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Let each of the positive integers $a_1 ,\dots, a_n$ be less than $m$ such that the least common multiple of any two of the positive integers $a_1 ,\dots, a_n$ is greater than the integer $m$.

Then what is the best possible upper bound for $\sum_{j=1}^n\frac 1{a_j}$?

I'm only aware of the upper bound $2$ but perhaps read somewhere that $6/5$ also serve as a bound (and would therefore like some clarity on what this actually is).

FD_bfa
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Souvik Dey
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    I made a brute force search for $m\le72$ and it seems like the best set is almost always ${1}$. The only exceptions I've found is ${2,3,5}$ for $m=5$, ${3, 4, 5, 7, 11}$ for $m=11$. See https://github.com/didgogns/number_theory/blob/master/211504.java for details. – didgogns Feb 03 '19 at 09:17
  • How could $a_i$ be less than $m$ and the LCM be greater ?! – Bastien Tourand Aug 28 '20 at 13:24
  • It's LCM. Not GCD – ARYAAAAAN Feb 07 '21 at 12:39
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    Hey I've been doing this problem and found two things so far. (I used m isn't equal to any $a_j$ or lcm). 1. The largest number I could find is .969022644023 which is from the set {3,5,7,8,11,13} so then m has to be 14. And 2. 2 is not an element of the set as the largest number a set can make using 2 is 0.84657359028 or $(1 + ln(2))/2$ – Evan May 05 '25 at 23:59
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    NVM just found a better one {3,7,8,10,11,13,17,19} that gives 0.980477752382 – Evan May 08 '25 at 11:28
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    4 is not an element as the best with 4 is 0.933400457696, and I'm starting to think that the smallest element isn't a power of two – Evan May 08 '25 at 12:12
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    OK I just came back, like without doing any research since my last comment, and spent an hour or so making this upper limit $$\frac{1+\left(x-1\right)\ln x}{x}$$ x is the smallest element or largest fraction in the set and this function gives you the largest value to get with x as smallest, so for x=3 i get 1.06574152578 which is pretty close to my 0.980477752382 but for x=5 i get an unrealistic 1.48755032995, this function is just smaller then the natural log so ya not a constant nor an answer to the original problem. the power of 2 thing is ridiculous. – Evan Jul 22 '25 at 00:19
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    I found a pattern that has convinced me that {3,7,8,10,11,13,17,19} gives the largest value for any set with lowest element value 3. that upper bound limit thing above is more like if i wanted an infinite sum of fractions. and the smallest element is most likely odd or prime – Evan Jul 29 '25 at 14:18
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    "I'm only aware of the upper bound 2" ... someone should provide a source for this assertion. – Greg Martin Sep 20 '25 at 06:25

2 Answers2

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This is the subject of Erdos Problems 542. In particular, the sum of reciprocals is always $\leq \frac{31}{30}=1.033\cdots$, as proved by Schinzel and Szekeres in 1959 [SS59].

The example $A=\{2,3,5\}$ shows that $31/30$ is the best possible in general.

In 1996 Chen [C96] improved this upper bound to $\frac{4699}{4620}=1.01709\cdots$ if $n>172509$.

Erdős conjectured (but I believe this is still open) that the upper bound should $\to 1$ as $n\to \infty$.

[SS59]: Schinzel, A. and Szekeres, G., Sur un probléme de M. Paul Erdős. Acta Sci. Math. (Szeged) (1959), 221-229.

[C96]: Chen, Yong-Gao, On a problem of Erdős. Acta Sci. Math. (Szeged) (1996), 101--114.

Thomas Bloom
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It says at https://artofproblemsolving.com/community/c7h2219971p16867627 that with $2$ as the upper bound it's a problem from the German TST of 2003.

At https://artofproblemsolving.com/community/c6h418619 there's a proof (with upper bound $2$):

Notice that for each $k$, there are $\lfloor N/x_k\rfloor$ multiples of $x_k$ not greater than $N$. Now, these multiples are all different because all the least common multiples are greater than $N$. Thus

$$ \sum_{i=1}^n \left \lfloor \frac{N}{x_i} \right \rfloor \leq N $$

Now, $\sum_{i=1}^n \left \{ \frac{N}{x_i} \right \} < N$ (where $\{ \}$ denotes fractional part), because each fractional part is less than $1$ and $n<N$ (we can't have $x_i = x_j$ for $i \neq j$). Adding the two inequalities we get $$ \sum_{i=1}^n \frac{N}{x_i} < 2N $$ and the result follows.

The answer goes on to say,

EDIT: It seems that the result holds with $3/2$ instead of $2$. Notice that each $x_i$ has a multiple $2^\alpha x_i$ such that $\frac N 2 <2^\alpha x_i \leq N$. Those multiples have to be different (for the lcm to be greater than $N$), so we get $n \leq \left \lceil \frac N 2 \right \rceil $. Thus $$ \sum_{i=1}^n \left \{ \frac{N}{x_i} \right \} < \left \lceil \frac N 2 \right \rceil < \frac N 2 + 1 $$ And then for the first inequality we can suppose that $x_i \neq 1$ for all $i$ (if not, the result is clear), and then $$ \sum_{i=1}^n \left \lfloor \frac{N}{x_i} \right \rfloor \leq N - 1 $$ and we finish as before.

A later post to AoPS says, "This problem, with the $\frac {3} {2}$ bound, has just been used at the 2011 Danube Cup," and gives links to two other posts (but those two links are broken, and I don't know how to repair them).

Gerry Myerson
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