A proof that three definitions of an integer sequence A055926
  (please see http://oeis.org/A055926 ) indeed yield the same sequence.

The definitions are:

  "Numbers n such that {largest m such that 1, 2, ..., m divide n} is
   different from {largest m such that m! divides n}";
  (The original one by Leroy Quet, Jul 16 2000)

  Two new definitions by Antti Karttunen, November 2013:

  "Numbers n which are either odd multiples of 12 or the largest m such
   that (m-1)! divides n is a composite number > 5."

  "Numbers n for which A232098(n) (largest m such that m! divides n^2) differs
   from (is greater than) A055881(n) (largest m such that m! divides n)."


The proof that the first two defintions are equivalent:


Case A1: For odd multiples of 6 (A017593), that is, for such n which
         are of the form (2k+1)*6, we see that A055881(n) is always 3, as 3!
         divides such numbers, but 4! not. On the other hand, A055874(n) is
         also always 3 for such numbers as there is only one 2 in their
         factorization. Thus none of these numbers appear in A055926
         according to the original definition.

Case A2: For odd multiples of 12 (A073762), that is, for such n which
         are of the form (2k+1)*12, we see that A055881(n) is always 3, as 3!
         divides such numbers, but 4! not. On the other hand, A055874(n) is
         at least 4 for such numbers. Thus the original definition implies
         that A055926 contains all the numbers of the form (2k+1)*12.
         The above two cases thus cover all the numbers whose factorial base
         representation ends with exactly two zeros, either as '...100' or
         '...300' (case A1) or '...200' (case A2).
         It is also easy to see that no numbers with less than two trailing
         zeros will occur in A055926 (i.e. numbers n with A055881(n)<3).

The numbers with more than two trailing zeros in their factorial base
representation are covered by the next two cases:

Case B1: For all those terms n of A055926 not in A073762, it
         must be true that A055881(n) = c-1 for some composite c > 5.
         We note that if A055881(n) were p-1 for some prime p, and
         A055874(n) > A055881(n) (which is required by the original
         definition of A055926), then it would mean that p would
         divide n, which would contradict our assumption that (p-1)!
         would be the highest factorial dividing n, as then also
         (p-1)!*p = p! is guaranteed to divide n, as that prime p could
         not have been obtained from the set 1, ..., p-1 by multiplying.
         Thus for n to occur here, A055881(n) can never be one less than
         prime, but instead, must be one of the terms of A072668 (one
         less than some composite).

Case B2: Whenever A055881(n) = c-1 for some composite c > 5, then n is
         a term of A055926 by the Quet's original definition. This is true
         because if c is composite then (c - 1)! is congruent to 0 (mod c);
         please see the section "Composite modulus" in Wikipedia's
         "Wilson's theorem" article:
           http://en.wikipedia.org/wiki/Wilsons_theorem
         In other words, there are enough factors among 1 .. c-1, that
         we can construct c (by multiplying) out of them. Thus A055874(n)
         is at least c, which is larger than A055881(n) (= c-1), and by
         the original definition belongs into A055926.

The following proves the equivalence of the first and third definitions:

Lemma C: A055926 gives also all the positions on which A232098(n)
         (largest m such that m! divides n^2) differs from A055881(n)
         (largest m such that m! divides n).

         For the numbers of the form (2k+1)*6 [A017593] we have
         ((2k+1)*6)^2 = (2m+1)*36, and while 3! divides that, 4! does not.

         For the numbers of the form (2k+1)*12 [A073762] we have
         ((2k+1)*12)^2 = (2m+1)*144, and 4!|144 (= 6*24).

         For the other numbers, the cases B1 & B2 from above apply also
         here with a slight modification: first, by squaring a number,
         we don't add any prime factors to it, so it is true also here
         (for this another alternative condition to hold) that A055881(n)
         must be c-1 for some composite, and into the other direction,
         we see that by squaring n, we have an "extra copy" of n from which
         to construct c, thus:

   A232098(n) = A055881(n^2) = A055881(c*(c-1)!*rest) >= c > A055881(n) = c-1.

         This means that whenever A055881(n) = c-1 for some composite c > 5,
         A232098(n) > A055881(n) and thus n belongs into A055926 also by
         the third definition.

We have also a corollary:

Lemma 2: A232099 is a subset of A055926.  It is clear that
           A055881(n) <= A055874(n) for all n,
         and likewise
           A055881(n) <= A232098(n).
         From the equivalence of the first and third definitions
         it also follows that
           A055874(n) > A055881(n)
         if and only if
           A232098(n) > A055881(n),
         thus it is not possible that A055874(n) = A055881(n) but that
         A232098(n) would differ from them.
         Thus, whenever A055874 differs from A055881, so does also A232098.


         Antti Karttunen, December 6 2013.

         If you spot mistakes in this proof or have any additional insight,
         then please contact me via my OEIS home page:
         http://oeis.org/wiki/User:Antti_Karttunen

         Thanks!