View source for 0F1(;r;z)0F1(;s;z)=2F1(r/2+s/2, r/2+s/2-1/2;r,s,r+s-1;4z)

==Theorem==
The following formula holds:
$${}_0F_1(;r;z){}_0F_1(;s;z)={}_2F_1 \left( \dfrac{r}{2} + \dfrac{s}{2}, \dfrac{r}{2}+\dfrac{s}{2}-\dfrac{1}{2};r,s,r+s-1;4z \right),$$
where ${}_0F_1$ denotes [[hypergeometric 0F1]] and ${}_2F_1$ denotes [[hypergeometric 2F1]].

==Proof==

==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=2F1(a,b;a+b+1/2;z)^2=3F2(2a,a+b,2b;a+b+1/2,2a+2b;z)|next=0F1(;r;z)0F1(;r;-z)=0F3(r,r/2,r/2+1/2;-z^2/4)}}: $4.2 (2)$ 

[[Category:Theorem]]
[[Category:Unproven]]