On the decomposition of $1$ as the sum of Egyptian fractions with odd denominators – Part II

computer-assisted-proofsegyptian-fractionselementary-number-theoryexamples-counterexamplesnumber theory

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. (See: e.g. here).

As the fraction $1/3$ can be further split as
$$\frac{1}{3} = \frac{1}{5}+\frac{1}{9}+\frac{1}{45},$$
here is my further question, continuing from this earlier one:

Questions

Must the fraction
$$\frac{1}{3e}$$
(for some odd $e \geq 1$) appear in each such decomposition? Is it possible to prove this? Or is there a counterexample?

Finally, note that is known that:

A general solution (to On A Splitting Equation of an Egyptian fraction to Egyptian fractions such that all produced fractions have odd denominators.) is for every positive positive integer $\ n\ $ :

  • If $n$ is odd , then $$\frac{1}{3n+2}+\frac{1}{6n+3}+\frac{1}{18n^2+21n+6}=\frac{1}{2n+1}$$ is a solution with odd denominators

  • If $n$ is even , then $$\frac{1}{3n+3}+\frac{1}{6n+3}+\frac{1}{6n^2+9n+3}=\frac{1}{2n+1}$$ is a solution with odd denominators

So, for every odd $\ k\ge 3\ $ we can write $\ \frac 1k\ $ with $\ 3\ $ distinct fractions with odd denominators.

Best Answer

It was not easy, but after some effort, I found a vector $v$ with the property that the entries are coprime to $6$ , there is no duplicate and the entries are reasonable small. It has $27$ entries. Here is the output from PARI/GP.

? print(v)
[5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 43, 47, 49, 53, 55, 65, 67, 71, 77, 79, 125345, 88690147, 48015826375, 1719853572313261, 905495183465410533418952375]
? Set(v)==v
%51 = 1
? select(m->gcd(m,6)==1,v)==v
%52 = 1
? sum(j=1,length(v),1/v[j])
%53 = 1
? length(v)
%54 = 27
?

The sum of the reciprocals of the entries of $v$ is in fact $1$.

Update : A somewhat longer vector, but the largest entry is much smaller :

? Set(v)==v
%226 = 1
? select(m->gcd(m,6)==1,v)==v
%227 = 1
? sum(j=1,length(v),1/v[j])
%228 = 1
? length(v)
%229 = 33
? print(v)
[5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 43, 47, 49, 53, 55, 61, 65, 67, 77, 121, 847, 1771, 7381, 7595, 16951, 75509, 424235, 472283, 499187, 681725, 5045971]
? length(digits(v[length(v)]))
%231 = 7
?

Second update :

? Set(v)==v
%481 = 1
? select(m->gcd(m,6)==1,v)==v
%482 = 1
? sum(j=1,length(v),1/v[j])
%483 = 1
? length(v)
%484 = 37
? print(v)
[5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 55, 65, 73, 77, 143, 205, 235, 637, 1001, 1247, 1927, 2021, 2117, 3565, 3689, 4015, 4945, 7975, 8177, 8729, 9367]
?
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