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COMMENT
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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.
Comment from T. D. Noe (noe(AT)sspectra.com), Jul 06 2005:
"Let c(p) be the smallest colossally-abundant 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"
Initially each term is divisible by the previous one. Is there a reason why this should always be true? - Santi Spadaro (santi_spadaro(AT)virgilio.it), 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 (noe(AT)sspectra.com), Jul 08 2005.
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EXTENSIONS
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More terms from Walter Nissen Apr 15 1997. Further terms from Devin Kilminster (devin(AT)maths.uwa.edu.au), Mar 10 2003
The term a(10) = 271#23#10! was apparently found independently by Bodo Zinser and Don Reble, circa Jul 05 2005
The next term, a(11) = 487#29#10!, was corrected by Don Reble, Jul 06 2005
a(12) = 857#37#11!42 from Don Reble, Jul 06 2005
a(13) = 1487#53#15!2 found by T. D. Noe and confirmed by Don Reble, Jul 07 2005
a(14)-a(17) found by T. D. Noe and and rechecked by him Oct 11 2005
a(15) corrected. The conjecture still fails at n=15 T. D. Noe (noe(AT)sspectra.com), Oct 13 2009
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