Peter wrote: An integer sequence $\{a_{n}\}_{n \ge 1}$ is defined by \[a_{1}=2, \; a_{n+1}=\left\lfloor \frac{3}{2}a_{n}\right\rfloor.\] Show that it has infinitely many even and infinitely many odd integers. Case 1: Let us suppose that the set of odd integers in $a_{n}$ is finite. It means that there is an $m$ so that for all $n\geq{m}$ $a_{n}$ is even. Then for all $n\geq{m}$ \[a_{n+1}=\frac{3}{2}a_{n}\] so $ a_{m+k}=(\frac{3}{2})^{k}a_{m}$. But if $r$ is the greatest number so that $2^{r}|a_{m}$ then $a_{m+r+1}$ is not an integer,contradiction! Case 2: Now let us suppose that the set of even numbers in $a_{n}$ is finite. That means we can find an $m$ so that for all $n\geq{m}$ $a_{n}$ is odd, which means that all the numbers $b_{n}=a_{n}-1$ are even integers. Then for all $n\geq{m}$ \[a_{n+1}=\frac{3a_{n}-1}{2}=\frac{3}{2}(a_{n}-1)+1\] and so $ b_{n+1}=\frac{3}{2}b_{n}$,but this case was discussed above.