Is it always possible to find 2^k-1
consecutive integers that have exactly k prime
factors, taking into account their multiplicities?
Let's define
Omega(m) = a1+a2+
...as, whenever the decomposition
of m, into distinct prime factors, is
m = p1^a1.p2^a2
... ps^as.
More precisely, the problem is to
find 2^k-1 consecutive integers (arranged here around their mean n): {n-a,
... ,n-1, n, n+1, ... , n+a}, with
a=2^(k-1)-1, for each of which Omega takes the value
k.
It is easy to see that this is best possible, since
any list of
2^k consecutive integers contains one integer that is
divisible by 2^k. Except for the special case 2^k
(easily dealt with), this particular member of the
list is clearly divisible by some other integer, hence its Omega value is larger than k.
A little more thought leads to the observation that
the central value n
(the mean) of this list of 2^k-1 integers should be
of the form 2^(k-1).p for some
prime p. In this case, we say that p is k-nice.
It is easy to see that if p is
k-nice, it is automatically j-nice for all j less or equal to k. This can be used to look for solutions in small cases.
An ancillary question is to characterize the density distribution of
k-nice primes. Among
the first 10,000 primes, there are 679 primes that are 2-nice, 3 that are 3-nice (the smallest being 52919), and none that are
4-nice.
Apparently, 12190527809 (found by
Brian Trial,
Ferndale, Michigan) is the smallest 4-nice prime.