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This is a follow question to the link: On the decomposition of $1$ as the sum of Egyptian fractions with odd denominators - Part II

Suppose we relax the condition that any term can be divisible by 3 with the largest term of the decomposition as $\frac{1}{5}$. What is the length of the optimal decomposition? and what exactly are the odd denominators?

I have manually generated the decomposition below of 1 with 33 terms with odd denominators:

$${5,7,9,11,15,17,21,23,31,33,35,45,47,51,63,69,93,99,105,135,231,347,465,561,693,1035,1041,1683,2079,3105,4371,240471,721413}$$

Is there a lesser length and larger smallest number other than the given above?

With the link above, the length of the decomposition is 27. Since we relax the condition, can there be a decomposition less than 27?

Keneth Adrian Dagal
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1 Answers1

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I managed to find a decomposition of length 19 with largest denominator equal to $209$:

$1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/15 + 1/17 + 1/19 + 1/21 + 1/33 + 1/35\\ + 1/39 + 1/45 + 1/77 + 1/117 + 1/133 + 1/153 + 1/187 + 1/209 = 1$

With length $21$ and largest denominator at most $209$ I found quite a few more ($77$ to be exact), two of which have largest denominator equal to $143$ (which I believe to be optimal):

$1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/15 + 1/21 + 1/23 + 1/33 + 1/35 + 1/39 + 1/45\\ + 1/55 + 1/63 + 1/65 + 1/69 + 1/77 + 1/91 + 1/99 + 1/115 + 1/143 = 1$

$1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/15 + 1/21 + 1/23 + 1/27 + 1/33 + 1/39 + 1/45\\ + 1/55 + 1/65 + 1/69 + 1/77 + 1/91 + 1/99 + 1/115 + 1/135 + 1/143 = 1$

Woett
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