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John McVey
  • Member for 6 years, 3 months
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Conjectured combinatorial non-equality
Well, that was exactly what I was looking to find. Thank you so much! The only remaining question (as I haven't yet figured out the etiquette for attribution from this site), should I actually get the publication (a) may I use this argument, and (b) what is the appropriate way to say "my thanks to Fedor Petrov for providing the following proof" (he says hoping that sentence to be the answer)?
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Conjectured combinatorial non-equality
@GerhardPaseman : I'm afraid I don't follow. Take $n=20$ and $d=1$, for example. The binomial ratio $\binom{20}{k+4}/\binom{20}{k+3}$ is larger than the polynomial ratio $[(20-k-5)(k+1)(k+2)]/[(20-k-6)(k+2)(k+3)]$ for $k\leq 6$ but not for $k\geq 7$. In fact, for these parameters, the binomial ratio decreases from ~$4.25$ to ~$.24$, while the polynomial ratio increases from ~$.36$ to ~$1.75$.
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Conjectured combinatorial non-equality
@GerhardPaseman: In fact, that monotonicity motivated my second bullet (regarding prime sets). Beyond (the possibility of) a prime dividing one and not the other, I didn't come up with anything. (And, yes, this appears very monotonic, strictly increasing on some interval $[0,k_0]$ and strictly decreasing on $[k_0,n−6]$, being my guess).
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Conjectured combinatorial non-equality
@AlexM.: I do mean the first of those: for every $k<\ell$. I'm pretty certain I already have personal notes written up showing the conjecture true at $\ell=k+1$.
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