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Let $n,k,\ell$ be integers for which $0\leq k<\ell \leq n-6$. For a fixed $n$, think of $k,\ell$ as being allowed to vary. I believe the values

$$(n-k-5)(k+1)(k+2)\binom n{k+3}~~~\text{and}~~~(n-\ell-5)(\ell+1)(\ell+2)\binom n{\ell+3}$$

are not equal. A proof they are not equal is the goal, but insight as to why they might not be equal would still be appreciated.

Motivation: The integer $(n-k-5)(k+1)(k+2)\binom n{k+3}$, when divided by $2(n-2)$, is the degree of the complex irreducible character of the symmetric group $S_n$ corresponding to the partition $(n-k-3,3,1^k)$. I am a group theorist working on a general conjecture on character degrees; these character degrees if they are distinct will prove my conjecture to be true in the case of both the symmetric and the alternating groups.

My work effort:

$\bullet$ I ran computer tests for $6\leq n\leq 7000$. For those $n$, the integers are distinct (modulo my ability to write code, anyway).

$\bullet$ I thought maybe the set of prime divisors could be used to differentiate the values for distinct $k,\ell$, but there are a large number of values for $n$ where this thought failed.

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  • $\begingroup$ 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. $\endgroup$ Aug 29, 2018 at 14:10
  • $\begingroup$ @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? $\endgroup$
    – Alex M.
    Aug 29, 2018 at 14:11
  • $\begingroup$ @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 $\ell=k+1$. $\endgroup$
    – John McVey
    Aug 29, 2018 at 14:22
  • $\begingroup$ @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). $\endgroup$
    – John McVey
    Aug 29, 2018 at 14:35
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    $\begingroup$ 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. $\endgroup$ Aug 30, 2018 at 8:18

1 Answer 1

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It looks like we may simply say which of them is greater.

Denoting $k+3=t$ and $f(t)=(n-k-5)(k+1)(k+2)\binom n{k+3}=(n-t-2)(t-1)(t-2)\binom n{t}$ we get $$\frac{f(t+1)}{f(t)}= \frac{t (n - t - 3) (n - t)}{(t - 2) (t + 1) (n - t - 2)}=\frac{n-t-1-\frac{2}{n-t-2}}{t-1-\frac2{t}}. $$ If $t<n/2$, this is greater than 1, if $t\geqslant n/2$, this is less than 1. Thus the function $f(t)$ increases up to $n/2$ and decreases after $n/2$. Therefore if $f(t)=f(s)$, $t<s$, we must have $t<n/2<s$. Let us compare $s$ and $n-t$, for this goal simplify the ratios $$ \frac{f(n-t)}{f(t)}=\frac{n-t-1}{t-1}>1 $$ $$ \frac{f(n-t+1)}{f(t)}=1- \frac{2 (n - 2) (n - 2 t + 1)}{(t - 2) (t - 1) (n - t - 2) (n - t + 1)}<1. $$ Thus by monotonicity we should have $n-t<s<n-t+1$, a contradiction.

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  • $\begingroup$ 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)? $\endgroup$
    – John McVey
    Aug 30, 2018 at 19:23
  • $\begingroup$ @John, when Fedor responds, of course respect his preferences. Until then, a sentence like yours is appropriate, and follow it up with a bibliographic reference that includes the question title , the year, and a URL like mathoverflow.net/q/309371 for the question. (If you want to refer to just the answer, click on 'share' above Fedor's photo for a URL/id for his answer, and cut off the /id part for the bibref.). Gerhard "Remember The StackExchange License Agreement" Paseman, 2018.08.30. $\endgroup$ Aug 30, 2018 at 23:51
  • $\begingroup$ @JohnMcVey of course you are welcome to use this argument and cite it on the manner suggested by License Agreement as suggested by Gerhard. $\endgroup$ Aug 31, 2018 at 6:53
  • $\begingroup$ To record for posterity, the result can be made stronger with a little less work. We can prove @Martin Rubey's observation directly (alternating (k), (n-k), (k+1), (n-k-1), etc. yields a strictly increasing sequence) using only the second and third computation. $\endgroup$
    – John McVey
    Sep 5, 2018 at 12:54

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