Timeline for Conjectured combinatorial non-equality
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 30, 2018 at 19:23 | vote | accept | John McVey | ||
| Aug 30, 2018 at 8:18 | comment | added | Martin Rubey | This may be easier to see in terms of cardinalities of standard Young tableaux. It seems that the following order is strictly increasing, eg. for $n=11$: $[8,3],[3,3,1^5],[7,3,1],[4,3,1^4],[6,3,1^2],[5,3,1^3]$. At least, this should give you some insight why this might be true. | |
| Aug 30, 2018 at 8:16 | answer | added | Fedor Petrov | timeline score: 9 | |
| Aug 30, 2018 at 8:11 | comment | added | Right | It seems that answer to your problem follows if we succeed to prove that equality $$2^l=\displaystyle \sum_{k=0}^{l }\dfrac{(n-(l+5))(k+3)(n-(k+3))!}{(n-(k+5))(l+3)(n-(l+3))!}$$ follows only if $k=l=0$, but I need to check some details to see is this enough. | |
| Aug 30, 2018 at 7:21 | history | edited | Fedor Petrov |
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| Aug 29, 2018 at 20:03 | comment | added | John McVey | @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 ~$1.75$. | |
| Aug 29, 2018 at 14:46 | comment | added | Gerhard Paseman | Try the following: let l =k+d, and start with d=1. Show that the ratio of the binomial terms is larger than the ratio of the polynomial terms except possibly at k=(n/2+2) (or near there, I'm guessing), and then increase d. You have your problem restricted to a set of pairs for each d, and then you can try prime factorization. Gerhard "First Find Where Ballpark Is" Paseman, 2018.08.29. | |
| Aug 29, 2018 at 14:35 | comment | added | John McVey | @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). | |
| Aug 29, 2018 at 14:22 | comment | added | John McVey | @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 $\ell=k+1$. | |
| Aug 29, 2018 at 14:11 | comment | added | Alex M. | @JohnMcVey: Are you trying to show that they are not equal for every $k$ and $l$, or rather that there exist $k$ and $l$ such that those are not equal? | |
| Aug 29, 2018 at 14:10 | comment | added | Gerhard Paseman | Your expression appears monotonic on large intervals of your domain. Have you used this to find restrictions on when equality could occur? Gerhard "Looks Rather Tractable To Me" Paseman, 2018.08.29. | |
| Aug 29, 2018 at 13:40 | review | First posts | |||
| Aug 29, 2018 at 14:12 | |||||
| Aug 29, 2018 at 13:39 | history | asked | John McVey | CC BY-SA 4.0 |