View source for 0F1(;r;z)0F1(;r;-z)=0F3(r,r/2,r/2+1/2;-z^2/4)

==Theorem==
The following formula holds:
$${}_0F_1(;r;z){}_0F_1(;r;-z)={}_0F_3\left( ;r, \dfrac{r}{2}, \dfrac{r}{2} + \dfrac{1}{2}; - \dfrac{z^2}{4} \right),$$
where ${}_0F_1$ denotes [[hypergeometric 0F1]] and ${}_0F_3$ denotes [[hypergeometric 0F3]].
==Proof==

==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=0F1(;r;z)0F1(;s;z)=2F1(r/2+s/2, r/2+s/2-1/2;r,s,r+s-1;4z)|next=2F0(a,b;;z)2F0(a,b;;-z)=4F1(a,b,a/2+b/2,a/2+b/2+1/2;a+b;4z^2)}}: $4.2 (3)$ 

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