The number $144$ can be factored in three different ways such that $$12\times12=9\times16=8\times18$$ with$$12+12=24;\quad 9+16=25; \quad8+18=26$$ the sum of the factor pairs being consecutive numbers, Similarly $180$ also has three such factor pairs, $(12,15), (10,18), (9,20)$.
Is there any algorithm or way to find such numbers which have three-factor pairs? Any help is appreciated.
I have a partial algorithm to give you a start.
You are looking for numbers with factors $a\cdot b = k$ such that $(1-a-b)^2-4ab = a^2+b^2-2(a+ab+b)+1$ is a perfect square. Additionally, you need $a+b-1+\sqrt{a^2+b^2-2(a+ab+b)+1}$ be divisible by 2. Finally, you want $k$ to be divisible by $\dfrac{a+b-1+\sqrt{a^2+b^2-2(a+ab+b)+1}}{2}$. Then, you will have found a pair factors. Extending this to a triple could follow a similar pattern. The pair would be:
Let $c = \dfrac{2k}{a+b-1+\sqrt{a^2+b^2-2(a+ab+b)+1}}$
Then you have: $c+\dfrac{a+b-1+\sqrt{a^2+b^2-2(a+ab+b)+1}}{2}+1 = a+b$ and $c\cdot \dfrac{a+b-1+\sqrt{a^2+b^2-2(a+ab+b)+1}}{2}=ab=k$