Labelling of a 2n-cycle with alternating coloured vertices in a specified fashion Take a 2n-cycle, colouring the vertices alternately blue and red.  For which n can we label the vertices 1 to 2n in such a way that the sum of the label of each blue vertex together with its two red neighbours is the same for every blue vertex?
 A: Well, there is a simple construction for odd $n=2k+1$'s. The construction is
$$k+1,4k+2,1,4k+1,k+2,4k,2,4k-1,\dots,2k,2k+4,k,2k+3,2k+1,2k+2$$
the smaller numbers are red, the larger are blue. The sum of label of each  blue vertex together with its two red neighbors is the same $5k+4$.
update:
Well, there is a not-that simple construction for evenly even $n=4k$. The construction for the red ones are:
$$1,6k,2,2k+2,3,2k+3,\dots ,k,3k,k+1,k+2,3k+1,k+3,3k+2,\dots,2k,4k-1,2k+1$$
(especially when $k=1$ the red ones are $1,6,2,3$)
These appears $1$ to $4k-1$ and a strange $6k$
The blue ones (in between) are
$$4k+1,4k,8k-2,8k-3,\dots 6k+2,6k+1,8k-1,6k-1,6k-2\dots,4k+2,8k$$
And the sum is $10k+2$.
update2:
Well, there is a really not-that simple construction for oddly even $n=4k+2$. The construction for the red ones are:
$$1,6k+3,2,2k+2,3,2k+3,\dots ,k,3k,k+1,k+2,3k+1,3k+2,k+3,\dots,4k,2k+1,4k+1$$
(especially when $k=1$ the red ones are $1,9,2,3,4,5$)
These appears $1$ to $4k+1$ and a strange $6k+3$
The blue ones (in between) are
$$4k+3,4k+2,8k+3,8k+2,\dots ,6k+7,6k+6,8k+4,6k+4,4k+4,6k+2,\dots,4k+6,4k+5,6k+5$$
And the sum is $10k+7$.
So, we proved that all the $n\ge 3$we can label in the way.
