TITLE: Given $k$, are there infinitely many $n$ so that $w(n) = w(n+k)$?
QUESTION [6 upvotes]: $w(n)$ denotes the number of distinct prime factors of $n$. I am wondering if any such result is known.

REPLY [2 votes]: Yes, this was proven in a stronger form by Goldston, Graham, Pintz and Yıldırım (2011).  Thanks to Gerry Myerson's answer here for the reference:
Daniel A. Goldston, Sidney W. Graham, Janos Pintz and Cem Y. Yıldırım, Small Gaps Between Almost Primes, the Parity Problem, and Some Conjectures of Erdős on Consecutive Integers, Int Math Res Not Volume 2011, Issue 7, Pp. 1439-1450, possibly available at http://imrn.oxfordjournals.org/content/2011/7/1439.short.
The preprint appears to be here: http://arxiv.org/abs/0803.2636. 
In particular, if one combines Theorems 9 and 12, then we have that for any $k \in \mathbb N$ and any prescribed $A\ge 6$, there exist infinitely many $n$ such that
$$\omega(n) = \omega(n+k) = A.$$
They also have similar results for most other "divisor-counting" arithmetic functions such as $\Omega(n)$ and $d(n)$.