2
$\begingroup$

Let $P(n,m)$ denote the set of all positive integer partitions of $n$ into parts that are pairwise distinct and bounded by $m$. Let $p(n,m) = |P(n,m)|$.

After some numerical experiments it appears

$$ p(n,m) + p(n+m,m) \geq p(n+k,m) $$

for all $1 \leq k \leq m$.

(An even sharper inequality may hold, but the above would be sufficient for the purposes of my investigation).

Would it be possible to explain this via an injection

$\sigma: P(n+k,m) \to P(n,m)\cup P(n+m,m)$?

Although less preferable, a proof (or reference to a proof) by other techniques would also be accepted.

$\endgroup$

1 Answer 1

2
$\begingroup$

At first, for any $x$ and $s\in \{1,2,\dots,m\}$ we have $p(x+s,m)+p(x-s,m)\geqslant p(x,m)$ by obvious injection (remove or add part equal to $s$). At second, numbers $p(x,m)$ when $m$ is fixed and $x$ varies increase upto $m(m+1)/4$ and decrease after that. This was discussed here. Well, now your claim follows: if the segment $[n,n+m]$ does not contain the point $m(m+1)/4$, we have even $p(n+k,m)\leqslant \max(p(n,m),p(n+m,m))$. If it does, suppose without loss of generality that $k\geqslant m/2$ (else symmetrize $n,n+k,n+m$ with respect to $m(m+1)/4$) we have $p(n+k,m)\leqslant p(n,m)+p(n+2k,m)\leqslant p(n,m)+p(n+m,m)$, since $n+2k\geqslant n+m\geqslant m(m+1)/4$.

$\endgroup$
1
  • $\begingroup$ I see, so my claim is very close to proving unimodality for this sequence for which there is yet no combinatorial proof. Thank you for your answer. $\endgroup$
    – user94267
    Jun 27, 2016 at 11:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.