Find $ a(n) $ in function of n, $ n\geq 2 $ where $$ a(2n+1)=(2n+1)a(n+1) $$ $$ a(2n)=2a(n), $$ and $ a(2)=1, $ $ a(3)=1. $
I have showed that $ a(n)=\frac{na(n+1)a(2n)}{a(2n+1)-a(n+1)} $, but it seems it doesn't help much, can anyone help me with this please?
Hint: Note that if $n$ is odd, then $$\frac{a(n+1)}{a(n)}=\frac{2}{n}$$ and if $n$ is even $$\frac{a(n+1)}{a(n)}=\left(\frac{n+1}{2}\right)\frac{a(n/2+1)}{a(n/2)}$$ now set: $n/2=k$.
Edit: A little more explanation
Set $n=2^{\alpha} m$ where $\alpha\ge 0$ is an integer and $m$ is odd. Hence $\alpha>0$ iff $n$ is even and $\alpha=0,n=m$ iff $n$ is odd. So we can write: $$\frac{a(2^\alpha m+1)}{a(2^\alpha m)}=\left(\frac{2^\alpha m+1}{2}\right)\frac{a(2^{\alpha-1} m+1)}{a(2^{\alpha-1} m)}\tag{*}\label{*}$$ for $\alpha>0$. And: $$\frac{a(n+1)}{a(n)}=\frac{a(m+1)}{a(m)}=\frac{2}{m}$$ for $\alpha=0$. Therefore, you may repeat the $\eqref{*}$ relationship $\alpha$ times to finally get: $$\frac{a(2^\alpha m+1)}{a(2^\alpha m)}=\left(\prod_{i=1}^{\alpha}\frac{2^i m+1}{2}\right)\frac{a(m+1)}{a(m)}=\left(\prod_{i=\color{red}1}^{\alpha}\frac{2^i m+1}{2}\right)\frac{2}{m}$$ Or more generally: $$\frac{a(n+1)}{a(n)}=\left(\prod_{\color{red}{i=0}}^{\alpha}\frac{2^i m+1}{2}\right)\frac{2}{m}\frac{2}{m+1}$$ for all $n$.