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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.

Mr Pie
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1 Answers1

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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.

José Carlos Santos
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  • Wow! How did you find that? $(+1)$ – Mr Pie Sep 07 '20 at 12:32
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    I used Mathematica to write$$\frac{\left(1-\dfrac1{an+b}\right)\left(1-\dfrac1{cn+d}\right)}{\left(1-\dfrac1{an+e}\right)\left(1-\dfrac1{cn+f}\right)}-\frac{\left(1+\dfrac1{an+b}\right)\left(1+\dfrac1{cn+d}\right)}{\left(1+\dfrac1{an+e}\right)\left(1+\dfrac1{cn+f}\right)}$$as a rational function. It's the quocient of two quartic polynomials, whose coefficients are functions of the numbers $a$, $b$, $c$, $d$, $e$, and $f$. Then I asked Mathematica to see for which values of those numbers is the numerator the null polynomial. There are lots of solutions (Mathematica found $82$ solutions!). – José Carlos Santos Sep 07 '20 at 12:53
  • Many of them are uninteresting (what I mean by this is that $a=0$ or $c=0$). Finally, among all those solutions, I searched (without the help of Mathematica) for a solution of which your equality was a particular case. – José Carlos Santos Sep 07 '20 at 12:54
  • @JoséCarlosSantos. Could you please provide the synthax ? I am curious. Thanks and cheers :-) – Claude Leibovici Sep 07 '20 at 14:30
  • Also, it works if $b=\cfrac{a+c+ad}{c}$ and $e=\cfrac{-a+c+ad}{c}$ – Mr Pie Sep 07 '20 at 15:19
  • @ClaudeLeibovici I did it in three steps. First step: to use the Together command to convert the expression from my first comment into a rational fraction. After that, CoefficientList[Numerator[%],n] to get the list of the coefficients of the numerator. Final step: Solve[%=={0,0,0,0,0,0},{a,b,c,d,e,f}]. – José Carlos Santos Sep 07 '20 at 16:28