Following a suggestion from Ed Pegg Jr, the sequence can be written in a more readable form as: 1!, 3!, 5!, 11# * 3! * 2, 17# * 5! * 2, 29# * 7! * 4, 53# * 7! * 12, 89# * 11! * 2, 157# * 17# * 8! * 6, 271# * 23# * 10!, 487# * 29# * 10!, 857# * 37# * 11! * 42, 1487# * 53# * 15! * 2, ..., where p# = primorial(p) = A034386.
From T. D. Noe, Jul 06 2005: (Start)
Let c(p) be the smallest colossallyabundant number having the prime factor p. See A073751 for info about computing these numbers.
Then the terms of this sequence can be expressed as
a(2) = c(3)
a(3) = c(5) * 2
a(4) = c(11) / 2
a(5) = c(17) / 3
a(6) = c(29) * 14
a(7) = c(53)
a(8) = c(89) * 4
a(9) = c(157) * 34
a(10) = c(271) * 23
a(11) = c(487) / 2
a(12) = c(857) / 2
a(13) = c(1487) * 212
a(14) = c(2621) * 710
a(15) = c(4567) * 2/21
a(16) = c(8011) / 2
a(17) = c(13999) * 1630. (End)
Initially, each term is divisible by the previous one. Is there a reason this should always be true?  Santi Spadaro, Aug 13 2002
The conjecture a(n)a(n+1) holds out to n=10.  Devin Kilminster (devin(AT)maths.uwa.edu.au), Mar 10 2003
The conjecture a(n)a(n+1) fails for n=15.  T. D. Noe, Jul 08 2005
We have a(n) = min{A007539(n), A134716(n)}, and clearly A007539(n) != A134716(n) for every n. For what values of n is the former lesser than the latter?  Jeppe Stig Nielsen, Jun 16 2015
