Nonvanishing of an entire quantum-binomial row and the characteristic digits

Reversing a binary word changes its inversion count from $v$ to $k(n-k)-v$, so $q^{-k(n-k)}{G}_{n,k}(q^2)$ is unchanged by $q\mapsto q^{-1}$. A symmetric integer Laurent polynomial is an integer combination of $1$ and $q^j+q^{-j}$: pair its opposite coefficients. Each $q^j+q^{-j}$ is a polynomial in $T=q+q^{-1}$ by the recurrence $s_{j+1}=T{s}_j-s_{j-1}$ with $s_0=2,s_1=T$. This proves existence of $B_{n,k}$. The highest Laurent power of a nonzero polynomial in $q+q^{-1}$ is its degree, with its leading coefficient unchanged, so substitution is injective and proves uniqueness.

Splitting words by their final letter gives $G_{n,k}=G_{n-1,k-1}+Q^k{G}_{n-1,k}$, with boundary values $G_{n,0}=G_{n,n}=1$. The rational product ${\prod }_{j=1}^k(1-Q^{n-k+j})\mathrm{/}(1-Q^j)$ has these same values and recurrence, as putting the two terms over their common denominator verifies. Induction therefore identifies it with $G_{n,k}$. The recurrence [F1] gives $[j{]}_q=q^{1-j}(1-q^{2j})\mathrm{/}(1-q^2)$ as an identity of rational functions, so canceling those factors using [F5] gives $B_{n,k}=[n]{!}_q\mathrm{/}([k]{!}_q[n-k]{!}_q)$ before specialization.

By [F4], positive characteristic $p$ is prime. Extend $K$, if necessary, to a field containing a root $q$ of $X^2-\delta X+1$, using [F3] if this polynomial is irreducible. Its constant term makes $q\mathrm{\ne }0$ and $\delta =q+q^{-1}$. Put $Q=q^2$. If $Q\mathrm{\ne }1$, the formula in step 2.1 shows that $[r{]}_{\delta }=0$ exactly when $Q^r=1$. Thus finite $\mathrm{\ell }$ is the order of $Q$. In positive characteristic this order is prime to $p$: if $\mathrm{\ell }=ph$, then $(Q^h-1{)}^p=Q^{\mathrm{\ell }}-1=0$ by [F2], contradicting minimality in a field. If $Q=1$, recurrence gives $[r{]}_{\delta }=r{q}^{r-1}$, so $\mathrm{\ell }=p$ in positive characteristic and $\mathrm{\ell }=\mathrm{\infty }$ in characteristic zero. Nonvanishing of the original $K$-values is unchanged by this field extension.

If $\mathrm{\ell }=\mathrm{\infty }$, every quantum factorial is nonzero in the field, so step 2.1 makes every $B_{n,k}(\delta )$ nonzero. Suppose henceforth that $\mathrm{\ell }<\mathrm{\infty }$. Expanding a product over the subsets of $\{0,\dots ,n-1\}$ gives ${\prod }_{j=0}^{n-1}(1+Q^{j}X)={\sum }_{k=0}^n{Q}^{k(k-1)\mathrm{/}2}{G}_{n,k}(Q)X^k\mathrm{.}$ Indeed, if the one positions of a word are $r_1<\cdots <r_k$, the selected subset $\{n-r_1,\dots ,n-r_k\}$ has sum $\mathrm{inv}(w)+k(k-1)\mathrm{/}2$. This proves the identity including both endpoint coefficients.

The length-$\mathrm{\ell }$ factor is ${\prod }_{j=0}^{\mathrm{\ell }-1}(1+Q^{j}X)=1-(-X{)}^{\mathrm{\ell }}$. When $Q\mathrm{\ne }1$, its powers are the distinct $\mathrm{\ell }$th roots of unity, so this follows by factoring $Y^{\mathrm{\ell }}-1$ and comparing constant terms. When $Q=1$, step 3.1 gives $\mathrm{\ell }=p$, and [F2] gives $(1+X{)}^p=1+X^p=1-(-X{)}^p$. Writing $n=A\mathrm{\ell }+b$ with $0\le b<\mathrm{\ell }$ and using $Q^{\mathrm{\ell }}=1$, the product in step 3.2 is therefore $(1-(-X{)}^{\mathrm{\ell }}{)}^A{\prod }_{j=0}^{b-1}(1+Q^{j}X)\mathrm{.}$ This is an equality in the polynomial ring over the extended field.

Write $k=C\mathrm{\ell }+d$ with $0\le d<\mathrm{\ell }$ and $0\le k\le n$. The coefficient of $X^k$ in step 4.1 is zero when $d>b$; otherwise it is $\left(\genfrac{}{}{0px}{}{A}{C}\right)$ times $G_{b,d}(Q)$ and powers of the units $Q$ and $-1$. There is only one such term, since the last product has degree $b<\mathrm{\ell }$. Every $G_{b,d}(Q)$ with $d\le b$ is nonzero by step 2.1, since all factorial indices are below $\mathrm{\ell }$. Consequently $B_{n,k}(\delta )\mathrm{\ne }0$ exactly when $d\le b$ and $\left(\genfrac{}{}{0px}{}{A}{C}\right)$ is nonzero in $K$. This conclusion keeps all scalar unit factors and asserts only their exact zero pattern.

If $n<\mathrm{\ell }$, step 5.1 already proves full-row nonvanishing. If $n\ge \mathrm{\ell }$ and $b<\mathrm{\ell }-1$, choosing $k=b+1\le n$ gives a zero. If $b=\mathrm{\ell }-1$, step 5.1 says the entire row is nonzero exactly when every ordinary $\left(\genfrac{}{}{0px}{}{A}{C}\right)$, $0\le C\le A$, is nonzero. In characteristic zero this always holds, giving precisely $n=(A+1)\mathrm{\ell }-1$.

In characteristic $p>0$, write $A={\sum }_j{a}_j{p}^j$. Repeated use of [F2] gives $(1+X{)}^A={\prod }_j(1+X^{p^j}{)}^{a_j}$. Uniqueness of base-$p$ digits makes the coefficient $\left(\genfrac{}{}{0px}{}{A}{C}\right)$ nonzero exactly when every digit $c_j$ of $C$ satisfies $c_j\le a_j$: the coefficient is ${\prod }_j\left(\genfrac{}{}{0px}{}{a_j}{c_j}\right)$, whose factors are nonzero because their factorial indices are below $p$. Thus the full ordinary row is nonzero exactly when every digit below the highest nonzero digit of $A$ is $p-1$. If a lower digit $a_j<p-1$, choose $C=(a_j+1)p^j\le A$ to get a zero; conversely the asserted full lower digits make every $C\le A$ digitwise admissible. This also includes $A=0$. Equivalently, $A+1=a{p}^s$ for some $s\ge 0$ and $1\le a<p$, absorbing a leading carry into $s$ if necessary. Substitution in step 6.1 gives $n=a\mathrm{\ell }{p}^s-1$, or $n=ap(i)-1$ with $i=s+1$. Together with steps 3.2 and 6.1 this proves every asserted branch.