Happy New Prime Year! 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.)
 A: As it was pointed out to me by Han Wu, 2010 wasn't that boring from the prime numbers point of view:
2010 = 2*3*5*(7+11+13+17+19)
A: I believe that the question you are asking is still open. It has only (relatively) recently been shown that $s(n)=s(n+1)=A$ has infinitely many solutions for $A\ge 3$. This was shown by Schlage-Puchta in 2003. This article by Goldston, Graham, Pintz, and Yildirim discusses this and related questions:
http://arxiv.org/abs/0803.2636
Remark: your arithmetic function $s(\cdot)$ is usually denoted $\omega(\cdot)$ nowadays, but was denoted $\nu(\cdot)$ by Ramanujan.
A: 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.
A: I decided to share this awk code to compute s(n), primarily because I like that it uses only addition and the distributive law, and not factoring, to compute s(n).  It also uses a bit of string processing and hash-table look up, but is a nice example of the use of associative arrays.  I also like it because it uses $O(\pi(n)\log(n))$ bytes of memory,
essentially one entry per prime number less than n. Apologies to sleepless in beantown:  I prefer obfuscated awk and nice algorithms to one-liners in Perl, so do not accept his challenge made in a comment on his answer.

BEGIN{  LIM = 10000 ; SEP = "," 
prev = count[1] = count[2] = count[3] = SENTINEL = 0
dir[1] = 1 ;        dir[2] = 0 ;        dir[3] = -1
str[1] = " / at " ; str[2] = " = at " ; str[3] = " \\ at "
notify[1] = notify[3] = 3; notify[2] = 6

for( n = 2 ; n < LIM ; n++ ) { # cmp means composite
  if (n in cmp) {  split(cmp[n], fl, SEP) ;  delete cmp[n] }
  else { # n is prime; make up factor list from scratch
     fl[1] = n ; fl[2] = SENTINEL }
  for(f = fl[j=1] ; f != SENTINEL ; f = fl[++j] ) {
     if ((nn = (n+f)) in cmp) cmp[nn] = f SEP cmp[nn]
     else cmp[nn] = f SEP SENTINEL }
  s = j - 1  

for (k in dir) { count[k] = (prev == (s - dir[k]))?(count[k] + 1): 1
    if (count[k] > notify[k]) print count[k] str[k] n ":" s }
  prev = s
} }

Sample output verifies the results of sleepless in beantown, plus shows that there are long runs where s is constant: 2=s(2302)=...=s(2308) .  It suggests that there is a function f(s) such that there are at most f(s) consecutive numbers with value s.  I
suspect f(1)=4, but do not yet have a proof.
Gerhard "Ask Me About System Design" Paseman, 2010.12.29
