OFFSET
0,2
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.
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.
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
EXAMPLE
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.
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}]
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Robert G. Wilson v, Feb 20 2002
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
STATUS
approved