Young-Fibonacci tableaux, content, and the Okada algebra Using the French convention, the content of the $i \times j$ box in the  Young diagram of a partition $\lambda$ is $i-j$. Now if 
$\lambda$ is partition of $n$ and $\sigma_\lambda: S_n \longrightarrow V_\lambda$ is the corresponding irreducible representation of the symmetric group $S_n$ then the sum of contents of all boxes in the young diagram of $\lambda$ equals
\begin{equation}  {\text{tr} \, \sigma_\lambda \big( t \big) \over 
{\text{dim} V_\lambda}} \cdot \big| T  \big|
\end{equation}
where $T$ is the conjugacy class consisting of all transpositions and $t$
is any choice of transposition. Moreover each Young tableau
$Y_\lambda$ encodes an eigenvector in $V_\lambda$ for the operator $\sigma_\lambda \big( J_k \big)$ with eigenvalue $c_k$ where
\begin{equation} J_k \, := \, \sum_{i=1}^{k-1} \, \big(i,k \big)
\, \in \Bbb{C} \big[ S_n \big]
\end{equation} 
and $c_k$ is the content of the box in $Y_\lambda$ labeled by $k$. 
Consider now the Young-Fibonacci lattice whose elements consist of words $w = a_1 \cdots a_d$ 
taken from the alphabet $\{1,2 \}$ which can be visualised by stacking boxes into adjacient vertical columns going from left to right such that the number of boxes in the $i$-th column is $a_i$. The rank $|w|$ of a word is simply the number of boxes in such a picture; equivalently $|w|$
equals $a_1 \, + \, \cdots \, + \, a_d$. 
I won't describe the covering relations that give the lattice structure --- but suffice it to say each word $w$ or rank $n$ encodes an irreducible representation $V_w$ of the Okada algebra $\mathcal{A}_n$ and each complete chain $Y_w$ ending at $w$ indexes a basis vector in $V_w$. 
Question: (1) Are there pairwise commuting operators $\tilde{J_1}, \dots, \tilde{J_n}$ within the Okada algebra $\mathcal{A}_n$ for which each complete chain $Y_w$ (viewed as a basis vector in $V_w$) is a simultaneous  eigenvector and (2) is there
a notion of content (a value for each covering relation in the Young-Fibonacci lattice) so that the $k$-th content $c_k$ of $Y_w$ (viewed as a complete chain) is the eigenvalue of $\tilde{J_k}$ corresponding to $Y_w$ and (3) will the sum of such contents along any complete chain $Y_w$ be constant ? 
regards, A. Leverkühn
 A: Liebe A. 
This is only an attempt to answer how one might compute the Fibonacci contents whose existence you ask about; presently I don't know how to determine the $\tilde{J_k}$ operators. Let's begin by noticing that the quantity 
\begin{equation}
\text{tr} \ \sigma_\lambda (t) \cdot \big| T \big|
\end{equation}
for $t \in T$ appears as the $\mu = 21^{n-2}$ contribution when expanding the Schur function $s_\lambda$ in terms of the power-symmetric functions $p_\mu$, specifically
\begin{equation}
\begin{array}{ll} 
\displaystyle s_\lambda 
&\displaystyle = \, {1 \over {n!}} \, \sum_{|\mu| = n} \, 
\text{tr} \, \sigma_\lambda \big(c_\mu \big)
\cdot \big| C_\mu \big| \, p_\mu \\ \\
&\displaystyle = \, \sum_{|\mu| = n} \, 
\text{tr} \, \sigma_\lambda \big(c_\mu \big)
\cdot {p_\mu \over {z(\mu)}} \end{array}
\end{equation}
where $c_\mu$ is any element in the conjugacy class $C_\mu$
indexed by the partition $\mu$ and 
\begin{equation}
z(\mu) \, := \, {n! \over {\big| C_\mu \big|}}
\end{equation}
You're probably aware that 
there are analogues of the Schur and power-symmetric functions
in Okada theory. Recall that for each word $w$ in the Young-Fibonacci
lattice a corresponding Okada-Schur function is constructed recursively using two infinite lists of variables $x_1, x_2, \dots$ and $y_1, y_2, \dots$
in the following way: For integers $n \geq 1$ define the following
$n \times n$ tridiagonal determinants
\begin{equation}
P_n \, := \, \det \begin{pmatrix} x_1 & y_1 & 0 & \cdots
\\ 1 & x_2 & y_2 & & \\ 
0 & 1 & x_3 & & \\ 
\vdots & & & \ddots \end{pmatrix}
\qquad \qquad
Q_{n-1} \, := \, \det \begin{pmatrix} y_1 & x_1y_2 & 0 & \cdots
\\ 1 & x_3 & y_3 & & \\ 
0 & 1 & x_4 & & \\ 
\vdots & & & \ddots \end{pmatrix}
\end{equation}
and by convention set $P_0 := 1$. 
The Schur-Okada function $s_w$ corresponding to $w$ is
defined by the recursion 
\begin{equation} s_w \, := \ \left\{ \begin{array}{ll} P_k &\text{if 
$w= \, 1^k \, $ and $k \geq 0$} \\  \\
Q_k \, \big[+ |v| \big] \cdot s_v &\text{if $w= \, 1^k \, 2 \, v \, $
and $k \geq 0$}
\end{array} \right\} \end{equation}
The notation $Q_k \big[ + |v| \big]$ means perform the substitutions 
$x_i \mapsto x_{i + |v|}$ and $y_i \mapsto y_{i + |v|}$ in the
expression for $Q_k$. Look in the paper by Goodman and Kerov (https://arxiv.org/pdf/math/9712266.pdf) for
a definition of the power-symmetric function analogues $p_w$. The crucial point is that
\begin{equation}
s_w \ = \ \sum_{|u| = n} \, X^w_u {p_u \over {z(u)}}
\end{equation} 
where the $X^w_u$ are Okada character values for the Okada algebra $\mathcal{A}_n$ and
\begin{equation}
z(u) \, := \, k_0! \cdots k_t! \, \big(2 + k_{1} \big) \, \cdots \, \big(2 + k_t \big) 
\end{equation}
where we have parsed $u$ as 
\begin{equation}
u \, = \ 1^{k_0} \, 2 \, 1^{k_1} \, 2 \cdots \, 2 \, 1^{k_t} 
\end{equation}
with $t$ equal to the number of occurrences of $2$ in $u$ and $k_j \geq 0$ for each $t \geq j \geq 0$. I would also point out that for $w$ with
$|w| = n$ the Okada character value $X^w_{1^n}$ equals the dimension of the irreducible $\mathcal{A}_n$-module $V_w$. So the correct analogue of  
\begin{equation}
{ \text{tr} \ \sigma_\lambda (t) \over {\text{dim} V_\lambda}} \cdot \big| T \big|
\end{equation}
(up to some uniform factor in $n$) ought to be
\begin{equation}
{X^w_{u(n)} \over {X^w_{1_n}}} \cdot {1 \over {z \big( u(n) \big)}}
\quad (\dagger) \end{equation}
for some family of words $u(n)$ with lengths $|u(n)| = n$. Having made a guess for this family, label the nodes of the Young-Fibonacci lattice by 
these values ($\dagger$) and then try to recursively solve for the contents associated to each covering relation in the lattice by taking
differences. 
your, Ines.
p.s. Here's a guess: $u(n) := \, 21^{n-2}$ for $n \geq 2$ with $u(1) := 1$.
p.p.s. According to Okada's paper $\displaystyle {|u|! \over {z(u)}} \in \Bbb{Z}$ and $\displaystyle \sum_{|u|=n} \, {n! \over {z(u)}} \, = \, n!$ so my guess is that the 
\begin{equation} \text{$\dagger \dagger$-values} \quad
{X^w_{u(n)} \over {X^w_{1_n}}} \cdot {n! \over {z \big( u(n) \big)}}
\end{equation}
be used instead of the previously defined $\dagger$-values
p.p.p.s Just to be clear let me elaborate with an example. Suppose we select $u(n) = 21^{n-2}$ for all $n\geq 2$ (this will be our choice to play the role of the conjugacy class consisting of all reflections in $S_n$ for $n\geq 2$). Suppose we want the fibonacci content associated to the covering relation (i.e. edge in the Hasse diagram) between $121$ and $1121$. So we need to compute the respective $\dagger \dagger$-values,
namely:
\begin{equation}
{X^{121}_{211} \over {X^{121}_{111}}} \cdot {4! \over {z(211)}} \, = \, 3
\quad \quad {X^{1121}_{2111} \over {X^{1121}_{1111}}} \cdot {5! \over {z(2111)}} \, = \, 4
\end{equation}
So the content ought to be $4-3 = 1$.
Continuing with the same example but for higher rank, please note that
for elements $u$ and $v$ of respective ranks $|u|=n-1$ and $|v|=n-2 one can use Okada's recursive formula to obtain
\begin{equation}
{X^{1u}_{21^{n-2}} \over {X^{1u}_{1^n}}} \, = \, 1 \quad \quad
{X^{2v}_{21^{n-2}} \over {X^{2v}_{1^n}}} \, = \, {1 \over {1-n}}
\end{equation}
In addition
\begin{equation}
{n! \over {z\big(21^{n-2} \big)}} \, = \, {n! \over {{(n-2)}! \, n }} \, = \, n-1
\end{equation}
There are four types of covering relations $u \sqsubset v$ between the elements within levels of rank $n$ and rank $n+1$ depending on whether the prefixes ($\text{pf}$ for short) of $u$ and $v$ are $1$ or $2$. I list them together with the corresponding contents in following table: 
\begin{equation} 
\begin{array}{cl} 
\text{contents:} 
&\text{types of covering relations:} \\ 
1 
&\text{if $\text{pf}(u) = 1 \, $ and $\, \text{pf}(v) = 1$} \\
-n 
&\text{if $\text{pf}(u) = 1 \, $ and $\, \text{pf}(v) = 2$} \\
n+1 
&\text{if $\text{pf}(u) = 2 \, $ and $\, \text{pf}(v) = 1$} \\
0
&\text{if $\text{pf}(u) = 2 \, $ and $\, \text{pf}(v) = 2$}
\end{array}
\end{equation}
A: Errors have been fixed
In this second response let me try to address what might 
play the role of the operator sum $\tilde{J_1}
+ \cdots + \tilde{J_n}$ which is the analogue of 
$\sum_{t \in T} \, t \ \in \Bbb{C}[S_n]$ 
where $T$ is the conjugacy class consisting of all 
transpositions. 
Given two sets of generic parameters $x_1, \dots, x_{n-1}$ and
and $y_1, \dots, y_{n-2}$ the associated 
Okada algebra $\mathcal{F}_n$ has a presentation 
given by generators $E_1, \dots, E_{n-1}$ subject to the
defining relations
\begin{equation}
\begin{array}{rll}
E_i^2 &= \ x_i \, E_i & \\
E_i \, E_j &= \ E_j \, E_i &\text{whenever $|i-j| >1$} \\
E_j \, E_i \, E_j &= \ y_i \, E_j &\text{whenever $j = i+1$}
\end{array}
\end{equation}
 Given a word $w$ the Okada power-symmetric functions $p_w$ is
defined by the recursion
\begin{equation}
p_w \ = \
\left\{
\begin{array}{ll}
\displaystyle x_1 \cdots \, x_k 
&\text{if $\, w = 1^k \, $ with $\, k \geq 0$} \\ 
\displaystyle q_k \cdot p_u \big[ + (k + 2) \big]
&\text{if $\, w= u  2 1^k \,$ with $\, k \geq 0$}
\end{array}
\right\}
\end{equation}
where 
\begin{equation}
q_k \ := \ (x_1 \cdots \, x_k) \, \Big( x_{k+1} x_{k+2} \, - \, (k+2) \,y_{k+1}  \Big)
\end{equation}
and $p_u \big[ +l \big]$ means perform the substitutions 
$x_i \mapsto x_{i+l}$ and $y_i \mapsto y_{i+l}$ for all $i \geq 1$.
For the purposes of this discussion, let's provisionally define the Okada power-symmetric function in terms of the Okada-Schur functions and the Okada characters values:
\begin{equation}
p_u \, = \ \sum_{|v| = |u|} \, X^v_u \, s_v
\end{equation}
Now let's define $\mathcal{F_n}$-valued versions $\Bbb{s}_w$ and $\Bbb{p}_w$
of the Okada-Schur and power-symmetric functions. For $n \geq k \geq 1$
define the following $k \times k$ tri-diagonal determinants whose values
are in the Okada algebra $\mathcal{F_n}$
\begin{equation}
\begin{array}{ll}
\mathcal{P}_k \, :=
&\det \, \begin{pmatrix} x_1 & x_2 E_1 & 0 & \cdots
\\ 1 & x_2 & x_3 E_2 & & \\ 
0 & 1 & x_3 & & \\ 
\vdots & & & \ddots \end{pmatrix}
\\ \\ \\
\mathcal{Q}_{k-1} \, :=
&\det \, \begin{pmatrix} x_2 E_1 & x_1 x_3 E_2 & 0 & \cdots
\\ 1 & x_3 & x_4 E_3 & & \\ 
0 & 1 & x_4 & & \\ 
\vdots & & & \ddots \end{pmatrix}
\end{array}
\end{equation}
where $\mathcal{P}_0 := 1$. 
Clearly an order must be observed
when tabulating the determinant --- following the conventions of
Kerov and Goodman, the $l$-th factor in the expansion will always
be selected from the $l$-th column. 
The values of the determinants $\mathcal{P}_k$ and 
$\mathcal{Q}_{k-1}$ are in fact independent of the order
in which products are taken in the Laplace expansion: This 
is because indices of the generators $E_1, \dots, E_{n-1}$
which participate in any given monomial in the expansion
of such a tri-diagonal determinant will always differ by 
at least two.
Employ the same recursion
above being mindful to place the accumulating $\mathcal{Q}$-factors
to the left and in order:
\begin{equation} \Bbb{s}_w \, := \ \left\{ \begin{array}{ll} \mathcal{P}_k &\text{if 
$w= \, 1^k \, $ and $k \geq 0$} \\  \\
\mathcal{Q}_k \, \big[+ |v| \big] \cdot \Bbb{s}_v &\text{if $w= \, 1^k \, 2 \, v \, $
and $k \geq 0$}
\end{array} \right\} \end{equation}
Once again $\mathcal{Q}_k \big[ + |v| \big]$ means
shift all indices by $|v|$. Play the same game and
define the 
$\mathcal{F}_n$-valued power-symmetric functions
by the expansion:
\begin{equation}
\Bbb{p}_u \, := \ \sum_{|v| = |u|} \, X^v_u \, \Bbb{s}_v
\end{equation}
I want to make use of Okada's trace
functional $\text{Tr}: \mathcal{F}_n \longrightarrow \Bbb{C}$
which is defined for an element $a \in \mathcal{F}_n$ 
using the Okada-Schur values by
\begin{equation}
\text{Tr}(a) \ = \
{1 \over {(x_1 \cdots \, x_n)}} \, \sum_{|v|=n} \, s_v \cdot \text{tr} \Big[ \sigma_v(a) \Big]
\end{equation} 
where $\sigma_v : \mathcal{F}_n \longrightarrow \text{End}(V_v)$
is the irreducible representation of $\mathcal{F}_n$ associated to 
a word $v$ with $|v|=n$ and where $\text{tr} \big[ \cdot \big]$ denotes the usual trace. Okada proves in his paper that
\begin{equation}
\begin{array}{ll}
\displaystyle \text{Tr} \, (1) &\displaystyle = \, 1 \\
\displaystyle \text{Tr} \, \big( ab \big) &\displaystyle = \, \text{Tr} \, \big(ba \big) \\
\displaystyle \text{Tr} \, \big( aE_i \big) &\displaystyle = \, {y_i \over {x_{i+1}}} 
\, \text{Tr} \, (a) \quad \text{when $a \in \mathcal{F}_i$} \quad
\left( { \scriptstyle 
\begin{array}{l} 
\text{As far as I can tell there seems to be} \\
\text{a missing $x_{i+1}$ in the denominator of} \\
\text{part (4) of Proposition 2.7 in Okada's} \\
\text{paper which I have tried to correct here.}
\end{array} }\right)
\end{array}
\end{equation}
Using these multiplicative properties, a simple induction on the length $|v|$ reveals that 
$\text{Tr} \, \big( \Bbb{s}_v \big) \, = \, s_v$ and consequently
$\text{Tr} \, \big( \Bbb{p}_u \big) \, = \, p_u$. Moreover 
\begin{equation}
\begin{array}{c}
\displaystyle
\text{tr} \, 
\Big[ \pi_v \big( \Bbb{p}_u \big)  \Big] \, = \, \big( x_1 \cdots x_n \big) \, X^v_u \\
\displaystyle \text{--- and ---} \\
\displaystyle
\text{tr} \, \Big[ \pi_v \big( \Bbb{s}_u \big) \Big] \, = \,
\big(x_1 \cdots x_n \big) \, \delta_{u,v}
\end{array}
\end{equation}
Isn't this observation an indication that $\Bbb{p}_u$ is an analogue of
a characteristic function of a conjugacy class in the group setting ? 
However, pursuing this analogy, it's not immediately clear 
whether the elements $\mathcal{p}_u$ are central in $\mathcal{F}_n$.
yours, Ines. 
p.s. In fact the order of products taken in the expansion of the $\mathcal{F}_n$-valued determinants 
$\mathcal{P}_k$ and $\mathcal{Q}_{k-1}$ is irrelevant --- this is because all monomials which occur involve $E$-generators whose subscripts differ by at least two.
