Writing a product as a factorial

Question

So I'm solving a differential equation with the Frobenius method but I'm stuck with the coefficients. I get that $a_n = \frac{(-2)^na_0}{n!(3n+1)(3n-2)(3n-5)...1}$ but I can't figure out how to write $(3n+1)(3n-2)(3n-5)...1$ as a factorial.

Answer 1

$$(3n+1)(3n-2)(3n-5)...(4)(1)=\prod_{k=0}^n(3k+1)=3^n\frac{\Gamma(n+\frac43)}{\Gamma(\frac43)}$$ $$a_n = \frac{(-2)^na_0}{n!(3n+1)(3n-2)(3n-5)...1}=\frac{(-2)^na_0}{n!3^n\frac{\Gamma(n+\frac43)}{\Gamma(\frac43)}}=\frac{(-\frac23)^n\Gamma(\frac43)a_0}{n!\Gamma(n+\frac43)}$$ I suppose that the function you are looking for is $\quad y(x)=\sum_{n=0}^\infty a_nx^n$ $$y(x) = a_0\sum_{n=0}^\infty \frac{(-1)^n\Gamma(\frac43)}{n!\Gamma\big(n+\frac43\big)}\left(\frac{2x}{3}\right)^n= a_0\:\frac{\Gamma\left(\frac43\right)J_{1/3}\left((\frac{2x}{3})^{1/2}\right)}{(\frac{2x}{3})^{1/6}}$$ $J\:$ is a Bessel function of first kind.