Product formula for the $(p,q,h)$ -deformed binomial coefficients

Let $m$ and $k$ be nonnegative integers, let $p$ and $q$ be parameters, and let $Z_p$ be the generator appearing in the $(p,q)$ -deformed generalized Weyl algebra. Write $[i]_{p,q}$ for the $(p,q)$ -integer and $\binom{m}{k}_{p,q}$ for the $(p,q)$ -binomial coefficient. The $(p,q,h)$ -deformed binomial coefficient is denoted by $\binom{m}{k}_{h|p;q}(Z_p)$ . Product formula. The $(p,q,h)$ -deformed binomial coefficients can be written as

\binom{m}{k}_{h|p;q}(Z_p)=\binom{m}{k}_{p,q}\prod_{i=0}^{k-1}(1+hp^{-k}[i]_{p,q}Z_p).

This is a product expansion obtained by combining the deformed binomial coefficients with the associated $(p,q)$ -deformed cycle numbers; the source does not indicate an unresolved status or provide further qualification of the result.

References

Primary source

Toufik Mansour, Lahcen Oussi and Matthias Schork, “Normal ordering in the (p,q)-deformed generalized Weyl algebra. III: The binomial formula”, arXiv:2607.11693 (2026).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A July 2026 paper presents the formula as a conjecture, while a posted degree-two calculation claims to disprove it; that calculation has not been independently checked.

Mansour, Oussi, and Schork introduced the product formula in their July 2026 paper on binomial formulas for the $(p,q)$ -deformed generalized Weyl algebra. The paper labels it Conjecture $3.37$ , not a theorem, and gives equivalent identities involving deformed cycle numbers.

Posted attempt

A posted calculation claims a counterexample at $m=2$ , $k=1$ : the defining relations give coefficient $1+q$ , whereas the conjectured product gives $[2]_{p,q}=p+q$ ; for $(p,q,h)=(2,3,1)$ these are $4$ and $5$ . It therefore claims the formula is false, but the calculation has not been independently verified.

Current status (as of August 2026): The paper's formula remains formally a conjecture, and a purported counterexample claims to settle it negatively, but no verified proof or disproof is recorded.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

Degree-two counterexample

The proposed product formula fails already for $m=2$ and $k=1$ .

The defining relations for the $(p,q)$ -deformed Jordan plane include

XY=qYX+hY^2Z_p, \qquad Z_pY=pYZ_p.

Therefore

\begin{aligned} (X+Y)^2 &=X^2+XY+YX+Y^2\\ &=X^2+(1+q)YX+Y^2+hY^2Z_p\\ &=X^2+(1+q)YX+(1+hp^{-2}Z_p)Y^2, \end{aligned}

where $Y^2Z_p=p^{-2}Z_pY^2$ was used in the last line. Hence the actual coefficient of $YX$ is

\binom21_{h|p;q}(Z_p)=1+q.

On the other hand, Conjecture 3.37 in the source predicts, for $k=1$ ,

\begin{aligned} \binom21_{h|p;q}(Z_p) &=\binom21_{p,q} \prod_{i=0}^{0}\left(1+hp^{-1}[i]_{p,q}Z_p\right)\\ &=\binom21_{p,q} =[2]_{p,q} =p+q, \end{aligned}

because $[0]_{p,q}=0$ .

Taking $p=2$ , $q=3$ , and $h=1$ , which lies in the source's generic regime, gives

1+q=4 \qquad\text{but}\qquad p+q=5.

Thus the proposed product formula is false.

The discrepancy is structural: the constant term of the actual normal-order coefficient satisfies

c_{m+1,k}=q^k c_{m,k}+c_{m,k-1}, \qquad c_{0,0}=1,

and is therefore the Gaussian coefficient $\binom{m}{k}_q$ , whereas the conjectured product has constant term $\binom{m}{k}_{p,q}$ .