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 

https://mathoverflow.net/questions/384591/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
$\mu^{(n)}_\mathrm{P}: \Bbb{YF}_n \longrightarrow \Bbb{R}_{>0}$ 
defined by

\begin{equation}
\begin{array}{ll}
\mu^{(n)}_\mathrm{P}(u) 
&\displaystyle := \ {1 \over {n!}} \, \dim^2(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}
\end{equation}

is a strictly positive probability distribution
on $\Bbb{YF}_n$ for each $n \geq 0$. Furthermore, these distributions are 
*coherent* in the sense that the ratios

\begin{equation}
\tilde{\mu}_\mathrm{P}(u \lhd v) \ := \
{\mu^{(n+1)}_\mathrm{P}(v) \over {\mu^{(n)}_\mathrm{P}(u)}}
\end{equation}

restrict to a probability distribution 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.

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



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

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{equation}
\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}
\end{equation}

These graphical conventions allows us to reformulate 
the value of the Plancherel measure as a product of
hook-lenghts, i.e.

\begin{equation}
\mu^{(n)}_\mathrm{P}(u) \ = \ \prod_{\Box \, \in \, u} \, {n! \over 
{\mathrm{h}^2(\Box)} }
\end{equation}

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{equation}
\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}
\end{equation} 

It will be convenient, when dealing with expansions into elementary 
symmetric polynomials, to make the 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 polynomial $E_k$ we
get

\begin{equation}
H^\vee_t(u) \ = \ \sum_{k=1}^n \, (-t)^{n-k} \, E_k \big( \mathrm{h}^2(\Box) \big)_{\Box \, \in \, u}
\end{equation}

and

\begin{equation}
F^\vee(z;t) \ = \ 
\sum_{k \geq 0} \, (-t)^{n-k} \, 
\overbrace{\sum_{n \geq 0} 
\, {z^n \over {n!}} \,
\langle E_k \rangle_n}^{F^\vee_k(z)}
\end{equation}

which effectively reduces the problem of calculating $F^\vee(z;t)$
to the problem of evaluating the expectation values 
$\langle E_k \rangle_n$.

**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{equation}
\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}
\end{equation} 

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

\begin{equation}
\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} \ + \
n(n-1) \langle E_{k-2} \rangle_{n-2}
\end{array}
\right.
\end{equation}

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

\begin{equation}
n^2\sigma_k(n) = 
\underbrace{\sigma_k(n-1) + \sigma_k(n-2)}_{\text{homogeneous part}}
+ 
\underbrace{n^2\sigma_{k-1}(n-1) 
+(n^2 +1)\sigma_{k-1}(n-2) + n^2\sigma_{k-2}(n-2)}_{\text{inductive heap of inhomogeneous junk}}
\end{equation}

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

\begin{equation}
z^2 \, {d^2 \over {dz^2}} \, F^\vee_k \ + \ 
z \, {d \over {dz}} \, F^\vee_k  \ + \
\big(z^2 + z \big) \, F^\vee_k 
\ = \
G_{\leq k}(z) \ + \ \big( \sigma_k(1) - \sigma_k(0) \big)z
\end{equation}

where $G_{\leq k}(z)$ is the generating function associated
to the *heap of inhomogeneous junk* which, by induction, 
will have been previously evaluated. The homogeneous ODE
has two nice independent solutions $Y_1(z) = e^z$ and 
$Y_2(z)= e^z \int z^{-1} e^{-2z} dz$ whose Wronskian is
just $W={z^{-1}}$. One starts the inductive engine beginning
with $F^\vee_0(z) = e^z$. For $k=1$ its not hard to see that
$\sigma_1(1)=1$ and $\sigma_1(0)=0$ while 

\begin{equation}
\begin{array}{ll}
\displaystyle G_{\leq 1}(z) 
&\displaystyle = \ {z^2 \over {1-z}} \ + \
2 \sum_{n \geq 2} \, n^2 z^n \\ 
&\displaystyle =  {z^2 \over {1-z}} \ + 
\ 2z \, \Bigg( {1+z \over {(1-z)^3}} \, - \, 1 \Bigg)
\end{array}
\end{equation}

so the ODE for $F^\vee_1(z)$ becomes

\begin{equation}
z^2 \, {d^2 \over {dz^2}} \, F^\vee_1 \ + \ 
z \, {d \over {dz}} \, F^\vee_1 \ + \
\big(z^2 + z \big) \, F^\vee_1 
\ = \
z + G_{\leq 1}(z) 
\end{equation}

By variation of parameters, a particular inhomogeneous solution is

\begin{equation}
\begin{array}{rl}
\displaystyle Y_\mathrm{particular}(z) 
&\displaystyle = \ V_1(z) \cdot e^z \ + \ V_2(z) \cdot 
e^z \overbrace{\int {dz \over z} \, e^{-2z} }^{\gamma(z)} \\ 
\displaystyle V_1(z)
&\displaystyle = \ -\int z \, e^z \, \gamma(z) \, 
\Big(z + G_{\leq 1}(z) \Big)  \, dz  \\ 
\displaystyle V_2(z)
&\displaystyle = \ \ \ \ \ \int z \, e^z \, 
\Big(z + G_{\leq 1}(z) \Big) \, dz
\end{array}
\end{equation}

> **Question.** Has the linear recurrence satisfied by $\sigma_k(n)$ or else the hierarchy of 2nd order inhomogeneous ODEs been studied?

thanks, ines.