through the $p$-cores appearing in the $p$-core tower of $\nu$). Since each such expression is positive by assumption, the overall sum is also positive.

Once Theorem \[thm.landaugeneral\] is proved, Theorem \[thm.multinomial\] is indeed trivial to prove:

Thanks to Theorem \[thm.landaugeneral\], we know we only need to show $${{h}_{s}\!\left(\lambda\right)} - t \cdot {{h}_{st}\!\left(\lambda\right)} \ge 0$$ for all $\lambda$, $s$ and $t$. This follows immediately from applying to each of the $s$-quotients of $\lambda$.

**Remark.** The same proof works for $v(\lambda;(x),\vec{\delta})$, as long as $x$ divides each of the $\delta_l$s. This case corresponds under $\Phi$ to a multinomial coefficient $u(n;(\sum_l \beta_l),\vec{\beta})$, but not all $u$ of that form map back to $v$ that answer Question \[q.hook\] positively. Consider $\vec{\beta} = (9,6,3,3,3)$, which then gives $v(\lambda;(3),(8,12,24,24,24))$, which is non-integral as soon as $\lambda$ is a 3-core with a hook of length 8 (by Theorem \[thm.landaugeneral\]), for instance $(6,4,2)$.

Explicit construction of non-integral ratios {#sec.nonintegral}
--------------------------------------------

The proof of Theorem \[thm.landaugeneral\] actually constructs example partitions leading to non-integral ratios: it translates a $\mu$ failing Condition  into a $\lambda$ failing Condition .

Consider for instance the parameters $(\gamma_k) = (1,30)$ and $(\delta_l) = (2,3,5)$. Take the partition $\mu = (6,6,6,6,6)$ and evaluate Condition  (maybe with 