A068069 a(n) is the least k which is the start of n consecutive integers each with a different number, 1 through n, of distinct prime factors.

1, 2, 5, 28, 417, 14322, 461890, 46908264, 7362724275, 4418626443462

Offset: 0

Author

Robert G. Wilson v, Feb 20 2002

Keywords

• nonn
• more

Comments

a(n) >= n!. If the canonical factorization of k is the product of p^e(p) over primes, then the number of distinct number of prime factors is simply the number of p's.

Examples

			a(1) = 2 because 2 has the single prime factor 2; a(2) = 5 because 5 = 5^1 & 6 = 2*3 which have 1 & 2 prime factors respectively; a(3) = 28 because 28 = 2^2*7^1, 29 = 29^1 & 30 = 2*3*5 which have 2, 1 & 3 prime factors respectively; a(4) = 417 because 417 = 3*139, 418 = 2*11*19, 419 = 419^1 & 420 = 2^2*3*5*7 which have 2, 3, 1 & 4 prime factors (distinct) respectively and this represents a record-breaking number.
		

Links

• P. Erdős, Megjegyzések a Matematikai Lapok két problémájához (Remarks on two problems, in Hungarian), Mat. Lapok 11 (1960), pp. 26-32.
• J.-M. De Koninck, J. B. Friedlander, and F. Luca, On strings of consecutive integers with a distinct number of prime factors, Proc. Amer. Math. Soc., 137 (2009), 1585-1592.

Crossrefs

Cf. A067665.

Programs

Mathematica

k = 3; Do[k = k - n; a = Table[ Length[ FactorInteger[i]], {i, k, k + n - 1}]; b = Table[i, {i, 1, n}]; While[ Sort[a] != b, k++; a = Drop[a, 1]; a = Append[a, Length[ FactorInteger[k]]]]; Print[k - n + 1], {n, 1, 7}]

Formula

Koninck, Friedlander, & Luca prove that a(n) > exp(2n + o(n)), but note that an earlier result of Erdős is "essentially equivalent". - Charles R Greathouse IV, Feb 04 2013

Extensions

One more term from Labos Elemer, May 26 2003

One more term from Donovan Johnson, Apr 03 2008

Corrected example and a(9) from Donovan Johnson, Aug 31 2010