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For $x$ a sum of fractions:

$$ x = \sum_{i=1}^{N}\frac{a_i}{b_i} $$ for all $a_i, b_i \in \mathbb{Z}$ with $ 0 < b_i \leq D$ and $N$ are non-zero positive integers, I know that the denominator of $x$ will not exceed the least common multiple of all positive integers less than or equal to $D$. See OEIS A003418.

So the denominator of $x$ will not grow without bound. My question is, what is the smallest fraction (closest to $0$) that such a sum can produce? Please provide a worked example assuming that $D = 12$. $N$ can be any integer greater than $1$.

Edit

I'd also like to know how to find sequences that produce the smallest value for a given $D$ and how many such sequences there are with minimal $N$ and fractions in lowest terms.

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  • $\begingroup$ Not sure what you mean by "fractions that sum to $0$ are disallowed". For example, with $D=2$ you have $$ \dfrac{1}{2} = \dfrac{1}{1} + \dfrac{-1}{2} = \dfrac{2}{1} + \dfrac{-3}{2} = \dfrac{3}{1} + \dfrac{-5}{2} = \ldots$$. Is anything disallowed? $\endgroup$ Feb 2, 2015 at 5:00
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    $\begingroup$ I've changed that to be "$N$ is minimal" instead to be clearer. $\endgroup$
    – hatch22
    Feb 2, 2015 at 5:04
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    $\begingroup$ OK, so for $D=3$ you have $$\dfrac{1}{6} = \dfrac{1}{2} - \dfrac{1}{3} = \dfrac{3}{2} - \dfrac{4}{3} = \dfrac{5}{2} - \dfrac{7}{3} = \ldots$$ all with $N=2$. $\endgroup$ Feb 2, 2015 at 5:19
  • $\begingroup$ So clearly there are still infinitely many solutions given that the numerators are unbounded. That does give me what I asked for. Is there some way of stating a condition so that the fractions involved are as small as possible? I prefer the $\frac{1}{2} - \frac{1}{3}$ case to the others because they are smallest. $\endgroup$
    – hatch22
    Feb 2, 2015 at 5:22

1 Answer 1

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The LCM of $1, \ldots, 12$ is $27720$. One example with sum $1/27720$ (there are infinitely many others) is $$ \dfrac{3}{7} + \dfrac{1}{8} + \dfrac{ 2}{9} + \dfrac{3}{10} + \dfrac{1}{11} -\dfrac{14}{12}$$

Hint: if $L(d)$ is the LCM of $1, \ldots, d$, then $$ \dfrac{1}{L(d)} = \dfrac{\gcd(d,L(d-1))}{d\; L(d-1)}$$

Now use Bezout's identity.

EDIT: There are two sets of $5$ integers $\le 12$ whose LCM is $27720$: $\{5,7,8,9,11\}$ and $\{7,8,9,10,11\}$. Let's try using five fractions with denominators $5,7,8,9,11$. From $3 \times 7 - 4\times 5 = 1$ we have $$\dfrac{3}{5} + \dfrac{-4}{7} = \dfrac{1}{35}$$ From $-13 \times 8 + 3 \times 35 = 1$ and this we have $$ \dfrac{1}{280} = \dfrac{-13}{35} + \dfrac{3}{8} = \dfrac{-39}{5} + \dfrac{52}{7} + \dfrac{3}{8}$$ From $-31 \times 9 + 1 \times 280 = 1$ and this we have $$ \dfrac{1}{2520} = \dfrac{-31}{280} + \dfrac{1}{9} = \dfrac{1209}{5} + \dfrac{-1612}{7} + \dfrac{ -93}{8} + \dfrac{1}{9}$$ Finally, from $-229 \times 11 + 1 \times 2520 = 1$ and this we have $$ \dfrac{1}{27720} = \dfrac{-276861}{5} + \dfrac{369148}{7} + \dfrac{21297}{8} + \dfrac{-229}{9} + \dfrac{1}{11}$$

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  • $\begingroup$ To clarify, I'd like to know how such sequences are found and if there is a way to know how many others there are. $\endgroup$
    – hatch22
    Feb 2, 2015 at 4:46
  • $\begingroup$ For $D \ge 2$ there are infinitely many solutions, because e.g. you can always add $\dfrac{1}{1} - \dfrac{2}{2}$. $\endgroup$ Feb 2, 2015 at 4:51
  • $\begingroup$ I was ignoring things like adding and subtracting $1$, assuming we are summing different fractions, but I failed to make that clear in the question and will edit it accordingly. Basically I'd like to minimize $N$. Thanks for the hint and illustration! $\endgroup$
    – hatch22
    Feb 2, 2015 at 4:53
  • $\begingroup$ OK I've edited the question so that I'm asking for sequences with fractions in lowest terms and minimal $N$. $\endgroup$
    – hatch22
    Feb 2, 2015 at 5:07

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