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.

Let Fibonacci's Greedy Algorithm be used to generate a sequence $S$ of Egyptian fractions for $\dfrac 3 n$.

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

Proof
We have that:

But then we have:

By Condition for 3 over n producing 3 Egyptian Fractions using Greedy Algorithm when 2 Sufficient, we are to find the smallest $n$ such that:
 * $n \equiv 1 \pmod 6$
 * $\exists d: d \divides n$ 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.