TITLE: A puzzle with some jumping frogs
QUESTION [31 upvotes]: (The following puzzle is ispired by this nice video of Gordon Hamilton on Numberphile)
In a pond there are $n$ leaves placed in a circle, for convenience they are numbered clockwise by $0,1,\ldots,n-1$. At the beginning, on each leaf there is a frog, so there are $n$ frogs. At each turn, the frogs can jump accordingly to the following rule: "If on leaf $j$ there are $k \geq 1$ frogs and if on leaf $(j + k) \bmod n$ there is at least one frog, then all the $k$ frogs on leaf $j$ can jump on leaf $(j + k) \bmod n$".
Is it true that the $n$ frogs can finally be all on the same leaf if and only if $n$ is a power of $2$?
It is quite easy to prove that if $n$ is a power of $2$ then there is a sequence of jumps that leads all the frogs on the same leaf. On the other hand, I checked by a brute force algorithm that no such sequence exists if $n \leq 14$ is not a power of $2$.
NOTE: I previously asked this question on MSE and I got non answers.
UPDATE 1: aorq checked that the answer is YES for all $n \leq 50$.
UPDATE 2: It might be worth trying to solve first this relaxed version of the problem: 
To each solution of the puzzle associate a directed graph on the vertexes $0,1,\ldots,n-1$, where a directed edge $i \to j$ means a jump of all the frogs from leaft $i$ to leaf $j$. Say that the solution has $\ell$ leader frogs if its associated directed graph has exactly $\ell$ sources.
For which $\ell$ is it true that the $n$ frogs can finally be all on the same leaf using a series of jumps with $\ell$ leader frogs, if and only if $n$ is a power of $2$?
We know that $\ell = 1$ does the job, since with only one leader frog the problem is equivalent to "For which $n$ are the first $n$ triangular numbers distinct modulo $n$" and the answer is "only for $n$ a power of $2$".

REPLY [3 votes]: Edit: As pertains to UPDATE 2, the post below gives references to settle the case of $\ell = 1$.

This is a bit long for a comment, but I thought it worthwhile in case the following is non-obvious:

It is quite easy to prove that if $n$ is a power of $2$ then there is a sequence of jumps that leads [to] all the frogs on the same leaf. 

(Maybe this really is "quite easy to prove," but I would not have known this to be the case at a glance without a familiarity with the references included below...)
If you begin with a frog and have it jump $1$ leaf, then have the newly formed duo jump $2$ leaves, then have the newly formed trio jump $3$ leaves, etc, then you move around the leaves by $+1, +2, +3, \ldots$ and, since you are jumping in a circle of $n$ leaves, the positions of frogs jumping together are at the triangular numbers modulo $n$. 
And so the assertion elsewhere in this thread, which (in particular) covers the quotation-pull above, is that this will be sufficient to reach all the leaves if $n$ is a power of $2$. Equivalently, the set of the first $n$ triangular numbers modulo $n$ is $\{0, 1, \ldots, n-1\}$ if $n$ can be written as a power of $2$.
In fact, the last statement is true with if replaced by iff, and this was the subject of a very nice write-up in the New York Times' Numberplay column as Daniel Finkel’s Circle-Toss Game. There is another proof of this proposition by induction here, which includes a pointer to an earlier proof by D Knuth in:

The Art of Computer Programming, Volume 3, Chapter 6.4. Excercise 20 in the second edition.

Of course, with respect to the main question:

Is it true that the $n$ frogs can finally be all on the same leaf if and only if $n$ is a power of $2$?

just the "if" is covered: the strategy above (as proved in the three different references given) shows how to get all the frogs to the same leaf for $n$ being a power of $2$. The "only if" part is what remains difficult, since the OP is proposed in greater generality than the circle-toss game: instead of the frogs needing to jump together in the triangular number pattern already described, one has access to lots of other patterns. I have no new suggestions for the "only if" component, although perhaps seeing the different proofs mentioned above (I especially enjoy the "multi-ball argument" in the first link) could serve as some sort of inspiration.