# prime gaps, sort of!

Consider the odd numbers: 1, 3, 5, ...

If we delete every multiple of 3, the largest gap between the remaining odd numbers is 4.

If we delete all multiples of 3 and 5, we get {1, 7, 11, 13, 17, 19, 23, 29...} The largest gap is 6.

If we delete multiples of all primes up to the nth prime starting from 3, can we tell what the largest gap (lg) will be between the remaining odd numbers? My hunch is that $$2p_{n-1}$$ <= lg < $$2p_n$$ where $$p_n$$ is the nth prime.

Can this be solved in a simple way? Is this a known result?

Thanks,

mmk.

• Your conjecture can be restated as follows: If $p$ is prime, then for every $k$, the set $\{k,k+1,k+2,\ldots,k+2p\}$ contains a number relatively prime to $p!$. (You can actually drop the assumption that $p$ is prime when stating it this way. I.e., For every $n$ and $k$, the set $\{k,k+1,k+2,\ldots,k+2n\}$ contains a number relatively prime to $n!$.) – Barry Cipra Dec 19 '18 at 15:58
• My previous comment was posted before the OP added $2p_{n-1}\le$ lg to the conjecture. – Barry Cipra Dec 19 '18 at 16:59
• I believe youtu.be/pp06oGD4m00 this video is relevant. – Zachary Hunter Dec 31 '18 at 7:05

Let $$p_{n}$$ be the $$n$$-th prime. If you delete all odd numbers which have a factor in $$\{3,5,\ldots,p_{n}\}$$ then the least number remaining will be the $$(n+1)$$-th prime $$p_{n+1}$$. Hence the gap will be $$p_{n+1}-1$$.
When you remove multiples of 3 the first number remaining is 5 and the gap is $$5-1=4$$. Similarly when you also remove multiples of $$5$$ the first number remaining is $$7$$ and the gap is $$7-1=6$$.