If the gamma function is given by
$$\Gamma(\alpha) = \int_0^{+\infty}t^{\alpha-1}e^{-t}\text dt$$
and the lower incomplete gamma function by
$$\gamma(\alpha,x) = \int_{0}^{x}t^{\alpha-1}e^{-t}\text dt$$
Is it possible to derive $\gamma(2\alpha,x)$ from $\gamma(\alpha,x)$, $\Gamma(\alpha)$, and $\Gamma(2\alpha)$? The reason why I believe a relationship like that must hold is that, when adding two random variables distributed according to a truncated $\Gamma(\alpha,\beta)$ distribution with support on $[0,w)$
$$f(x;\alpha,\beta)=\mathbb{1}_{0\leq x < w}\frac{\beta^\alpha x^{\alpha-1} e^{-x\beta}}{\gamma(\alpha,w\beta)}$$
I get a term that goes like
$$\mathbb{1}_{0\leq x < w}\frac{\beta^{2\alpha} x^{2\alpha-1} e^{-x\beta}}{\gamma(\alpha,w\beta)^2} \frac{\Gamma(\alpha)^2}{\Gamma(2\alpha)}$$
which looks a lot like the truncated gamma function with parametre $2\alpha$
$$\mathbb{1}_{0\leq x < w}\frac{\beta^{2\alpha} x^{2\alpha-1} e^{-x\beta}}{\gamma(2\alpha,w\beta)}$$