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Suppose that we decompose $1$ as a sum of Egyptian fractions with odd denominators.

I noticed (from a cursory view) that the fraction $$\frac{1}{3}$$ appears in each of such decompositions.

Questions

Must the fraction $1/3$ appear in each such decomposition? Is it possible to prove this? Or is there a counterexample?

Jose Arnaldo Bebita Dris
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  • $1=\frac 12+\frac 14+\frac 15+\frac 1{20}$. – lulu Mar 22 '20 at 19:04
  • Thank you for the comment, @lulu. But perhaps I was not clear enough: I meant that all denominators should be odd. – Jose Arnaldo Bebita Dris Mar 22 '20 at 19:07
  • Ah, you did say that...I'll try for an example with that in mind. – lulu Mar 22 '20 at 19:08
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    Do you have an expression that works using $\frac 13$? If so, perhaps you can use $\frac 13=\frac 15+\frac 19+\frac 1{45}$ to remove the $\frac 13$. Of course the usual rules for Egyptian fractions requires that you avoid using a fraction more than once, so your expression can't have the denominators $5,9,45$ for this particular trick to work. – lulu Mar 22 '20 at 19:15
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    There are similar substitutions that might work, depending on your examples: for instance $\frac 13= \frac 1{11}+\frac 1{13}+\frac 1{15}+\frac 1{17}+\frac 1{25}+\frac 1{91163} + \frac 1{16621293975}$. – lulu Mar 22 '20 at 19:20
  • Thank you very much for your time and attention, @lulu! =) – Jose Arnaldo Bebita Dris Mar 22 '20 at 23:48

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$1 = \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \frac{1}{13} + \frac{1}{15} + \frac{1}{17} + \frac{1}{19} + \frac{1}{21} + \frac{1}{23} + \frac{1}{25} + \frac{1}{27} + \frac{1}{33} + \frac{1}{611} + \frac{1}{265525} + \frac{1}{97544139723} + \frac{1}{8457652617058141652925} $

Robert Israel
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