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Recall the constructions $[n]_q=\frac{1-q^n}{1-q}, [n]!_q=[1]_q[2]_q\cdots[n]_q$ with $[0]!_q:=1$ and the $q$-binomials (Gaussian polynomials) $$\binom{n}k_q=\frac{[n]!_q}{[k]!_q[n-k]!_q}.$$ Given two polynomials $f(q)$ and $g(q)$, we write $f(q)\geq g(q)$ provided that $f(q)-g(q)$ is a polynomial having non-negative coefficients.

I would like to ask:

QUESTION. Suppose $n<m, j<k, 2j<m$ and $2k<n$. Is it true that whenever $\binom{n}k_q\geq\binom{m}j_q$ then $\binom{n}{k-1}_q\geq\binom{m}{j-1}_q$?

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  • $\begingroup$ I think you mean $n>m$ instead of $n<m$. Then for $k\ge j$, condition and consequence always hold by induction/recursion. But it seems that as soon as $k<j$, we have e.g. that $\binom{n}k_q-\binom{m}{k+1}_q$ has at least one negative term $-q^{k+1}$, so the condition can never hold. Unless I am missing something... $\endgroup$
    – Wolfgang
    Commented Oct 3, 2022 at 15:45
  • $\begingroup$ @Wolfgang: I really mean $n<m$. $\endgroup$
    – T. Amdeberhan
    Commented Oct 3, 2022 at 19:37

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