Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$W(n)$ is the function that counts number of distinct prime divisors of $n$. I have been able to prove for any $m$ consecutive integers starting with $1+a$ with the condition $a\leq (m^2-4m)/4$ , there exist a number $n$ in that sequence with the property $W(n)\leq 2$.

Is it worth to publishing? Is it some thing new?

share|cite|improve this question
Before publishing anything, you should manage to even state your claim properly. I used to think the same as @vrugtehagel, but now I think, that you mean the following: If $m > 4$, given any sequence of $m$ consecutive integers, there is at least one integer with at most two distinct prime factors within that sequence. – MooS Mar 16 at 8:41
@jack but $\omega(2)=1$ right? So for any $m\geq 0$ the sequence $2,3,\ldots,2+m$ contains at least one number $k$ with $\omega(k)=1$ (take $k=2$). Am I missing something? – Surb Mar 16 at 8:42
Well, this is completely trivial as @Surb pointed out. – MooS Mar 16 at 8:51
@jack "Bertrand said it before, and I'll say it again: there is always a prime between $n$ and $2n$". Bertrand's postulate. – Patrick Stevens Mar 16 at 9:10
@jack what you say is a weaker version of Bertrand's postulate. A lot weaker, in fact, since your bound is larger, and Bertrand found an $n$ with $\omega(n)=1$, while you only have $\omega(n)\leq 2$ – vrugtehagel Mar 16 at 9:17

Short answer: no, don't publish this. If you want to publish anything, you should first make sure you've stated the theorem properly.

As has been discussed in the comments, the theorem was a little unclear. But you've explained what theorem you actually meant, so let's state it once more to avoid any confusion.

Option 1. For any integer $m>4$, there exists a sequence of $m$ consecutive integers such that at least one number in that sequence has at most $2$ distinct prime factors.

This is trivial: I can give you any sequence starting at a prime, for example, $$23,24,\cdots,23+m-1$$ and that is such a sequence (since the first number of the sequence, in this case, $23$, has $\omega(23)=1$).

However, let's state the other two options here.

Option 2. For any integer $m>4$, there exists a sequence of $m$ consecutive integers all having at most $2$ distinct prime factors.


Option 3. For any sequence of $m>4$ consecutive integers, there is a number in that sequence with at most $2$ prime factors.

The third option is disproved by MooS and Patrick Stevens by counterexamples (see MooS's answer or Patrick Stevens' comment).

Option 2 is also disproved by Patrick Stevens, by cleverly noting that any sequence of $30$ consecutive integers contains at least one multiple of $30$, and so at least one number in that sequence has at least $3$ prime factors.

share|cite|improve this answer
Option 2 is false. Let $m$ be $31$. Then there is a multiple of $30$ in every range of $m$ integers. – Patrick Stevens Mar 16 at 9:24
Damn, Option 2 is unproven so easily, I feel bad, I havent seen this :( – MooS Mar 16 at 9:25
Furthermore we cannot weaken Option $2$ by replacing '2 distinct prime factors' by a higher amount - say $n$. Because we can just replace $30$ by the product of the first $n+1$ primes to disprove it. – MooS Mar 16 at 9:28
This is easy to answer, since the sequence $1,2, \dotsc, 29$ does it :) – MooS Mar 16 at 9:31
This thread is a prime example of the old edict, If you want an answer from the internet quickly, post the wrong answer and wait to be corrected. I wonder if this person just wanted to know if his theory was correct. :) – Dan Mar 16 at 17:52

After investigating a big list of sequence A001221, I found $$\omega(30684)=\omega(30685)=\omega(30686)=\omega(30687)=\omega(30688)=3,$$


$$\omega(n)=3 \text{ for } 99843 \leq n \leq 99850,$$

hence Option 3 of the other answer turns out to be false and there is little evidence that increasing $m$ - say $m>8$ - might really help us.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.