28. (GBR 5) The sequence $\left\{a_{n}\right\}$ of integers is defined by $a_{1}=2, a_{2}=7$, and  $$ -\frac{1}{2}<a_{n+1}-a_{n}-\frac{3}{2}<\frac{1}{2} $$ for all $n \geq 1$.
28. The sequence is uniquely determined by the conditions, and $a_{1}=2, a_{2}=7, a_{3}=25, a_{4}=89, a_{5}=317, \ldots$; it satisfies $a_{n}=3 a_{n-1}+2 a_{n-2}$ for $n=3,4,5$. We show that the sequence $b_{n}$ given by $b_{1}=2, b_{2}=7$, $b_{n}=3 b_{n-1}+2 b_{n-2}$ has the same inequality property, i.e., that $b_{n}=a_{n}$ : $b_{n+1} b_{n-1}-b_{n}^{2}=\left(3 b_{n}+2 b_{n-1}\right) b_{n-1}-b_{n}\left(3 b_{n-1}+2 b_{n-2}\right)=-2\left(b_{n} b_{n-2}-b_{n-1}^{2}\right)$ for $n>2$ gives that $b_{n+1} b_{n-1}-b_{n}^{2}=(-2)^{n-2}$ for all $n \geq 2$. But then  $$ \left|b_{n+1}-\frac{b_{n}^{2}}{b_{n-1}}\right|=\frac{2^{n-2}}{b_{n-1}}2^{n-1}$$ for all $n$. It is obvious that $a_{n}=b_{n}$ are odd for $n>1$.