Smallest n for which 3 over n produces 3 Egyptian Fractions using Greedy Algorithm when 2 Sufficient

Theorem

Consider proper fractions of the form $\dfrac 3 n$ expressed in canonical form.

The smallest $n$ for which $S$ consists of $3$ terms, where $2$ would be sufficient, is $25$.

We have that:

$\ds \frac 3 {25}$

$=$

$\ds \frac 1 9 + \frac 2 {225}$

as $\ceiling {25 / 3} = \ceiling {8.333\ldots} = 9$

$\ds $

$=$

$\ds \frac 1 9 + \frac 1 {113} + \frac 1 {25 \, 425}$

as $\ceiling {225 / 2} = \ceiling {112.5} = 113$

But then we have:

$\ds \frac 3 {25}$

$=$

$\ds \frac 6 {50}$

$\ds $

$=$

$\ds \frac 5 {50} + \frac 1 {50}$

$\ds $

$=$

$\ds \frac 1 {10} + \frac 1 {50}$

$n \equiv 1 \pmod 6$

$\exists d: d \divides n$ and $d \equiv 2 \pmod 3$

The first few $n \ge 4$ which satisfies $n \equiv 1 \pmod 6$ are:

$7, 13, 19, 25$

of which $7, 13, 19$ are primes, so they do not have a divisor of the form $d \equiv 2 \pmod 3$.

We see that $5 \divides 25$ and $5 \equiv 2 \pmod 3$.

Hence the result.

$\blacksquare$