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This is really a comment, but it's getting a bit long for the comment box.

I want to point out that in addition to what you call the "greedy process," there's another obvious attempt, which could be called the "stingy process."

Let a target $\mu$ be given. There's basically one rule, namely don't squander unused space unless absolutely forced to. More formally, if I have already chosen points $0\le x_1, \ldots , x_N\le a$, then these define $N$ intervals $I_j$. I must now place the next point such that the minimum distance is $\ge\mu/(N+1)$ (I will keep the whole sequence $\ge\mu$, not just the $\liminf$). If I have an interval of length at least twice this distance, then I put my point inside such an interval (let's say inside the smallest one that works and I will also make one distance equal to $\mu/(N+1)$, though that might not be optimal). If not, then I grudgingly add an interval of that length to the right end of the current configuration.

This gives me intervals that get subdivided, and every once in a while a new interval gets added at the current right endpoint. The whole procedure will be a success if the limit of these right endpoints is $\le 1$.

I fooled around some with this; let's take $\mu=1$. Then the first few points are $$ 0, 1/2, 1/2+1/3, 1/4, 1/4+1/3+1/5, 1/2+1/6, \ldots $$ It seems that $\lim a_N =1/2+1/3+1/5+1/7+1/11+1/14+\ldots \simeq 1.4$. If this is correct (but I'm not making any strong claims it is), then it means that we can get $\mu\simeq 1/1.4\simeq 0.7$, which is better than the golden mean.

In any event, it should be easy (for someone more computer savvy) to do the numerics and get a more reliable estimate.

This is really a comment, but it's getting a bit long for the comment box.

I want to point out that in addition to what you call the "greedy process," there's another obvious attempt, which could be called the "stingy process."

Let a target $\mu$ be given. There's basically one rule, namely don't squander unused space unless absolutely forced to. More formally, if I have already chosen points $0\le x_1, \ldots , x_N\le a$, then these define $N$ intervals $I_j$. I must now place the next point such that the minimum distance is $\ge\mu/(N+1)$ (I will keep the whole sequence $\ge\mu$, not just the $\liminf$). If I have an interval of length at least twice this distance, then I put my point inside such an interval (let's say inside the smallest one that works and I will also make one distance equal to $\mu/(N+1)$, though that might not be optimal). If not, then I grudgingly add an interval of that length to the right end of the current configuration.

This gives me intervals that get subdivided, and every once in a while a new interval gets added at the current right endpoint. The whole procedure will be a success if the limit of these right endpoints is $\le 1$.

I fooled around some with this; let's take $\mu=1$. Then the first few points are $$ 0, 1/2, 1/2+1/3, 1/4, 1/2+1/3+1/5, 1/2+1/6, \ldots $$ It seems that $\lim a_N =1/2+1/3+1/5+1/7+1/11+1/14+\ldots \simeq 1.4$. If this is correct (but I'm not making any strong claims it is), then it means that we can get $\mu\simeq 1/1.4\simeq 0.7$, which is better than the golden mean.

In any event, it should be easy (for someone more computer savvy) to do the numerics and get a more reliable estimate.

This is really a comment, but it's getting a bit long for the comment box.

I want to point out that in addition to what you call the "greedy process," there's another obvious attempt, which could be called the "stingy process."

Let a target $\mu$ be given. There's basically one rule, namely don't squander unused space unless absolutely forced to. More formally, if I have already chosen points $0\le x_1, \ldots , x_N\le a$, then these define $N$ intervals $I_j$. I must now place the next point such that the minimum distance is $\ge\mu/(N+1)$ (I will keep the whole sequence $\ge\mu$, not just the $\liminf$). If I have an interval of length at least twice this distance, then I put my point inside such an interval (let's say inside the smallest one that works and I will also make one distance equal to $\mu/(N+1)$, though that might not be optimal). If not, then I grudgingly add an interval of that length to the right end of the current configuration.

This gives me intervals that get subdivided, and every once in a while a new interval gets added at the current right endpoint. The whole procedure will be a success if the limit of these right endpoints is $\le 1$.

I fooled around some with this for $\mu=1/2$. The first few points are $$ 0, 1/4, 1/4+1/6, 1/8, 1/4+1/6+1/10, 1/4+1/12, \ldots $$ It seems that $\lim a_N =1/4+1/6+1/10+1/14+1/22+1/38+\ldots \simeq 0.7$ (though I wouldn't be surprised either if that's wrong). This means that we can afford a bigger $\mu$ than $1/2$, which is not exciting since we already know that we can do better than that.

But one can now check how far one can go with this. It should be easy (for someone more computer savvy) to get a good numerical estimate.

This is really a comment, but it's getting a bit long for the comment box.

I want to point out that in addition to what you call the "greedy process," there's another obvious attempt, which could be called the "stingy process."

Let a target $\mu$ be given. There's basically one rule, namely don't squander unused space unless absolutely forced to. More formally, if I have already chosen points $0\le x_1, \ldots , x_N\le a$, then these define $N$ intervals $I_j$. I must now place the next point such that the minimum distance is $\ge\mu/(N+1)$ (I will keep the whole sequence $\ge\mu$, not just the $\liminf$). If I have an interval of length at least twice this distance, then I put my point inside such an interval (let's say inside the smallest one that works and I will also make one distance equal to $\mu/(N+1)$, though that might not be optimal). If not, then I grudgingly add an interval of that length to the right end of the current configuration.

This gives me intervals that get subdivided, and every once in a while a new interval gets added at the current right endpoint. The whole procedure will be a success if the limit of these right endpoints is $\le 1$.

I fooled around some with this; let's take $\mu=1$. Then the first few points are $$ 0, 1/2, 1/2+1/3, 1/4, 1/4+1/3+1/5, 1/2+1/6, \ldots $$ It seems that $\lim a_N =1/2+1/3+1/5+1/7+1/11+1/14+\ldots \simeq 1.4$. If this is correct (but I'm not making any strong claims it is), then it means that we can get $\mu\simeq 1/1.4\simeq 0.7$, which is better than the golden mean.

In any event, it should be easy (for someone more computer savvy) to do the numerics and get a more reliable estimate.

This is really a comment, but it's getting a bit long for the comment box.

I want to point out that in addition to what you call the "greedy process," there's another obvious attempt, which could be called the "stingy process."

Let a target $\mu$ be given. There's basically one rule, namely don't squander unused space unless absolutely forced to. More formally, if I have already chosen points $0\le x_1, \ldots , x_N\le a$, then these define $N$ intervals $I_j$. I must now place the next point such that the minimum distance is $\ge\mu/(N+1)$ (I will keep the whole sequence $\ge\mu$, not just the $\liminf$). If I have an interval of length at least twice this distance, then I put my point inside such an interval (let's say inside the smallest one that works, though that might not be optimal). If not, then I grudgingly add an interval of that length to the right end of the current configuration.

This gives me intervals that get subdivided, and every once in a while a new interval gets added at the current right endpoint. The whole procedure will be a success if the limit of these right endpoints is $\le 1$.

I fooled around some with this for $\mu=1/2$. The first few points are $$ 0, 1/4, 1/4+1/6, 1/8, 1/4+1/6+1/10, 1/4+1/12, \ldots $$ It seems that $\lim a_N =1/4+1/6+1/10+1/14+1/22+1/38+ \simeq 0.7$ (though I wouldn't be surprised either if that's wrong). This means that we can afford a bigger $\mu$ than $1/2$, which is not exciting since we already know that we can do better than that.

But one can now check how far one can go with this. It should be easy (for someone more computer savvy) to get a good numerical estimate.

This is really a comment, but it's getting a bit long for the comment box.

I want to point out that in addition to what you call the "greedy process," there's another obvious attempt, which could be called the "stingy process."

Let a target $\mu$ be given. There's basically one rule, namely don't squander unused space unless absolutely forced to. More formally, if I have already chosen points $0\le x_1, \ldots , x_N\le a$, then these define $N$ intervals $I_j$. I must now place the next point such that the minimum distance is $\ge\mu/(N+1)$ (I will keep the whole sequence $\ge\mu$, not just the $\liminf$). If I have an interval of length at least twice this distance, then I put my point inside such an interval (let's say inside the smallest one that works and I will also make one distance equal to $\mu/(N+1)$, though that might not be optimal). If not, then I grudgingly add an interval of that length to the right end of the current configuration.

This gives me intervals that get subdivided, and every once in a while a new interval gets added at the current right endpoint. The whole procedure will be a success if the limit of these right endpoints is $\le 1$.

I fooled around some with this for $\mu=1/2$. The first few points are $$ 0, 1/4, 1/4+1/6, 1/8, 1/4+1/6+1/10, 1/4+1/12, \ldots $$ It seems that $\lim a_N =1/4+1/6+1/10+1/14+1/22+1/38+\ldots \simeq 0.7$ (though I wouldn't be surprised either if that's wrong). This means that we can afford a bigger $\mu$ than $1/2$, which is not exciting since we already know that we can do better than that.

But one can now check how far one can go with this. It should be easy (for someone more computer savvy) to get a good numerical estimate.