Well with 2 points it is simple - poles apart.

With 3 it is easy too - an equilateral triangle

4 gets us thinking about the vertices of a regular tetrahedron

6 makes us think of the mid-points of its edges

But what about 5?

Well if we try one at each pole of a sphere and 4 on its equator, then the equator ones are nearer together than is any one of them to the poles.

So lets try a little trick:

REMOVE one of the points from the equispaced 6!

Ah, yes, so we have a "semiconductor crystal" with one vacancy - one "hole" for an electron to move into

There are 6 alternative positions for the vacancy - but NOT 6 alternative configurations.

Each of the 6 is the SAME, as can be shown by rotating it.

So did this trick solve the question "5 equispaced points"

Well, it depends what equispaced is.

They ARE equispaced in as much as EACH of the 5 has a nearest neighbour at distance d AND d is the same for every one of them!

OK so how about 7 points

Addding one to 6 is tough - where should we try to put it?

So remove one from 8!

8 points evenly spaced are the vertices of a cube (and 12 are the mid points of its edges)

So we remove one of the 8

Once again for each point remaining we can find a nearest neighbour the same distance away as for all other points.

Once again every one of the 8 "alternative" shapes is the SAME - a mere rotation of itself!

How much longer can we keep doing this "vacancy" trick

Can we remove 2, 3?

And HOW MANY alternative different "crystals" are there for each N?

How about N= 17?