While in New Zealand I met Eamonn O'Brien, who's one author of a paper that, among other things, tabulates the number of groups of each order up to 2000. This has led me to find quite q few interesting sequences, and given quite a lot of fun. First, some comments on the numbers. For n = 1 to 32, the number of groups of order n is 1,1,1,2,1,2,1,5,2,2,1,5,1,2,1,14,1,5,1,5,2,2,1,15,2,2,5,4,1,4,1,51. Certain numbers (such as 1,2,5,15) are much more common than others. The numbers of groups of orders 1,2,4,8,16,32,64,128,256,512,1024 are 1,1,2,5,14,51,267,2328,56092,10494213,49487365422. The number of groups of order at most 2000 but not 1024 is only 423164062, so the ones of order 1024 outnumber the others 10-to-1! Remarkably, the number of groups of order p^n (p prime) is known quite closely - it has the form p^{ (2n^3/27) + O(n^{5/2}) }, the 5/2 here improving on an earlier bound of 8/3. The smallest N for which the number of groups of order N is 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 is 1 4 75 28 8 42 375 510 308 90 140 88 56 16 24 100 675 156 1029 820 1875 For n = 22, we must have N > 2000. This raises the interesting question of whether every positive integer is the number of groups of some order? O'Beirne told me that Keith Devlin has worked on this, but I didn't get the impression he'd got very far. Lenstra and I pushed it quite a long way, though with some mistakes that I've since been rectifying. I'll report on this later. I propose to use GROUPS(N) for the number of groups of order N, but will abbreviate it to G(N) in this letter. Let's call N group-deficient or group-abundant according as G(N) < N or G(N) > N, and group-perfect if G(N) = N. Then it seems almost certain that 1 is the only group-perfect number, and that almost all numbers are group-deficient. The best we've proved in this direction, though, is that all squarefree numbers (and so most numbers) are group-deficient. Here's the elegant proof that was found by Lenstra and simplified a bit by me: Any group of squarefree order N has a presentation of the form 1 = a^d = b^(N/d), ab = ba^k, where d is some divisor of n, and k is prime to d. So the number of such groups is at most the sum of phi(d) over d dividing N, which is well-known to be N. However, if N > 1 this counts the cyclic group twice (for d = 1 and d = N), and so then the number of groups of order N is strictly less than N. One of the conjectures above can be put in form A(N)/N -> 0, where A(N) is the number of group-abundant numbers up to N. However, the convergence is slow. For example, if N = 49487365422 and n = 1024, then every multiple of n below N is group-abundant, and so A(N)/N exceeds 1/n. We can replace n by any 2^m > 16 and N by GROUPS(n), which is approximately 2^{(2n^3)/27}, to see that this argument proves that we have 1 A(N)/N > ------------------------- roughly, 3 times cuberoot(logN /2) for an infinity of N, the logarithm being taken to base 2. Here's how the sequence of group-abundant numbers starts: 1,32,48,64,96,128,144,160,192,256,288,320,384,432,448,480,512,576,640,648, 672,704,720,768,800,832,864,896,960,1024,1088,1152,1216,1248,1280,1296, 1344,1408,1440,1458,1536,1600,1664,1728,1792,1920,1944 Lenstra's theorem tells us that none of these can be squarefree - in fact so far they've always been divisible by some fourth power, which suggests that perhaps that theorem can be improved. Most of my thinking has been on the universality question - "is every positive integer the number of groups of some order?". As I said, I'll report on that later. It gives rise to some more interesting sequences by the way, including 7,11,19,29,31,47,49,... . REgards to all - JHC