IV OBM

An induction shows that $a_k=3^k{2}^{n-k}-1$ for $k\le n-1$. It is certainly true for $k=0$. Suppose it is true for $k<n-1$. Then $3a_k+1=3^{k+1}{2}^{n-k}-2$. Since $n-k>1$, the odd part is $3^{k+1}{2}^{n-(k+1)}-1$, so the result is true for $k+1$. That gets us as far as $a_{n-1}=3^{n-1}2-1$.

Now we want the odd part of $2(3^n-1)$. Certainly $3^n-1$ is even. We have $3^n-1\equiv (-1{)}^n-1\equiv 2(\mathrm{mod}4)$ for $n$ odd, so for $n$ odd it is not divisible by 4. Hence for $n$ odd we have $a_n=\frac{3^{n-1}}{2}$.