0,2

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 count of the number of p's.

a(0) = 1 because 1 have no prime factors; a(1) = 2 beacuse 2 have the single prime factor of one; a(2) = 5 because 5 = 5^1 & 6 = 2*3 which have 1 & 2 prime factors respectively; a(3) = 28 because 28 = 2^1*19^1, 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.

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}]

Cf. A067665.

Adjacent sequences: A068066 A068067 A068068 this_sequence A068070 A068071 A068072

Sequence in context: A019043 A009635 A138293 this_sequence A105787 A110497 A000472

more,nonn

Robert G. Wilson v (rgwv(AT)rgwv.com), Feb 20 2002

One more term from Labos E. (labos(AT)ana.sote.hu), May 26 2003

One more term from Donovan Johnson (donovan.johnson(AT)yahoo.com), Apr 03 2008