# Conjectured combinatorial non-equality

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.

• 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 14:10
• @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 14:11
• @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 14:22
• @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 14:35
• 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 8:18

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.