A064709 Initial term of run of (exactly) n consecutive numbers with just 2 distinct prime factors.

6, 14, 20, 33, 54, 91, 323, 141

Offset: 1

Author

Robert G. Wilson v, Oct 13 2001

Keywords

• nonn
• hard
• fini
• full

Comments

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

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's 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. - 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

Eggleton and MacDougall prove that no terms exist beyond a(9) and conjecture that a(9) does not exist. - Jason Kimberley, Jul 08 2017

Examples

			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.
		

Links

• Roger B. Eggleton and James A. MacDougall, Consecutive integers with equally many principal divisors, Math. Mag. 81 (2008), 235-248. [_T. D. Noe_, Oct 13 2008]
• Carlos Rivera, Prime Puzzle 427

Crossrefs

Cf. A064708.

Programs

Mathematica

Function[s, Function[t, Map[t[[First@ FirstPosition[t[[All, -1]], k_ /; k == #] ]] &, Range[0, Max@ t[[All, -1]] ] ][[All, 1]] ]@ Join[{{First@ s, 0}, {#[[1, 1, 1]], 1}}, Rest@ Map[{#[[1, 1]], Length@ # + 1} &, #, {1}]] &@ SplitBy[Partition[Select[#, Last@ # == 1 &][[All, 1]], 2, 1], Differences] &@ Map[{First@ #, First@ Differences@ #} &, Partition[s, 2, 1]]]@ Select[Range[10^5], PrimeNu[#] == 2 &] (* Michael De Vlieger, Jul 17 2017 *)
With[{t=Table[If[PrimeNu[n]==2,1,0],{n,350}]},Flatten[Table[ SequencePosition[ t,Join[{0},PadRight[{},k,1],{0}],1],{k,8}],1]][[All,1]]+1 (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 16 2020 *)