The Okada algebra $\mathfrak{O}_n$ over a field $K$ has generators
$E_1,\dots,E_{n-1}$ and relations $E_i^2=x_iE_i$,
$E_{i+1}E_iE_{i+1}=y_i E_{i+1}$, and $E_iE_j=E_jE_i$ for $|i-j|\geq
2$, where $x_i,y_i\in K$. Okada (Algebras
associated to the Young–Fibonacci lattice,
*Trans. Amer. Math. Soc.* 346 (1994), 549–568) shows that
$\mathfrak{O}_n$ is semisimple of dimension $n!$ when the $x_i$'s and
$y_i$'s are generic. I computed (if I didn't make an error) that when
$n=3$, the discriminant of $\mathfrak{O}_3$ is a constant times $x_1^2
y_1^4(x_1x_2-y_1)^4$. Can this result be extended to larger $n$?

**Addendum.** I managed to compute the discriminant of
$\mathfrak{O}_4$. It is a constant times
$$ x_1^{20} y_1^{10} y_2^{24} (x_1 y_2 - x_3 y_1)^6
(x_1 x_2 - y_1)^{10} (x_1 x_2 x_3 - x_1 y_2 - x_3 y_1)^6. $$

disriminanthere? Is this related to the Gram determinant of the Markov trace on the Okada algebra? If so, it should factor into a product of (specialized)cloneSchur functions. The examples that you give bear this out. I have code that calculates this. $\endgroup$shiftsof clone Schur functions (i.e. where the parameter indices of a clone Schur function are shifted by a common integer). $\endgroup$cloneSchur function (a polynomial expressed in the parameters $x_1, \dots, x_n$ and $y_1, \dots, y_{n-1}$ determined by $w$). Okada's pairing has the property that it is a Markov trace. $\endgroup$tracefunctional $\varphi: \frak{O}_n \rightarrow \Bbb{C}$ together with any basis of $\frak{O}_n$. In general the determinant of this Gram matrix ought to be some product of specialized clone Schur functions because the simultaneous non-vanishing of these polynomials $s_w$ for all $|w| < n$ is equivalent to the semi-simplicity of $\frak{O}_n$. Florent Hivert and I are trying to find a closed formula for this Gram determinant in the case of Okada's Markov trace. $\endgroup$