \, \underbrace{\sum_{n \geq 0} \, (zt)^n \, {\langle E_k \rangle_n \over {n!}}}_{\text{= \, F^\vee_k(zt) see below}} \end{array}$$$$

Evaluating expectation values: Fibonacci words $$u \in \Bbb{YF}_n$$ with $$n \geq 2$$ can be separated into two disjoint groups: Those of the form $$u=1v$$ for $$v \in \Bbb{YF}_{n-1}$$ and those of the form $$u=2v$$ for $$v \in \Bbb{YF}_{n-2}$$. Depending on whether the prefix of $$u$$ is $$1$$ or $$2$$ we can write down a recursive formula for the value of $$E_k(u) := E_k \big( \mathrm{h}^2(\Box) \big)_{\Box \, \in \, u}$$ by analyzing the hook length(s) of the box(es) in the left-most column, specifically:

$$$$\begin{array}{lll} E_k(1v) &= E_k(v) + n^2E_{k-1}(v) &\text{if} \ |v| = n-1 \\ E_k(2v) &= E_k(v) + (n^2+1)E_{k-1}(v) + n^2E_{k-2}(v) &\text{if} \ |v| = n-2 \end{array}$$$$

Using the observation that $$\dim(1v) = \dim(v)$$ and $$\dim(2v) = (|v| + 1)^2 \dim(v)$$ we may conclude

$$$$\langle E_k \rangle_n = \left\{ \begin{array}{l} \displaystyle {1 \over n} \langle E_k \rangle_{n-1} \ + \ {n-1 \over n} \langle E_k \rangle_{n-2} \\ \\ \displaystyle + \ n \langle E_{k-1} \rangle_{n-1} \ + \ {(n-1)(n^2+1) \over n} \langle E_{k-1} \rangle_{n-2} \\ \\ \displaystyle + \ n(n-1) \langle E_{k-2} \rangle_{n-2} \end{array} \right.$$$$

If we set $$\sigma_k(n) := {1 \over {n!}} \, \langle E_k \rangle_n$$ then the above recursion can be rewritten as:

$$$$(\dagger) \ \ \left\{ \begin{array}{l} \displaystyle n^2\sigma_k(n) \ = \ \underbrace{\sigma_k(n-1) \ + \ \sigma_k(n-2)}_{\text{homogeneous part}} \ + \ \gamma_{

all of which can be converted, using the usual yoga of generating functions, into the following second order inhomogeneous ODE for $$F^\vee_k(x) := \sum_{n \geq 0} \sigma_k(n) x^n$$.

$$$$\begin{array}{c} \displaystyle x^2 \, {d^2 \over {dx^2}} F^\vee_k(x) \ + \ x {d \over {dx}} F^\vee_k(x) \ - \ \big(x^2 + x \big) F^\vee_k(x) \\ \displaystyle \ = \ \\ \displaystyle G_{

which, after setting $$F^\vee_k(x) := e^x J^\vee_k(x)$$, can be rewritten as

$$$$(\dagger \dagger) \ \ \left\{ \begin{array}{c} \displaystyle x {d^2 \over {dx^2}} J^\vee_k(x) \ + \ \big(2x + 1 \big) {d \over {dx}} J^\vee_k(x) \\ = \\ \displaystyle {1 \over x} \Big[ G_{

where the generating function $$G_{ will have been evaluated earlier by induction on $$k \geq 0$$. The associated homogeneous ODE of $$(\dagger \dagger)$$ has two nice independent solutions $$Y_1(x) = 1$$ and $$Y_2(x)= \int x^{-1} e^{-2x} dx$$ whose Wronskian is $$W=x^{-1} e^{-2x}$$. One starts the inductive engine beginning with $$F^\vee_0(x) = e^x$$ or, equivalently with $$J^\vee_0(x) = 1$$. For $$k=1$$ clearly $$\sigma_1(0)=0$$ and $$\sigma_1(1)=1$$ while

$$$$\begin{array}{ll} \displaystyle G_{<1}(x) &\displaystyle = \ \sum_{n \geq 2} \, {n^3 + n -1 \over {(n-1)!}} \, x^n \\ &\displaystyle = \ \big(x + 8x^2 + 6x^3 + x^4 \big) \, e^x \ - \ x \end{array}$$$$

so the ODE for $$J^\vee_1(x)$$ becomes

$$$$\begin{array}{c} \displaystyle x {d^2 \over {dx^2}} J^\vee_1(x) \ + \ \big(2x+1\big) {d \over {dx}} J^\vee_1(x) \\ \displaystyle = \\ \underbrace{\big(x + 8x^2 + 6x^3 + x^4 \big) \, e^x}_{G_{<1}(x) \ + \ x } \end{array}$$$$

By variation of parameters, a particular inhomogeneous solution is

$$$$\begin{array}{rl} \displaystyle Y_\mathrm{particular}(x) &\displaystyle = \ v_1(x) \cdot Y_1(x) \ + \ v_2(x) \cdot Y_2(x) \\ \displaystyle v_1(x) &\displaystyle = \ -\int xe^{2x} \, Y_2(x) \, \Big(x + G_{<1}(x) \Big) \, dx \\ \displaystyle v_2(x) &\displaystyle = \ \ \ \ \ \int xe^{2x} \, Y_1(x) \, \Big(x + G_{<1}(x) \Big) \, dx \end{array}$$$$

After solving $$J^\vee_1(x)$$ (and thus for $$F^\vee_1(x)$$) we repeat the process for $$k > 1$$. At each stage we solve by variation of parameters, using the two homogeneous solutions $$Y_1(x)$$ and $$Y_2(x)$$, the $$(\dagger \dagger)$$-ODE whose inhomogeneous term is itself computed from the data obtained in the previous layer of computation.

Question. Does anyone know how to solve either the $$(\dagger)$$-hierarchy of linear recurrences, the $$(\dagger \!\dagger)$$-hierarchy of 2nd order inhomogeneous ODEs, or equivalently the $$(\dagger \! \dagger \! \dagger)$$-ODE explained in the second answer/response below? By solve I mean to express the solution in terms of elementary functions, continued fractions, or else by some nice class of special functions (e.g. hypergeometric).

thanks, ines.

• Hi. The guideline for writing questions usually do not encourage the use of formulas like "Thank you", "Thanks in advance" or similar, since questions and answers are considered as mini blog-posts and one wants to minimize noise and clutter. I tend to agree with this interpretation. Instead of saying "thanks" (that is implicit) one can upvote the answers that helped them, and accept the answer that helped them the most. See the related post meta.stackexchange.com/questions/115694/… Feb 23 at 6:53
• For a fixed valued of $k \geq 0$ the ODE corresponding to the generating function of the expectation values $\langle E_k \rangle_n$ is first order and (hopefully) easier to solve. If its solution can be obtained, it may then be possible to recover the generating function of the $\sigma_k(n)$ by the inverse Laplace transform. I'll work on this (at least for small values of $k$). Feb 26 at 4:37
• I'm referring to this trick from the theory of Borel re-summation: $\begin{array}{ll} \displaystyle \sum_{n \geq 0} \, \langle E_k \rangle_n \, z^n &\displaystyle = \ \int_0^\infty \, dt \, e^{-t} \sum_{n \geq 0} \, {\langle E_k \rangle_n \over {n!}} \, (zt)^n \\ &\displaystyle = \ \int_0^\infty \, dt \, e^{-t} \, F_k(zt) \\ &\displaystyle = \ z^{-1} \text{$\frak{L}$} \big\{ F_k \big\}(z^{-1}) \quad \text{Laplace transform} \end{array}$ Feb 26 at 6:19
• See the section OGF $\displaystyle \longleftrightarrow$ EGF conversion formulas in the Wikipedia entry en.wikipedia.org/wiki/… Feb 26 at 6:24
• Any explanation for the minus vote ? Jul 4 at 20:48

I want to propose a (recursive) method to obtain $$F_k(x)$$ by exploiting a formula well known from the theory of Borel re-summation. In the present context it looks like this:

$$$$\begin{array}{ll} \displaystyle \sum_{n \geq 0} \, \langle E_k \rangle_n \, z^n &\displaystyle = \ \int_0^\infty \, dt \, e^{-t} \sum_{n \geq 0} \, {\langle E_k \rangle_n \over {n!}} \, (zt)^n \\ &\displaystyle = \ \int_0^\infty \, dt \, e^{-t} \, F_k(zt) \\ &\displaystyle = \ \Big( {1 \over z} \Big) \cdot \,\text{\frak{L}} \big\{ F_k \big\} \Big({1 \over z} \Big) \quad \text{Laplace transform} \end{array}$$$$

I'll illustrate the approach for $$k=1$$. Let $$E_1(x) := \sum_{n \geq 0} \, \langle E_1 \rangle_n \, x^n$$. In this case the recurrence relation, as indicated the initial post, is

$$$$\underbrace{n \langle E_1 \rangle_n}_{xE_1'(x) - x} \ = \ \left\{ \begin{array}{l} \displaystyle \underbrace{\langle E_1 \rangle_{n-1}}_{xE_1(x)} \ + \ \underbrace{(n-1) \langle E_1 \rangle_{n-2}}_{x^3E_1'(x) \ + \ x^2E_1(x)} \ + \ \\ \displaystyle \underbrace{\ n^2 \langle E_0 \rangle_{n-1} \ }_{\sum_{n \geq 2} n^2 x^n} \ + \ \underbrace{(n-1)(n^2+1)\langle E_0 \rangle_{n-2}}_{\sum_{n \geq 2} (n-1)(n^2+1) \, x^n} \end{array} \right.$$$$

for $$n \geq 2$$ with the initial values $$\langle E_1\rangle_0 = 0$$ and $$\langle E_1 \rangle_1 = 1$$. I've "underbraced" the terms of the recurrence by their respective contributions to the $$1$$-st order non-homogeneous ODE for the generating function $$E_1(x)$$ which is, upon simplification,

$$$$E'_1(x) \ + \ {1 \over {x-1}} \, E_1(x) \ = \ {x^3 -x^2 + 5x +1 \over {(1+x)(1-x)^5}}$$$$

Mathematica tells me that the general solution is of the form

$$$$\ {\text{const} \over {1-x}} \ + \ {1 \over 4} {x^2+x+2 \over {(1-x)^4}} \ + \ {3 \over 8} {1 \over {(1-x)}} \log \Big( {1-x \over {1+x}} \Big)$$$$

The value of the constant term is forced to be $$\text{const} = - {1 \over 2}$$ since $$\langle E_1\rangle_0 = 0$$. After performing the requisite inverse transforms we should get (excuse me for mimicking Wikipedia)

$$$$\begin{array}{ll} \displaystyle F_k(zt) &\displaystyle = \ \sum_{n \geq 0} \, {\langle E_1 \rangle_n \over {n!}} \, (zt)^n \\ &\displaystyle = \ {1 \over {2\pi}} \, \int_{-\pi}^{\pi} \, d\theta \, E_1 \big(zt e^{-i\theta} \big) \exp \big( e^{i\theta} \big) \\ &\displaystyle \ \ \ \ \ \text{(Not sure what this is yet.)} \end{array}$$$$

ines.

• Clearly the trick should be the write down and solve the $1$-st order ODE for $\tilde{E}_k(x) := {1 \over {(1-x)}} \, E_k(x)$ where $E_k(x) := \sum_{n \geq 0} \, \langle E_k \rangle_n \, x^n$. Feb 28 at 18:38

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

$$$$\begin{array}{lll} \displaystyle H_t(1v) &\displaystyle = \, \big(n^2 - t \big) H_t(v) &\text{if |v|=n-1} \\ \displaystyle H_t(2v) &\displaystyle = \, \big( 1 - t \big) \big( n^2 - t \big) H_t(v) &\text{if |v|=n-2} \end{array}$$$$

and consequently

$$$$\langle H_t \rangle_n \ = \ \left\{ \begin{array}{ç} \displaystyle \ \ \ \, {1 \over n} \big( n^2 - t \big) \langle H_t \rangle_{n-1} \ \ + \\ \displaystyle {n-1 \over n} \big( 1 - t\big) \big(n^2 - t\big) \langle H_t \rangle_{n-2} \end{array} \right.$$$$

or, after setting $$\Bbb{f}_t(n) := {1 \over {n!}} \langle H_t \rangle_n$$

$$$$n^2 \Bbb{f}_t(n) \ = \ \left\{ \begin{array}{c} \displaystyle \big(n^2 - t \big)\Bbb{f}_t(n-1) \\ + \\ \displaystyle \big(1 - t \big)\big(n^2 - t \big) \Bbb{f}_t(n-2) \end{array} \right.$$$$

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:

$$$$\begin{array}{c} (\dagger \! \dagger \! \dagger) \ \ \displaystyle A(z) {d^2 \over {dz^2}} F(z;t) \ + \ B(z) {d \over {dz}} F(z;t) \ + \ C(z) F(z;t) \ = \ 0 \\ \text{where} \\ \begin{array}{l} A(z) \ = \ (1-t)z^3 + z^2 - z \\ B(z) \ = \ 5(1-t)z^2 + 3z - 1 \\ C(z) \ = \ (1-t)(4-t)z + (1-t) \end{array} \end{array}$$$$

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.

$$$$\begin{array}{l} \displaystyle F(z;0) &\displaystyle = \ 1 + z + 2z^2 + 3z^3 + 5z^4 + \cdots \\ &\displaystyle = \ {1 \over {(1 - z -z^2)}} \end{array}$$$$

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

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

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:

$$$$\begin{pmatrix} 1-t & 1 - {1 \over 4} t & 0 & 0 & \\ t-1 & 1 - {1 \over 4} t & 1 - {1 \over 9} t & 0 & \\ 0 & t-1 & 1 - {1 \over 9} t & 1 - {1 \over 16}t & \\ 0 & 0 & t-1 & 1 - {1 \over 16}t & & \\ & & & & \ddots & \end{pmatrix}$$$$

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

$$$$\sum_{n \geq 0} \, \det (T_n) \, z^n$$$$

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

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

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

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

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

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

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

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

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.

$$$${1 \over {1 - 1 \cdot z \ - \ {\displaystyle 1 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\displaystyle 1 - 0 \cdot z \ - \ {0 \cdot z^2 \over {\ddots}}}}}}}}$$$$

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.

• Please bear in mind that every edit comes at the expense of every other question on MathOverflow, bumping the post to the top and pushing others down the stack. That can't be helped, but please be considerate of others and combine edits as economically as you can. Mar 14 at 11:34
• I did not down vote, but it's worth bearing in mind that many take the view that MO is not and should not be a "blog" style website for sharing mathematical thoughts, and constantly updating your answers to your own question can be seen as too much in that blog style. Jul 5 at 0:09
• @SamHopkins This is not a blog posting. It is, rather, a question (stated in two or three different ways) explained in the scope of a larger problem . And indeed many users take the view that questions which are submitted to MO should provided a context. I'm also aware that there is a fair amount of "spite" down-voting which goes on, as well as an ample dose of condescension. I'll probably be down-voted just for making this remark. Jul 5 at 1:07