Quantum Schubert polynomials $\mathfrak{S}_u^q(x)$ indexed by $S_\infty$ are polynomials in the polynomial ring $\mathbb{Z}[x,q]$ in infinitely many variables that form a basis of this ring over $\mathbb{Z}[q]$. The structure constants, which are polynomials in the $q$ variables, are usually written as $$\mathfrak{S}_u^q(x)\mathfrak{S}_v^q(x)=\sum_{w,d}q^dc_{u,v}^{w,d}\mathfrak{S}_w^q(x)$$ where $c_{u,v}^{w,d}$ is an integer indexed by the three permutations $u,v,w$ and a weak composition $d$ that indicates the powers in the $q$-monomial. These are the Gromov-Witten invariants of the complete flag variety.
I have deduced a new formula that I can't find anywhere and I'm trying to determine whether it is new. Suppose $d$ is a weak composition with exactly one nonzero value, which is equal to $1$ and is at index $i$. Then $c_{u,v}^{w,d}=0$ unless $\ell(us_i)<\ell(u)$ and $\ell(vs_i)<\ell(v)$, in which case $$c_{u,v}^{w,d}=c_{us_i,vs_i}^w$$ where $c_{us_i,vs_i}^w$ is the ordinary Littlewood-Richardson coefficient of the corresponding ordinary Schubert polynomials and $s_i$ is the adjacent transposition $(i,i+1)$. That is to say, the coefficient of the monomial $q_i$ is $c_{us_i,vs_i}^w$. The formula also works for double quantum Schubert polynomials, and the equation is identical.
My proof of the formula entirely uses things that apparently only I know, but it's such a direct relation between the coefficients that it seems highly probable that someone has found it already, perhaps proved in a different way. Is this formula known?
