Having some rational number $0<r<1$, such that $r=\frac{b-2}{b}$ and $b$ is odd, it is expected to find some general lower bound in terms of $k$ for the maximum denominator $n_k$ when we decompose $r$ in egyptian fractions in the following way: $$\frac{b-2}{b}=\frac{1}{n_1}+\frac{1}{n_2}+...+\frac{1}{n_k}$$

Concretely, it would be great to show that $$n_k>\frac{kb}{2}$$

Thanks in advance!


As there has been no answer yet, I have checked that if we soften the restriction of the numerator being equal to $b-2$ and we let it be some positive integer $a$, then we can build an egyptian fraction such that $n_k=\frac{b}{p_i}$, being $p_i$ some prime number such that $p_i\mid b$, noticing that $$\frac{p_i+\frac{b}{p_i}}{b}=\frac{1}{\frac{b}{p_i}}+\frac{1}{p_i}$$

For example, setting $b=35$, we could derive that $$\frac{12}{35}=\frac{1}{7}+\frac{1}{5}$$

However, it is impossible to build $\frac{b-2}{b}$ following this method, as $b-2$ is odd and this algorithm yields only even numerators.

  • $\begingroup$ maybe math.stackexchange.com/questions/3581188/… can be helpful $\endgroup$
    – Peanut
    Commented Mar 18, 2020 at 10:10
  • $\begingroup$ Thanks @Dude! But the link suggested does not address the problem of the lower bound; it provides an upper bound. The upper bound problem has been deeply studied by Bleicher and Erdös (1976), and Yokota and Tenenbaum (1988) $\endgroup$ Commented Mar 18, 2020 at 10:54


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