On Ramanujan's curious equality for $\sqrt{2\,(1-3^{-2})(1-7^{-2})(1-11^{-2})\cdots} $

Solution 1:

It can be transformed to next equation.

$$\displaystyle m=\frac{n^2}{n^2-1}\frac{a+1}{a-1}\frac{b+1}{b-1}\frac{c+1}{c-1}$$

so, function m decrease monotonously ,when all values are bigger than 2. If whatever m is, this equation has only finite solutions(when all are integer), $2\leq m\leq12(n=2,a=2,b=3,c=5)$.

Solution 2:

I was able to find a general equation for your second relation with the $4n+1$ primes. For the first relation, discovered by Ramanujan, a general formula can be found here where the variable is $a=3$.

Here is my general solution:

$$\sqrt{\cfrac{n}{n-4}\bigg(1-\cfrac{1}{n^2}\bigg)\bigg\{1-\cfrac{1}{(n-3)^2}\bigg\}\bigg\{1-\cfrac{1}{(2n-3)^2}\bigg\}\bigg\{1-\cfrac{1}{(2n+1)^2}\bigg\}}=\bigg(1+\cfrac{1}{n-3}\bigg)\bigg(1+\cfrac{1}{2n-3}\bigg)\bigg(1+\cfrac{1}{2n+1}\bigg)$$ in which the relation you presented is obtained from $n=8$.

Often Ramanujan's formulae are derived from his incredible intuition. Likewise, I used my intuition and let $a=a_0n+a_1$, $b=b_0n+b_1$ and $c=c_0n+c_1$. I then matched these values to your relation to evaluate each variable around $n$ and solving for $m$ through some heavy algebra, lots of things cancelled out and outputted $m=n/(n-4)$.

If you want a nicer result, substitute $n\mapsto n+2$ so $m=(n+2)/(n-2)$, but then your relation would be obtained by letting $n+2=8\Rightarrow n=6$ instead. Also, in this generalisation, $n=3$ would yield prime denominators.

:)

Found some more general solutions through trial and error:

$$\sqrt{\cfrac{3n+2}{3n-4}\bigg(1-\cfrac{1}{n^2}\bigg)\bigg\{1-\cfrac{1}{(2n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(3n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(6n-7)^2}\bigg\}}$$ $$=\bigg(1+\cfrac{1}{2n+1}\bigg)\bigg(1+\cfrac{1}{3n+1}\bigg)\bigg(1+\cfrac{1}{6n-7}\bigg)$$

Even when $m$ is a square number! $\;\style{display: inline-block; transform: rotate(90deg)}{\Rsh}$

$$\bigg(1+\cfrac 1{n-1}\bigg)\sqrt{\bigg(1-\cfrac{1}{n^2}\bigg)\bigg\{1-\cfrac{1}{(2n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(3n-1)^2}\bigg\}\bigg\{1-\cfrac{1}{(6n-5)^2}\bigg\}}$$ $$=\bigg(1+\cfrac{1}{2n+1}\bigg)\bigg(1+\cfrac{1}{3n-1}\bigg)\bigg(1+\cfrac{1}{6n-5}\bigg)$$ and $$\bigg(1+\cfrac 1{n-1}\bigg)\sqrt{\bigg(1-\cfrac{1}{n^2}\bigg)\bigg\{1-\cfrac{1}{(2n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(3n-2)^2}\bigg\}\bigg\{1-\cfrac{1}{(6n-1)^2}\bigg\}}$$ $$=\bigg(1+\cfrac{1}{2n+1}\bigg)\bigg(1+\cfrac{1}{3n-2}\bigg)\bigg(1+\cfrac{1}{6n-1}\bigg)$$

as well as an outlier.

$$\sqrt{3\bigg(3+\frac{4}{n-1}\bigg)\bigg(1-\cfrac{1}{n^2}\bigg)\bigg\{1-\cfrac{1}{(2n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(3n)^2}\bigg\}\bigg\{1-\cfrac{1}{(6n-1)^2}\bigg\}}$$ $$=\bigg(3-\frac 1n\bigg)\bigg(1+\cfrac{1}{2n+1}\bigg)\bigg(1+\cfrac{1}{3n}\bigg)\bigg(1+\cfrac{1}{6n-1}\bigg)$$

And last one, most similar to the main one in the link:

$$\sqrt{\cfrac{n+1}{n-1}\bigg(1-\cfrac{1}{n^2}\bigg)\bigg\{1-\cfrac{1}{(2n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(3n+1)^2}\bigg\}\bigg\{1-\cfrac{1}{(6n+5)^2}\bigg\}}$$ $$=\bigg(1+\cfrac{1}{2n+1}\bigg)\bigg(1+\cfrac{1}{3n+1}\bigg)\bigg(1+\cfrac{1}{6n+5}\bigg)$$

The magical factor common among all of these solutions: $1+\frac 1{2n+1}$