Search: id:A064709
Results 1-1 of 1 results found.

%I A064709
%S A064709 6,14,20,33,54,91,323,141
%N A064709 Initial term of run of (exactly) n consecutive numbers with just 2 distinct 
               prime factors.
%C A064709 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
%C A064709 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
%C A064709 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
%C A064709 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).
%C A064709 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.
%C A064709 The two arguments limit the maximum length of a run to 8. (End)
%C A064709 Wenders' argument is incomplete because the consecutive even numbers 
               can have the form 2^a p^b. As stated in the paper by Eggleton and 
               MacDougall, it is still a conjecture that 9 consecutive omega-2 numbers 
               do not exist. [From T. D. Noe (noe(AT)sspectra.com), Oct 13 2008]
%D A064709 Roger B. Eggleton and James A. MacDougall, Consecutive integers with 
               equally many principal divisors, Math. Mag 81 (2008), 235-248. [From 
               T. D. Noe (noe(AT)sspectra.com), Oct 13 2008]
%H A064709 Carlos Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_427.htm">
               Prime Puzzle 427</a>
%e A064709 6 = 2*3; 14 = 2*7 and 15 = 3*5; 20 = 2^2*5, 21 = 3*7 and 22 = 2*11; 33 
               = 3*11, 34 = 2*17, 35 = 5*7 and 36 = (2*3)^2; etc.
%Y A064709 Cf. A064708.
%Y A064709 Sequence in context: A101567 A123267 A064708 this_sequence A118129 A046712 
               A162823
%Y A064709 Adjacent sequences: A064706 A064707 A064708 this_sequence A064710 A064711 
               A064712
%K A064709 nonn,easy,fini,full
%O A064709 1,1
%A A064709 Robert G. Wilson v (rgwv(AT)rgwv.com), Oct 13 2001

Search completed in 0.001 seconds