OFFSET
1,1
COMMENTS
For n>=2 there cannot be more than 2^n - 1 consecutive n-almost primes. Is it known whether there always exists such a run of length 2^n - 1? If not, I conjecture so. This is confirmed to be true for terms through a(4). Terms here equal the last terms of corresponding finite sequences: a(3) = A067813(6). a(4) was computed by Don Reble as A067814(14). a(5) >= A067820(12).
a(4) is smaller than the number 488995430567765317569 found by Forbes. [From T. D. Noe, Oct 29 2008]
LINKS
Tony Forbes, Fifteen consecutive integers with exactly four prime factors, Math. Comp. 71 (2002), 449-452. [From T. D. Noe, Oct 29 2008]
EXAMPLE
a(2) = 33 because 33, 34, 35 is the least run of three consecutive 2-almost primes (semiprimes).
CROSSREFS
KEYWORD
hard,nonn
AUTHOR
Rick L. Shepherd, Apr 05 2006
STATUS
approved