The kissing number in a certain number of dimensions is the greatest number of spheres that can be brought to touch (or "kiss") a central sphere if all the spheres are the same size. in one dimension the kissing number is 2. In two dimensions the problem is slightly less trivial but hardly difficult. It will probably not come as much of a surprise that the best arrangement in 2 dimensions is the familiar hexagonal arrangement.
A little thought will show that all the angular space around the central circle is taken up. Since the centers of 3 equally sized spheres mutually in contact with each other form an equilateral triangle.
So the minimum angle between the centers of any two circles touching the central circle is just Pi/3, the angle of an equilateral triangle. The sum of the angles between the centers of all of the circles touching the central circle must add up to a full 2*pi and each of these angles must be at least Pi/3. So the most circles we could possibly get to touch the central circle is 2*Pi/(Pi/3) = 6 circles. Since the hexagonal arrangement actually achieves this maximum we have our proof.
In three dimensions a good solution is found by taking the optimal 2 dimensional hexagonal arrangement of spheres around a central one in a plane but now it is possible to add spheres above and below the plane that touch the central sphere. Three spheres can be placed above and three spheres below the plane of the hexagonaly arranged spheres while still contacting the central sphere. The arrangement gives 6 + 3 + 3 = 12 as a lower bound for the kissing number in three dimensions. The centers of the spheres arrayed around the central one now form the vertices of one of the Archimedean solids, the Cuboctahedron.
Is this the optimal arrangement or is there possibly some more clever arrangement which could fit an extra sphere or two in? Generalizing our tactic for the two dimensional case we can ask what is the minimum angular space taken up by each sphere. A calculation of the solid angle taken up by a sphere touching the central sphere gives. Pi*(2-sqrt(3)) = 0.841 steradians. Dividing the total number of steradians by this value gives 4*pi/(Pi(2-sqrt(3)) = 14.92 So this gives us a definite upper limit of 14 spheres that we can fit around a central sphere.
But this bound ignores the fact that spheres do not fit flush with each other so we can't hope to fill all the angular space around the central sphere. Consider an arrangement of equal sized spheres placed at the vertices of a regular tetrahedron such that all the spheres touch each other. If all the spheres around the central sphere could be made to form such tetrahedral arrangements such that the lines between the centers of all touching exterior spheres form equilateral triangles clearly this arrangement would be optimal. I worked on a proof for a bit but already in three dimensions the problem is quite hard and I am not quite sure how to approach it. Nevertheless such tetrahedral arrangements are not possible in 3 dimensions or in any higher dimension either. The triangular packing of 2 dimensions is unique.