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Ramanujan had an interesting method for coming up with different hypergeometric identities. I'll provide a brief followthrough of how:

Ramanujan's Method of Morley's Identity:

Start with the product of two binomials $\displaystyle(1+u)^{y+n}\left(1+1/u\right)^x$. The coefficient of $u^n$ (denoted as $[u^n]$) is given$$\displaystyle\begin{align*}[u^n](1+u)^{y+n}\left(1+\frac 1u\right)^x & =[u^n]\sum\limits_{k=0}^{\infty}\binom xku^{-k}\sum\limits_{r=0}^{\infty}\binom{y+n}ru^r\\ & =\sum\limits_{k=0}^{\infty}\binom xk[u^{n+k}]\sum\limits_{r=0}^{\infty}\binom{y+n}ru^r\\ & =\sum\limits_{k=0}^{\infty}\binom xk\binom{y+n}{n+k}\end{align*}$$ And we also have the coefficient of $u^n$ from $(1+u)^{x+y+n}u^{-x}$$$\displaystyle\begin{align*}[u^{n+x}](1+u)^{x+y+n} & =[u^{n+x}]\sum\limits_{k=0}^{\infty}\binom{x+y+n}{k}u^k=\binom{x+y+n}{n+x}\end{align*}$$ Since they are equal, the hypergeometric identity follows soon afterward with a bit of manipulation.

Questions:

  1. Since $(1+u)^{y+n}(1+1/u)^x$ gives $_2F_1$, how would you get larger sequences, such as $_5F_4$?
  2. Using that, is it possible to prove$$_5F_4\left[\begin{array}{c c}\frac 12n+1,n,-x,-y,-z\\\frac 12n,x+n+1,y+n+1,z+n+1\end{array}\right]=\frac {\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(x+z+n+1)}$$

I've spent some time, but I have no idea what to do. I believe that the RHS can be represented by$$\binom{x+y+z+n}r$$

For $r=$ something. However, I am neither sure what the LHS is, and what $r$ is.

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  • $\begingroup$ Alternatelly he top formulation can almost certainly be simply calculated using the Riordan transformation. Since I have been working with these I can supply the process but it's the standard one. For instance from BouldBk.pdf and other places. More extensive and less elementary sequences can be done also. Can you fill out "hypergeometric identity follows"; either by stating it or giving a reference? Although I have to say this particular identity is nice :) $\endgroup$ – rrogers Apr 22 '17 at 13:15
  • $\begingroup$ @rrogers The identity is$$_2F_1(-x,-y;n+1)=\frac {\Gamma(n+1)\Gamma(x+y+n+1)}{\Gamma(x+n+1)\Gamma(y+n+1)}$$For $\text{Re }(x+y+n+1)>0$ $\endgroup$ – Crescendo Jul 29 '17 at 17:55
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    $\begingroup$ Good! Nice problem! Now I see what is going on. Basically, you think you can break up $[u^{n-m+n...}] (1+u)^{(a+b+c+...n)}$ into pieces and then set the pieces equal to the coefficient indicated? In combinatorial functionality the only problem with this is called "diagonalization"; that is the "n" occurring in the selector and exponent. This has a standard solution which is sometimes useful, but I am not sure it's even a problem here. BTW: the reference I gave should have been GouldBK.pdf, not BouldBk.pdf. $\endgroup$ – rrogers Jul 31 '17 at 13:30
  • $\begingroup$ @rrogers Great to know! I'll do a bit more reading on this. $\endgroup$ – Crescendo Jul 31 '17 at 15:06
  • $\begingroup$ @rrogers My only problem is that I'm not sure how to get larger functions. The product of $(1+u)$ and $(1+1/u)$ gives the standard hypergeometric function. But how do you get $_3F_2$ or $_5F_4$? $\endgroup$ – Crescendo Jul 31 '17 at 20:08
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The following is all done in the domain of the Method of Coefficients. Some of the indexing restrictions can be alleviated if needed. The results are a little tedious but simple. It's nice to have generalized techniques on hand.
Lemma 1. For $b>a$ and $(b-a)\in\mathbb{Z}$ the generating function for $\left(\begin{array}{c} n+a\cdot k\\ m+b\cdot k \end{array}\right)$ is $\left[t_{0}^{n}\right]\frac{t_{0}^{m}}{\left(1-t_{0}\right)^{m+1}}\cdot\frac{1}{1-\left(\frac{t_{0}^{b-a}}{\left(1-t_{0}\right)^{b}}\right)\cdot s}$
where s is the indexing variable. i.e.
The $\left[t_{0}^{n}\right]$ expression is an operator transforming ${\displaystyle \sum\left(\frac{t_{0}^{i}}{\left(1-t_{0}\right)^{j}}\right)^{k}}s^{k}$ to $ {\displaystyle \sum}\left(\begin{array}{c} n+a\cdot k\\ m+b\cdot k \end{array}\right)s^{k}$

$Proof. \left[t_{0}^{n}\right]\frac{t_{0}^{m}}{\left(1-t_{0}\right)^{m+1}}\cdot\frac{1}{1-\left(\frac{t_{0}^{b-a}}{\left(1-t_{0}\right)^{b}}\right)\cdot s}=\left[t_{0}^{0}\right]\frac{t_{0}^{m-n}}{\left(1-t_{0}\right)^{m+1}}{\displaystyle \sum_{k}}\left(\frac{t_{0}^{b-a}}{\left(1-t_{0}\right)^{b}}\right)^{k}s^{k}$
For a particular k we evaluate the term selected by $\left[t_{0}^{0}\right]$ (the $CT()$ type of term).
$\left[t_{0}^{0}\right]\frac{t_{0}^{m-n}}{\left(1-t_{0}\right)^{m+1}}{\displaystyle \sum_{k}}\left(\frac{t_{0}^{b-a}}{\left(1-t_{0}\right)^{b}}\right)^{k}=\left[t_{0}^{0}\right]\frac{t_{0}^{m-n+\left(b-a\right)\cdot k}}{\left(1-t_{0}\right)^{m+1+b\cdot k}}$
$ =\left[t_{0}^{0}\right]t_{0}^{m-n+\left(b-a\right)\cdot k}{\displaystyle {\displaystyle \sum_{i}\left(\begin{array}{c} m+b\cdot k+i\\ i \end{array}\right)t_{0}^{i}}}$
Setting $m-n+\left(b-a\right)\cdot k+i=0$ gives $i=n-m-\left(b-a\right)\cdot $
$ =\left(\begin{array}{c} m+b\cdot k+n-m-\left(b-a\right)\cdot k\\ n-m-\left(b-a\right)\cdot k \end{array}\right) =\left(\begin{array}{c} n+a\cdot k\\ n-m-\left(b-a\right)\cdot k \end{array}\right)$

$ =\left(\begin{array}{c} n+a\cdot k\\ m+b\cdot k \end{array}\right)$
Lemma 2. ${\displaystyle \sum_{k=0}}{\displaystyle \prod_{i=0}^{j}\left(\begin{array}{c} n_{i}+a_{i}\cdot k\\ m_{i}+b_{i}\cdot k \end{array}\right)s^{k}=\left(\prod_{i=0}^{j}\left(\left[t_{i}^{n_{i}}\right]\frac{t_{i}^{m_{i}}}{\left(1-t_{i}\right)^{m_{i}+1}}\right)\right)\cdot\frac{1}{1-s\cdot{\displaystyle \prod_{i=0}^{j}\left(\left(\frac{t_{i}^{\left(b_{i}-a_{i}\right)}}{\left(1-t_{i}\right)^{b_{i}}}\right)\right)}}}$
Proof. This is just the iterative application of Lemma 1 since the terms are isolated.
Lemma 3. ${\displaystyle \sum_{k=0}}{\displaystyle \prod_{i=0}^{j}\left(\begin{array}{c} n_{i}+a_{i}\cdot k\\ m_{i}+b_{i}\cdot k \end{array}\right)=\left(\prod_{i=0}^{j}\left(\left[t_{i}^{n_{i}}\right]\frac{t_{i}^{m_{i}}}{\left(1-t_{i}\right)^{m_{i}+1}}\right)\right)\cdot\frac{1}{1-{\displaystyle \prod_{i=0}^{j}\left(\left(\frac{t_{i}^{\left(b_{i}-a_{i}\right)}}{\left(1-t_{i}\right)^{b_{i}}}\right)\right)}}}$
Proof. This is just the droping of s , or setting s=1 if you prefer but that has the implication of evaluation. Which is not necessary since the purpose is to generate coefficients not function values.
Example 4. ${\displaystyle \sum_{k}\left(\begin{array}{c} x\\ k \end{array}\right)\left(\begin{array}{c} y+n\\ n+k \end{array}\right)=\left[t_{0}^{x}\right]\left[t_{1}^{y+n}\right]\frac{t_{0}^{0}}{\left(1-t_{0}\right)^{1}}\cdot\frac{t_{1}^{n}}{\left(1-t_{1}\right)^{n+1}}\cdot\frac{1}{1-\frac{t_{0}^{1}}{\left(1-t_{0}\right)^{1}}\cdot\frac{t_{1}^{1}}{\left(1-t_{1}\right)^{1}}}}$
$=\left[t_{0}^{0}\right]\left[t_{1}^{0}\right]\frac{t_{0}^{-x}}{\left(1-t_{0}\right)^{1}}\cdot\frac{t_{1}^{-y}}{\left(1-t_{1}\right)^{n+1}}\cdot\frac{1}{1-\frac{t_{0}^{1}}{\left(1-t_{0}\right)^{1}}\cdot\frac{t_{1}^{1}}{\left(1-t_{1}\right)^{1}}}$
$=\left[t_{0}^{0}\right]\left[t_{1}^{0}\right]\frac{t_{0}^{-x}}{\left(1-t_{0}\right)^{0}}\cdot\frac{t_{1}^{-y}}{\left(1-t_{1}\right)^{n}}\cdot\frac{1}{\left(1-t_{0}-t_{1}\right)}=\left[t_{0}^{0}\right]\left[t_{1}^{0}\right]t_{0}^{-x}\frac{t_{1}^{-y}}{\left(1-t_{1}\right)^{n+1}}{\displaystyle \cdot\frac{1}{\left(1-\frac{t_{0}}{\left(1-t_{1}\right)}\right)}}$
$ =\left[t_{0}^{0}\right]\left[t_{1}^{0}\right]t_{0}^{-x}\frac{t_{1}^{-y}}{\left(1-t_{1}\right)^{n+1}}\cdot{\displaystyle \sum_{i}\left(\frac{t_{0}}{\left(1-t_{1}\right)}\right)^{i}=\left[t_{0}^{0}\right]\left[t_{1}^{0}\right]{\displaystyle \sum_{i}}\frac{t_{0}^{i-x}\cdot t_{1}^{-y}}{\left(1-t_{1}\right)^{n+1-i}}}$
Evaluating the $t_{0}$ term we have i=x
$=\left[t_{1}^{0}\right]{\displaystyle \cdot}\frac{t_{1}^{-y}}{\left(1-t_{1}\right)^{n+1+x}}=\left[t_{1}^{0}\right]\cdot t_{1}^{-y}{\displaystyle \sum_{j}\left(\begin{array}{c} n+x+j\\ j \end{array}\right)^{j}t_{1}^{j}}$
Setting j-y=0 gives j=y
$ \left(\begin{array}{c} n+x+y\\ y \end{array}\right)=\left(\begin{array}{c} n+x+y\\ n+x \end{array}\right)$
Which is your first answer
Remark If wanted I can carry out the three term evaluation in the comments but it gets a little tedious. This was a good problem for me because I have been meaning to organize this for ages.

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