Okay, so I am working on a proof of the upper bound of the odd perfect numbers. The following is what I have so far, and now I am stuck.
If $n$ is an odd perfect, it satisfies the following property:
$$\sum\limits_{j=0}^{⌊n/2⌋}\binom{2n-2j}{n}\left(\binom{2n}{j}-\sum_{d|n,d<n}\binom{2d}{j}\right)=J(n)(\sigma(n)-2)$$
Where $J$ is somewhat "well-defined". They are the coefficients of a $2n$ degree polynomial $P$ $$ P_{2n}(z)=\sum\limits_{k=0}^{2n}J(k)z^k=\prod\limits_{i=1}^{2n}(z-r_i);J(0)=1 $$
such that $$ \sum\limits_{i=1}^{2n}(1-\frac{1}{r_i})^{-1}=0$$ $$\sum\limits_{i=1}^{2n}(1-\frac{1}{r_i})=0$$ $$\prod\limits_{i=1}^{2n}(1-\frac{1}{r_i})^{-1}=1$$
Is there a closed form solution to the following expression from above:
$$\sum\limits_{j=0}^{⌊n/2⌋} \binom{2n-2j}{n}\binom{2n}{j}$$
I have researched various papers on binomial sums and identities and cannot seem to find anything that can help besides hypergeometric functions.
It doesn't seem to be in the OEIS. Maple gives me a hypergeometric expression
$$ {2\,n\choose n} {\mbox{$_3$F$_2$}(-2\,n,-n/2,-n/2+1/2;\,-n,-n+1/2;\,-1)} $$