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A064709 Initial term of run of (exactly) n consecutive numbers with just 2 distinct prime factors. 6
6, 14, 20, 33, 54, 91, 323, 141 (list; graph; refs; listen; history; text; internal format)



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, 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, 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, Jan 05 2008

Comments from Vim Wenders, 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)

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, Oct 13 2008]

For a(10) to exist, one of the consecutive terms must be in A033846. Also, the sequence cannot have any terms for n > 14. If a(15) exists, one term has to be of the form A = 2^n*5^m. Then, there also must be two other terms divisible by 5, excluding A. Call them B and C.

       Case 1: If A is the smallest of these terms, then B = 5*(2^n*5^(m-1)+1) and C = 10*(2^(n-1)*5^(m-1)+1). However, for large enough values of m and n, either B/5 or C/10 is divisible by 3 and another prime > 5.

       Case 2: If A is the middle term, then B = 5*(2^n*5^(m-1)-1) and C = 5*(2^n*5^(m-1)+1). For large enough values of m and n, either B/5 or C/5 is a multiple of 3 and another prime > 5.

       Case 3: If A is the highest term, B = 5*(2^n*5^(m-1)-1) and C = 10*(2^(n-1)*5^(m-1)-1). Again, for large enough values of m and n, either B/5 or C/10 is a multiple of 3 and another prime > 5.

       Thus, if there are more terms in the sequence, they can only be a(9), a(10), a(11), a(12), a(13) or a(14).

     Other comments: The only numbers that satisfy a(8) are 141 and 212. The reason a(9) wasn't satisfied in either of these is because one end of the run of numbers was a prime and the other end stopped at a multiple of 10. I believe it is possible to show that a(10) can never exist because there cannot be a multiple of 10 in the run of consecutive numbers, perhaps because there cannot be two multiples of 5. - Derek Orr, May 24 2014


Roger B. Eggleton and James A. MacDougall, Consecutive integers with equally many principal divisors, Math. Mag 81 (2008), 235-248. [From T. D. Noe, Oct 13 2008]


Table of n, a(n) for n=1..8.

Carlos Rivera, Prime Puzzle 427


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.


Cf. A064708.

Sequence in context: A123267 A236928 A064708 * A118129 A046712 A162823

Adjacent sequences:  A064706 A064707 A064708 * A064710 A064711 A064712




Robert G. Wilson v, Oct 13 2001



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Last modified May 26 05:25 EDT 2017. Contains 287077 sequences.