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Here is a rewriting of the proof of ${(\frac{n}{\log n})}^3$ lower bound which seems to allow some leeway to play with. Let $p$ be a prime around and less than $\frac{n}{3\log n}$. Then there must be $3\log n$ numbers in the same residue class mod $p$. For three of these, say $ap+r < bp+r < cp+r$, we have $c/a < 2$. The LCM is at least $\frac{a}{c}p^3$ (the case where a fraction of the numbers have r=0 is easy).

The main advantage seems to be the choice of many primes.

Two possibilities:

1. We want our prime to be around $\frac{n}{10}$ (say) rather than $\frac{n}{3\log n}$. but the problem is we cannot claim $>> \log n$ residues for such a prime. Is it feasible to prove that for one of the many primes we have in the range, there must be a residue which appears many times? Looks hard to me though.

2. Instead of looking for the same residues, for a pair $ap+r,bp+s$, we can look to minimize $as-br$ since the gcd of this pair must divide this. So we can look at numbers $a/r$. Also we can set up relations -for example there must be three (many?) numbers $ap+r,bp+s,cp+t$ such that $\frac{a+1}{r}=\frac{b+1}{s}=\frac{c+1}{t}$ (mod p).

We can assume that we have at least $\frac{n}{6}$ of the numbers between $\frac{2n}{3}$ and $\frac{n^2}{4}$.

Let's consider primes between $\frac{n}{2}$ and $\frac{2n}{3}$.

No number gives the same remainder for two different primes and if each remainder occurs at most $k$ times for any prime ($k$ is maximal), then we have at least $\frac{n}{k}$ distinct remainders for any prime and a total of at least $\frac{n^2}{ck\log n}$ different remainders, all less than $\frac{2n}{3}$. ( c is some constant)

This means that $k$ is at least $\frac{n}{d\log n}$ (d is another constant) so there is a prime for which a remainder occurs these many times. This proves the result.

Here is a rewriting of the proof of ${(\frac{n}{\log n})}^3$ lower bound which seems to allow some leeway to play with. Let $p$ be a prime around and less than $\frac{n}{3\log n}$. Then there must be $3\log n$ numbers in the same residue class mod $p$. For three of these, say $ap+r < bp+r < cp+r$, we have $c/a < 2$. The LCM is at least $\frac{a}{c}p^3$ (the case where a fraction of the numbers have r=0 is easy).

The main advantage seems to be the choice of many primes.

Two possibilities:

1. We want our prime to be around $\frac{n}{10}$ (say) rather than $\frac{n}{3\log n}$. but the problem is we cannot claim $>> \log n$ residues for such a prime. Is it feasible to prove that for one of the many primes we have in the range, there must be a residue which appears many times? Looks hard to me though.

2. Instead of looking for the same residues, for a pair $ap+r,bp+s$, we can look to minimize $as-br$ since the gcd of this pair must divide this. So we can look at numbers $a/r$. Also we can set up relations -for example there must be three (many?) numbers $ap+r,bp+s,cp+t$ such that $\frac{a+1}{r}=\frac{b+1}{s}=\frac{c+1}{t}$ (mod p).

We can assume that we have at least $\frac{n}{6}$ of the numbers between $\frac{2n}{3}$ and $\frac{n^2}{4}$.

Let's consider primes between $\frac{n}{2}$ and $\frac{2n}{3}$.

No number gives the same remainder for two different primes and if each remainder occurs at most $k$ times for one any prime ($k$ is maximal), then we have at least $\frac{n}{k}$ distinct remainders for that any prime and a total of around at least $\frac{n^2}{ck\log n}$ different remainders, all less than $\frac{2n}{3}$. ( c is some constant)

This means that $k$ is at most least $\frac{n}{d\log n}$ (d is another constant) so there is a prime for which a remainder occurs these many times. This proves the result.

2 added 662 characters in body; added 10 characters in body; deleted 1 characters in body

Here is a rewriting of the proof of ${(\frac{n}{\log n})}^3$ lower bound which seems to allow some leeway to play with. Let $p$ be a prime around and less than $\frac{n}{3\log n}$. Then there must be $3\log n$ numbers in the same residue class mod $p$. For three of these, say $ap+r < bp+r < cp+r$, we have $c/a < 2$. The LCM is at least $\frac{a}{c}p^3$ (the case where a fraction of the numbers have r=0 is easy).

The main advantage seems to be the choice of many primes.

Two possibilities:

1. We want our prime to be around $\frac{n}{10}$ (say) rather than $\frac{n}{3\log n}$. but the problem is we cannot claim $>> \log n$ residues for such a prime. Is it feasible to prove that for one of the many primes we have in the range, there must be a residue which appears many times? Looks hard to me though.

2. Instead of looking for the same residues, for a pair $ap+r,bp+s$, we can look to minimize $as-br$ since the gcd of this pair must divide this. So we can look at numbers $a/r$. Also we can set up relations -for example there must be three (many?) numbers $ap+r,bp+s,cp+t$ such that $\frac{a+1}{r}=\frac{b+1}{s}=\frac{c+1}{t}$ (mod p).

We can assume that we have at least $\frac{n}{6}$ of the numbers between $\frac{2n}{3}$ and $\frac{n^2}{4}$.
Let's consider primes between $\frac{n}{2}$ and $\frac{2n}{3}$.
No number gives the same remainder for two different primes and if each remainder occurs at most $k$ times for one prime, then we have at least $\frac{n}{k}$ distinct remainders for that prime and a total of around $\frac{n^2}{ck\log n}$ different remainders, all less than $\frac{2n}{3}$. ( c is some constant)
This means that $k$ is at most $\frac{n}{d\log n}$ (d is another constant) so there is a prime for which a remainder occurs these many times. This proves the result.