1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Robert Israel Feb 2 '15 at 5:00
  • 1
    $\begingroup$ I've changed that to be "$N$ is minimal" instead to be clearer. $\endgroup$ – hatch22 Feb 2 '15 at 5:04
  • 1
    $\begingroup$ 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 $N=2$. $\endgroup$ – Robert Israel Feb 2 '15 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 '15 at 5:22
3
$\begingroup$

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 $$ \dfrac{1}{27720} = \dfrac{-276861}{5} + \dfrac{369148}{7} + \dfrac{21297}{8} + \dfrac{-229}{9} + \dfrac{1}{11}$$

$\endgroup$
  • $\begingroup$ 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 '15 at 4:46
  • $\begingroup$ For $D \ge 2$ there are infinitely many solutions, because e.g. you can always add $\dfrac{1}{1} - \dfrac{2}{2}$. $\endgroup$ – Robert Israel Feb 2 '15 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 '15 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 '15 at 5:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.