TITLE: Happy New Prime Year!
QUESTION [38 upvotes]: It happens that next year 2011 is prime, while outgoing 2010 is
highly composite in the sense that the number of its distinct prime factors
is 4, maximal possible for a year $< 2310$.
Let me denote by $s(n)$ the number of distinct prime factors of $n$
and note that $s(2011)=1$, $s(2012)=2$ and $s(2013)=3$. I wonder
whether there is a rigorous argument or some heuristic considerations
to show that, for each $k\ge1$, there exist (infinitely many integers)
$n$ satisfying $s(n+1)=1$, $s(n+2)=2$, $\dots$, $s(n+k)=k$.
This can be thought as a generalization of the infiniteness of primes
($k=1$), but I ask this question for curiosity only.
Happy New Prime Year 2011! (Please do not count the exclamation mark as factorial.)

REPLY [4 votes]: Here is another pattern I learned from Bharath Kumar Annamaneni in his buzz post.
2011= 157 + 163 + 167 + 173 + 179 + 181 + 191 + 193 + 197 + 199 + 211 .
2011, Being A Prime Number Itself, Is Also A Sum Of 11 Consecutive Prime Numbers .
Wow.