An obvious lower bound for 1 is
It is not hard to see that
Graphically for $k = 3$:
So the answers are
When all vertices can be multiply occupied, as @KyleParsons pointed out in the comments, no edge can be traversed by two people at a time (I had overlooked this in an earlier version).
It is easy to see that any two-way split S=A+B of the sum S=1+..+n=n(n+1)/2 can be achieved by forming two disjoint subsets. Let A=ac,B=bc with a,b relatively prime AB is a perfect square precisely if a and b are. Therefore
this will be the case
The theorem doesn't rule out even powers of primes of the form 4k+3 but that is only ...
I'm going against some of the things you suggested here but one possibility I've found that could work is the following (where each space must be filled by a unique digit between 1 and 9)
Framing it in this way makes it look like a single equation. I checked using a computer and this has exactly two solutions
which look like yours but with the end pairs ...
It's grouped like this
(2), (4,5,3), (10,11,13,7,5), (28,29,31,35,?,?,?)
For $i$th group , it is $3^i+2^j$, with $j$ increases from 0 to $i$
and then $3^j+2^i$, with $j$ decrease from $i-1$ to 0
There is the write down
$\ \ 2=3^0+2^0 \\
\ \ 4=3^1+2^0,\ \ 5=3^1+2^1,\ \ 3=3^0+2^1 \\
10=3^2+2^0, 11=3^2+2^1, 13=3^2+2^2,\ \ 7=3^1+2^2, 5=3^0+2^2 \\
The technique I found helpful was
First of all, notice that
Now tweak (1) and (2) by swapping two numerators and fixing up:
A similar switcheroo gets us
This one we get in a slightly different way. (Maybe there's an easier way, but this is the first thing I found.)
and we're done.
Imagine coloring the vertex opposite the black face red. As we roll the tetrahedron around, we keep track of the projection of the red vertex onto the plane. It is easy to see that the red vertex will always be constrained to the set of red points below, which form the vertices of a hexagonal grid. In particular, the red vertex will never end up at any of ...