

A117969


Start of least run of maximal length of consecutive nalmost primes.


0




OFFSET

1,1


COMMENTS

For n>=2 there cannot be more than 2^n  1 consecutive nalmost 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

Table of n, a(n) for n=1..4.
Tony Forbes, Fifteen consecutive integers with exactly four prime factors, Math. Comp. 71 (2002), 449452. [From T. D. Noe, Oct 29 2008]


EXAMPLE

a(2) = 33 because 33, 34, 35 is the least run of three consecutive 2almost primes (semiprimes).


CROSSREFS

Cf. A067813, A067814, A067820, A067821, A067822.
Sequence in context: A083459 A034173 A132519 * A003820 A112980 A109336
Adjacent sequences: A117966 A117967 A117968 * A117970 A117971 A117972


KEYWORD

hard,nonn


AUTHOR

Rick L. Shepherd, Apr 05 2006


STATUS

approved



