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.


1 Answer 1


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 $n+2k\geqslant n+m\geqslant m(m+1)/4$.

  • $\begingroup$ 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

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