Vtgyjfy

I stumbled upon this relation while trying to answer this post. I was trying to find a relation between the two generalized hypergeometric functions,

$$A=,_3F_2left(color{blue}{tfrac12,tfrac12},tfrac12;color{red}{tfrac32,tfrac32};color{fuchsia}{tfrac12}right)$$

$$B=,_3F_2left(tfrac32,tfrac32,tfrac32;tfrac52,tfrac52;tfrac12right)$$

It seems,

$$A+tfrac1{18}B = ,_2F_1left(tfrac12,tfrac12;tfrac32;tfrac12right) =frac{pi}{2sqrt2}$$

Note that from a $_3F_2$, the sum reduces to a $_2F_1$, and $tfrac1{18}= color{blue}{tfrac12tfrac12} color{red}{tfrac23tfrac23} color{fuchsia}{tfrac12} $.

Question: In general, let

$$p=q+1\c_n = a_n+1\d_n = b_n+1$$

where $a_n, b_n$ are arbitrary but the last pair must satisty $a_p+1=b_q$. Is it true that,

$$
{}_pF_qleft(left.begin{array}{c} a_1,a_2,dots ,a_p\ b_1,b_2,dots ,b_q end{array}right| zright)+z,frac{a_1a_2dots a_{p-1}}{b_1b_2dots b_q}{}_pF_qleft(left.begin{array}{c} c_1,c_2,dots ,c_p\ d_1,d_2,dots ,d_q end{array}right| zright)\={}_{p-1}F_{q-1}left(left.begin{array}{c} a_1,a_2,dots ,a_{p-1}\ b_1,b_2,dots ,b_{q-1} end{array}right| zright)\
{}
\
$$

(Note: The pair $a_p,b_q$ disappears in the $text{RHS}$.)

We first use the differentiation formula for the generalized hypergeometric function
begin{equation}
frac{a_1a_2dots a_{p}}{b_1b_2dots b_q}{}_pF_qleft(left.begin{array}{c} c_1,c_2,dots ,c_p\ d_1,d_2,dots ,d_q end{array}right| zright)=frac{d}{dz}{}_pF_qleft(left.begin{array}{c} a_1,a_2,dots ,a_p\ b_1,b_2,dots ,b_q end{array}right| zright)
end{equation}
Then, the LHS of the proposed identity can be written as
begin{equation}
_pF_qleft(left.begin{array}{c} a_1,a_2,dots ,a_p\ b_1,b_2,dots ,b_q end{array}right| zright)+z,frac{a_1a_2dots a_{p-1}}{b_1b_2dots b_q}{}_pF_qleft(left.begin{array}{c} c_1,c_2,dots ,c_p\ d_1,d_2,dots ,d_q end{array}right| zright)=left( 1+frac{z}{a_p}frac{d}{dz} right){} _pF_qleft(left.begin{array}{c} a_1,a_2,dots ,a_p\ b_1,b_2,dots ,b_q end{array}right| zright)tag{1}label{eq1}
end{equation}
To differentiate the hypergeometric function, we use the Euler's integral transform
begin{align}
& _pF_qleft(left.begin{array}{c} a_1,a_2,dots ,a_p\ b_1,b_2,dots ,b_q end{array}right| zright)\
&=frac{Gamma(b_q)}{Gamma(a_p)Gamma(b_q-b_p)} int_0^1t^{a_p-1}left( 1-t right)^{b_q-a_p-1}{}_{p-1}F_{q-1}left(left.begin{array}{c} a_1,a_2,dots ,a_{p-1}\ b_1,b_2,dots ,b_{q-1} end{array}right| tright),dt
end{align}
Here $b_q=a_p+1$, then
begin{align}
_pF_qleft(left.begin{array}{c} a_1,a_2,dots ,a_p\ b_1,b_2,dots ,b_q end{array}right| zright)&=
a_p int_0^1t^{a_p-1}{}_{p-1}F_{q-1}left(left.begin{array}{c} a_1,a_2,dots ,a_{p-1}\ b_1,b_2,dots ,b_{q-1} end{array}right| ztright),dt\
&=frac{a_p}{z^{a_p}} int_0^zu^{a_p-1}{}_{p-1}F_{q-1}left(left.begin{array}{c} a_1,a_2,dots ,a_{p-1}\ b_1,b_2,dots ,b_{q-1} end{array}right| uright),du
end{align}
Then
begin{align}
frac{d}{dz}&,{} _pF_qleft(left.begin{array}{c} a_1,a_2,dots ,a_p\ b_1,b_2,dots ,b_q end{array}right| zright)\
&=frac{a_p}{z},{}_{p-1}F_{q-1}left(left.begin{array}{c} a_1,a_2,dots ,a_{p-1}\ b_1,b_2,dots ,b_{q-1} end{array}right| zright)-frac{a_p}{z} ,{}_pF_qleft(left.begin{array}{c} a_1,a_2,dots ,a_p\ b_1,b_2,dots ,b_q end{array}right| zright)
end{align}
Plugging this expression in eq. eqref{eq1} we find theRHS of the proposed identity.