I was playing around with numbers and I appear to have discovered a very fascinating fraction:
For all $n>0$, $$\cfrac{\bigg(1-\cfrac 1{3n+2}\bigg)\bigg(1-\cfrac 1{6n+1}\bigg)}{\bigg(1-\cfrac 1{3n+1}\bigg)\bigg(1-\cfrac 1{6n+5}\bigg)}=\cfrac{\bigg(1+\cfrac 1{3n+2}\bigg)\bigg(1+\cfrac 1{6n+1}\bigg)}{\bigg(1+\cfrac 1{3n+1}\bigg)\bigg(1+\cfrac 1{6n+5}\bigg)}$$
How did I derive this?
I was looking at my answer here and decided to play around with those mixed fractions (by playing around, I mean multiplying some, dividing others, merely because I was bored). It was greatly to my surprise that I found this (by sheer accident too, due to my interest in conjugates).
Anybody know how to derive this rigorously, and if other such fractions exist?
Thanks.
There are many ways of generalizing that equality. For instance, if you take three natural numbers $a$, $c$, and $f$, and you define$$b=\frac{a-c+af}c,\ d=f-2\frac ca\text{, and }e=\frac{af-a-c}c$$then$$(\forall n\in\Bbb N):\frac{\left(1-\dfrac1{an+b}\right)\left(1-\dfrac1{cn+d}\right)}{\left(1-\dfrac1{an+e}\right)\left(1-\dfrac1{cn+f}\right)}=\frac{\left(1+\dfrac1{an+b}\right)\left(1+\dfrac1{cn+d}\right)}{\left(1+\dfrac1{an+e}\right)\left(1+\dfrac1{cn+f}\right)}.$$If $a=3$, $c=6$, and $f=5$, this will give you the equality from your question.