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

==Theorem==
The following formula holds:
$${}_1F_1(a;2a;z){}_1F_1(b;2b;-z)={}_2F_3 \left( \dfrac{a}{2} + \dfrac{b}{2}, \dfrac{a}{2} + \dfrac{b}{2} + \dfrac{1}{2}; a+\dfrac{1}{2}, b+\dfrac{1}{2}, a+b; \dfrac{z^2}{4} \right),$$
where ${}_1F_1$ denotes [[hypergeometric 1F1]] and ${}_2F_3$ denotes [[hypergeometric 2F3]].

==Proof==

==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=1F1(a;r;z)1F1(a;r;-z)=2F3(a,r-a;r,r/2,r/2+1/2;z^2/4)|next=findme}}: $4.2 (6)$ 

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