6

$w(n)$ denotes the number of distinct prime factors of $n$. I am wondering if any such result is known.

Bobson Dugnutt
  • 10,912
  • 2
    Goldston, Graham, Pintz and Yildirim have shown that for at least one of $k=2,4,6$ the equation $\omega(n)=\omega(n+k)=2$ holds infinitely often. This is a much stronger equality but not as universal in $k$. – Erick Wong Feb 09 '16 at 17:40
  • Interesting, thanks! –  Feb 09 '16 at 17:51
  • 1
    I feel like this shouldn't be true for $k=1$ –  Feb 09 '16 at 17:57
  • 2
    Actually I found a 2011 citation that builds on the prior result to prove this for $k=1$ (if so, then likely it extends to other values of $k$ as well). Will post the details later. – Erick Wong Feb 09 '16 at 18:01
  • That's awesome as my initial question was about k=1 but I didn't want to ask that in case I wouldn't get any answer. –  Feb 09 '16 at 18:47

1 Answers1

2

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 $d(n)$.

Community
  • 1
Erick Wong
  • 25,868
  • 3
  • 43
  • 96
  • Thanks! It appears too complicated for me to understand, could you say what the idea of the proof is? –  Feb 10 '16 at 07:19
  • @user336-iactuallychosethis Maybe you could post a new question which describes the amount of detail you're looking for specifically? – Erick Wong Feb 11 '16 at 01:57
  • never mind, I'll get back to the proof when I know more. Thanks for the answer! –  Feb 11 '16 at 09:58
  • @user336-iactuallychosethis Take a look at the the bottom of Page 4 and the continuation of Page 5. It's a very readable example of how the basic construction works at a very high level of discussion. – Erick Wong Feb 12 '16 at 23:48