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COMMENT
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The given terms up to a(8) = 141 are the only terms less than 10^18. To speed the search, note that any string of 6 or more consecutive numbers contains a multiple of 6, and hence must contain a number of the form 2^a * 3^b. Conjecture: 141 is the last term, because numbers with only two different prime factors get pretty rare, so having several in a row near a number of the form 2^a * 3^b is pretty unlikely. - Joshua Zucker (joshua.zucker(AT)stanfordalumni.org), May 05 2006
Sequence cannot have any terms for n > 29, since a run of 30 or more consecutive numbers must contain a multiple of 30, divisible by at least 3 primes. - Franklin T. Adams-Watters (FrankTAW(AT)Netscape.net), Oct 23 2006
I searched numbers of the form n=2^a * 3^b through 10^700 and could not find any solution where even 4 numbers (n+2, n-2, n+3, n-3) had omega=2. The last such number through 10^700 is only 169075682574336=2^33 * 3^9. So a full set of 9 numbers seems quite unlikely - Fred Schneider (frederick.william.schneider(AT)gmail.com), Jan 05 2008
Comments from Vim Wenders (vim(AT)gmx.li), Apr 02 2008: (Start) The sequence is complete. The argument of Franklin T. Adams-Watters is easily extended: if 2^a.3^b, a,b, >=1 is a term then omega(2^a.3^b+-6) > 2 (because the exponents of 2 and 3 follow a ruler like sequence). So the last possible term would be a(11).
Also, if 2.p, p prime, is in the run of an initial value to check, then p+2, p+4, ... has to be prime too, (for the values 2p+4=2(p+2),2p+8=2(p+4) ...), which is impossible for obvious reason.
The two arguments limit the maximum length of a run to 8. (End)
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