The sequence is finite. For any multiple of 32, there are more than 32 groups of that order. Thus, the sequence 1,2,...,32 can't appear in
A000001, and this sequence is at most 31 terms long.
The sequence is either 6 or 7 terms long. This can be shown by first showing every entry of
A373650 is congruent to 24 mod 48. It then follows that if n is such that
A000001(n+i) = i for i=1,2,3,4, then n+8 is a multiple of 16. But then
A000001(n+8) >= 14, so we can't have
A000001(n+i) = i for i=1,2,3,4,8.
From a(2) onwards, each entry is a multiple of 24, but not a multiple of 48.
a(7) > 223000000 if it exists.
Each entry is congruent to 0, 2 or 4 modulo 5.
