A242492 For any integer m > 1, the m-th term of the sequence is the minimal squarefree integer x with exactly m prime divisors such that x+1 and x+2 are also squarefree integers with exactly m prime divisors.

33, 1309, 203433, 16467033, 1990586013, 41704979953, 102099792179229

Offset: 2

Author

Daniel Constantin Mayer, May 16 2014

Keywords

• hard
• nonn
• more

Comments

The five terms for m = 2,3,4,5,6 were computed with the aid of PARI/GP. But it seems to be rather difficult to compute higher terms, if they exist at all.

The distribution of squarefree integers with exactly m prime factors is given in the book by Montgomery and Vaughan, Multiplicative Number Theory, but I do not have access to it and do not know whether it also addresses the problem of three consecutive numbers of this kind.

Examples

			33 = 3*11, 34 = 2*17, 35 = 5*7;
1309 = 7*11*17, 1310 = 2*5*131, 1311 = 3*19*23;
203433 = 3*19*43*83, 203434 = 2*7*11*1321, 203435 = 5*23*29*61;
16467033 = 3*11*17*149*197, 16467034 = 2*19*23*83*227, 16467035 = 5*13*37*41*167; (CPU time 48 seconds)
1990586013 = 3*13*29*67*109*241, 1990586014 = 2*23*37*43*59*461, 1990586015 = 5*11*17*19*89*1259. (CPU time 2 hours and 34 minutes)
		

References

• Hugh L. Montgomery and Robert C. Vaughan: "Multiplicative Number Theory: 1. Classical Theory", Cambridge studies in advanced mathematics, vol. 97, Cambridge University Press (2007)

Links

• Math Overflow, Asymptotics of special square-free numbers, Mar 20 2014
• Daniel Constantin Mayer, PARI/GP script "AdjacentSquareFree.gp"
• Daniel Constantin Mayer, PARI/GP script "SquareFreeTriplets.gp"
• Prime puzzles & problems connection, Square-free triples

Crossrefs

Cf. A007675 (any m), A039833 (m=2), A066509 (m=3), A176167 (m=4), A192203 (m=5), A068088 (sextets with gap).

Cf. A242605-A242608 for start of triples of consecutive squarefree numbers with m=2,...,5 prime factors, A242621 for the analog of the present sequence in that spirit.

Programs

PARI

{default(primelimit,2M); lb=2; ub=2*10^9; m=1; i=0; j=0; loc=0; while(m<6, m=m+1; for(n=lb,ub, if(issquarefree(n)&&(m==omega(n)), loc=loc+1; if(1==loc, i=n; ); if(2==loc, if(i+1==n, j=n; ); if(i+1
				

Formula

a(n) = A093550(n)-1. - M. F. Hasler, May 20 2014