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.
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 $p_{n+1}-1$.
This can be seen in your example;
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 $7-1=6$.