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$W(n)$ 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?

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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
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@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
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Well, this is completely trivial as @Surb pointed out. – MooS Mar 16 at 8:51
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@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
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@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.

or

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.

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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
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Damn, Option 2 is unproven so easily, I feel bad, I havent seen this :( – MooS Mar 16 at 9:25
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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
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This is easy to answer, since the sequence $1,2, \dotsc, 29$ does it :) – MooS Mar 16 at 9:31
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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 $$\omega(30684)=\omega(30685)=\omega(30686)=\omega(30687)=\omega(30688)=3,$$

and

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

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.

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