Bulgarian Solitaire: Size of root loops
This is probably covered in one of the papers that show up in Gerry's search, but since I spent a little time on it I will post a solution anyway. The solution is also relatively short.
El'endia's conjecture is correct. A cycle of the prescribed length can be constructed as follows. We first recall the fixed point partition $p_k=(k,k-1,k-2,\ldots,2,1)$ of the triangle number $n=T_k=k(k+1)/2$. Suppose that instead the total number of stones is $n=k(k+1)/2-m$, where $0<m<k$. I claim that in this case we get a cycle of length $k$ by removing one stone from the $m$ smallest piles of the partition $p_k$. In other words, let us study the partition $$p_k^*=(k,k-1,k-2,\ldots,m+1,m^*,(m-1)^*,...,2^*,1^*),$$ where I use the funny notation $\ell^*$ to denote a pile of $\ell-1$ stones, because that pile really wants to have $\ell$ stones, but it feels a little bit left out. Therefore $1^*$ really is an empty pile.
Let $p_k^{*,m,i}$ denote the partition gotten from $p_k$ by marking $m$ consecutive entries starting from the $i$-pile with an asterisk. If $i<m$, then we move the remaining asterisks to the beginning of the partition in the usual cyclic fashion (so the above partition $p_k^{*}=p_k^{*,m,m}$). For example, $$p_5^{*,3,2}=(5^*,4,3,2^*,1^*)=(4,4,3,1,0)=(4,4,3,1).$$ The key observation is that the Bulgarian solitaire move maps the partition $p_k^{*,m,i}$ to the partition $p_k^{*,m,i-1}$, where we interpret $i-1=0$ as $k$ (or decrement $i$ by one modulo $k$). This is easy to see, because the entries of all these partitions are in descending order. The only slightly tricky thing is that the empty pile $1^*$ corresponds to $k^*$ after the Bulgarian solitaire move, because the presence of an empty pile means that the new pile loses one of its stones, and gets only $k^*$ stones instead of the usual $k$.
As an example, let $n=25=28-3=T_7-3$. Then we get a cycle of 7 partitions starting with $(7,6,5,4,3^*,2^*,1^*)=(7,6,5,4,2,1)$. The cycle consists of the partitions $$(7^*,6,5,4,3,2^*,1^*)=(6,6,5,4,3,1),$$ $$(7^*,6^*,5,4,3,2,1^*)=(6,5,5,4,3,2),$$ $$(7^*,6^*,5^*,4,3,2,1)=(6,5,4,4,3,2,1),$$ $$(7,6^*,5^*,4^*,3,2,1)=(7,5,4,3,3,2,1),$$ $$(7,6,5^*,4^*,3^*,2,1)=(7,6,4,3,2,2,1),$$ $$(7,6,5,4^*,3^*,2^*,1)=(7,6,5,3,2,1,1),$$ and back to $$(7,6,5,4,3^*,2^*,1^*)=(7,6,5,4,2,1).$$