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.

for every$k$ and $l$, or rather thatthere exist$k$ and $l$ such that those are not equal? $\endgroup$for every$k<\ell$. I'm pretty certain I already have personal notes written up showing the conjecture true at $\ell=k+1$. $\endgroup$3more comments