@MattSamuel , indeed. Which bound was J. Zhang (or is Derek Holt) writing about?

Mistakes will be made. If we are forgiving of ourselves, we can forgive others, and in such a cooperative spirit the mistakes made in this forum will be fixed. May the original poster collectively forgive us for getting it wrong and thank us for getting it right.

Shouldn't the number of Sylow groups for S_2 be 2 then, strictly speaking? Not to mention 1 for S_1 ?

You can do it by hand, but it would take a while. Factor 2*2953 + 1 as 3 * 11 * 179, and consider those terms which are multiples of 3 or 11 or 179 separately. Or as Robert Israel commented, work modulo each prime.

@GHfromMO, that might have been your reading of the original post. Mine (and I imagine Todd's) is that the idea "2p+1 is prime" is a conjectured consequence. Of course, if 2p+1 is an assumption, then your observation holds, and Wilson is not needed to get the congruence. Then again, Robert Israel's example shows Wilson is of little use here.

This may be in Joe Roberts's calligraphic text on number theory.

How does one get that 2p+1 is prime from your observation?

Oops. I flipped a sign bit. I no longer know if one-sided functions make sense, much less help with F_D. It might help to dichotomize into x2 bigger or less than D/2, but I am unsure now if that helps.

Suppose you consider one-sided basic D functions. These are those functions in which x2 and p2 are both positive. Then consider G_D as any function which is a sum of one sided functions. Then f in F_D should be a difference of two functions in G_D. While decomposing f may be hard, dealing with functions in G_D should be easy. Is it? Does it help to decompose f into functions in to F_D' which in turn may be decomposed into functions of G_D for D' bigger than D?

In spherical coordinates, have rho be (0,2,4,6,8)*2pi/10. I don't know what theta is, but it should be largest so that the 6 spheres reflected and then rotated in rho by 2pi/10 give an arrangement of 12 spheres kissing, with 2 central spheres. There are several variations you can play. For example, theta=60 gives five spheres each touching the two central spheres. Pack 10 more spheres around these five, and then see if there is room for four more. If not, move the arrangement (decrease theta) "to the right" to make it asymmetric, and try adding a sphere on the right and three on the left

For small values of k/N, the sum is well approximated by (N+1) choose k, and for slightly larger values by a multiplicative correction of the form (1/(1 - k/N)) . For k near the size of N/c for c an integer not too large, you can use a recursion where the recursive step involves just addition and runs in O(k) time (O(k^2) if you are measuring bit operations).

Actually, if you can push the two rings of five close enough together, you may be able to get two more rings of five on the outside.

You might investigate packing which involve 10 spheres in two rings of five around the contact point of the central spheres. (Note there are several ways, but probably fewer than 20, to bunch them up on the different rings.) You might be able to place another 9 spheres with a nonsymmetric arrangement on the two rings.

There can't be three primtive triples, but there can be two with an additional triple satisfying the relations.

I think A_4^n being big means the answer to the question is no.

I think there is a rational transformation which shows the outside version reduces to the inside. The relevant picture may e en be in Guy's book.