**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) = a_{1}+a_{2}+
...a_{s}, whenever the decomposition
of m, into distinct prime factors, is
m = p_{1}^a_{1}.p_{2}^a_{2}
... p_{s}^a_{s}.

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.