For every $n \geq 0$ there is an inclusion of the ordered set $\{0<1<...<n\}$ into the product $\{0<1\}^{\times n}$ sending $i$ to the increasing sequence $(0 < ...<0<1<...1)$, in which 1 appears $i$ many times. In n-category theory the totally ordered set with $n+1$ elements and the powerset of the set with n elements admit the following categorifications called the $n$-th oriental $O(n)$ and the lax n-cube $\boxtimes^n$. These are n-categories, whose 1-truncations are the ordered set with $n+1$ elements and the powerset of the set with n elements, respectively. Is there a n-functor $O(n) \to \boxtimes^n$ that sends $i$ to the increasing sequence $(0 < ...<0<1<...1)$, in which 1 appears $i$ many times?