Evaluating Double Sum is there any chance to calculate the double series
$$\sum\limits_{k_1,k_2=0}^\infty \frac{\Gamma(2(k_1+k_2)+a)}{\Gamma(k_1+b)\Gamma(k_2+b)} \frac{x^{k_1}}{k_1!}\frac{y^{k_2}}{k_2!} $$ with $x,y$ sufficiently small and $a,b>0$ or even for some specific values of $a$ and $b$? Or is there a considerable estimation? Maybe even some helpful book/reference. Anything highly appreciated.
Thanks in advance
 A: Proposition:
$$\sum_{k_1,k_2=0}^\infty \frac{\Gamma(2(k_1+k_2+\gamma))}{\Gamma(k_1+\alpha) \Gamma(k_2+\beta)} \frac{x^{k_1}}{k_1!} \frac{y^{k_2}}{k_2!}= \frac{\Gamma(2 \gamma)}{ \Gamma(\alpha) \Gamma(\beta)} F_4 \left(\gamma,\gamma+\frac{1}{2};\alpha,\beta;4x,4y \right),$$
where $F_4$ denotes the fourth Appell function.
Proof:
Lagrange's duplication formula gives
$$\Gamma(2(k_1+k_2+\gamma))=\frac{ 2^{2(k_1+k_2+\gamma)-1}}{\sqrt{\pi}} \Gamma \left(k_1+k_2+\gamma \right) \Gamma \left(k_1+k_2+\gamma+\frac{1}{2} \right).$$
It follows that
$$\frac{\Gamma(2(k_1+k_2+\gamma))}{\Gamma(k_1+\alpha) \Gamma(k_2+\beta)} \frac{x^{k_1}}{k_1!} \frac{y^{k_2}}{k_2!}=\frac{ 2^{2(k_1+k_2+\gamma)-1}}{\sqrt{\pi}} \frac{(\gamma)_{k_1+k_2} \Gamma(\gamma) \left(\gamma+\frac{1}{2} \right)_{k_1+k_2} \Gamma\left( \gamma+{1\over2} \right)}{(\alpha)_{k_1} \Gamma(\alpha) (\beta)_{k_2} \Gamma(\beta)} \frac{x^{k_1}}{k_1!} \frac{y^{k_2}}{k_2!},$$
and summing up we get
$$\sum_{k_1,k_2=0}^\infty \frac{\Gamma(2(k_1+k_2+\gamma))}{\Gamma(k_1+\alpha) \Gamma(k_2+\beta)} \frac{x^{k_1}}{k_1!} \frac{y^{k_2}}{k_2!}=\frac{2^{2\gamma-1}}{\sqrt{\pi}} \frac{\Gamma(\gamma) \Gamma \left( \gamma+\frac{1}{2} \right)}{\Gamma(\alpha) \Gamma(\beta)} \sum_{k_1,k_2=0}^\infty \frac{(\gamma)_{k_1+k_2} \left( \gamma+\frac{1}{2} \right)_{k_1+k_2}}{(\alpha)_{k_1} (\beta)_{k_2}} \frac{(4x)^{k_1}}{k_1!} \frac{(4y)^{k_2}}{k_2!}. $$
Finally, one can use Lagrange's duplication formula in the other direction to get
$$\frac{\Gamma(2 \gamma)}{ \Gamma(\alpha) \Gamma(\beta)} F_4 \left(\gamma,\gamma+\frac{1}{2};\alpha,\beta;4x,4y \right).$$
