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#enumerativecombinatorics

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Mark Gritter<p>I _think_ this is all simple paths on a 4x3 rectangular lattice, up to symmetry (including swapping start and end.)</p><p>But it took me long enough to get the symmetry conditions right that I should probably double-check by enumerating them w/o symmetry and then compute the orbits directly.</p><p><a href="https://mathstodon.xyz/tags/Combinatorics" class="mention hashtag" rel="nofollow noopener" target="_blank">#<span>Combinatorics</span></a> <a href="https://mathstodon.xyz/tags/MathArt" class="mention hashtag" rel="nofollow noopener" target="_blank">#<span>MathArt</span></a> <a href="https://mathstodon.xyz/tags/EnumerativeCombinatorics" class="mention hashtag" rel="nofollow noopener" target="_blank">#<span>EnumerativeCombinatorics</span></a></p>
Don_Rubiel<p>Now, our techniques to solve this combinatorics problem:<br>We studied an operad (the operad of posets), and then we studied power series indexed by posets.</p><p>Since the power series are indexed by posets, endomorphisms of posets can be extended to endomorphisms of those power series. That is, if we have a series $F(P)$ for every poset $P$, and we have $Q$ a function taking $P_1,...,P_n$ posets to a poset $Q(P_1,...P_n)$, then we define $Q(F(P_1),...,F(P_n)):=F(Q(P_1,...P_n))$.</p><p>If we work with sets, then we dont have to worry too much about the properties of this induced functions. They are just set theoretical functions.</p><p>Now, the correct language here is the language of <a href="https://mathstodon.xyz/tags/operads" class="mention hashtag" rel="nofollow noopener" target="_blank">#<span>operads</span></a>. Since we work with the operad of posets, this are topological operads. It is just that the topological spaces are finite but with a topology distinct from the discrete topology!.</p><p>Returning to the problem of enumerating shuffles between a tree and a linear tree. If we consider the series $F(P)$ whose $n$ coefficient enumerates shuffles between a poset $P$ and the linear order with $n$ cards, here $P$ is the poset of vertices of the input tree,<br>then we have an action of the operad of posets of the form $Q(F(P_1),...,F(P_n)):=F(Q(P_1,...P_n))$.<br>Now, the difficult part is to compute the actions of the posets. But in the paper <a href="https://doi.org/10.1007/s10801-025-01386-7" rel="nofollow noopener" translate="no" target="_blank"><span class="invisible">https://</span><span class="ellipsis">doi.org/10.1007/s10801-025-013</span><span class="invisible">86-7</span></a> we were able to compute the action of the posets $\{x,y\}$ and $\{x&lt;y\}$, this is enough to compute the series of any series parallel poset, which includes trees.<br> <br>Sometimes, the action of the operad preserves the combinatorial / enumerative property of the series. That is, the operad sends series solving a problem to series solving the same problem. <a href="https://mathstodon.xyz/tags/EnumerativeCombinatorics" class="mention hashtag" rel="nofollow noopener" target="_blank">#<span>EnumerativeCombinatorics</span></a> 2/3</p>
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