Pinko wrote: The sequence $\{x_n\}_{n=0}^\infty$ is defined by the following equations: $x_n=\sqrt{x_{n-1} x_{n-2}+\frac{n}{2}}$ ,$\forall$ $n\geq 2$, $x_0=x_1=1$. Prove that there exist a real number $a$, such that $an<x_n<an+1$ for each natural number $n$. Choosing $a=\frac 1{\sqrt 6}$, rather simple inductions solve the problem : Starting conditions are immediate. About induction steps : 1) If $x_{n-1}>\frac{n-1}{\sqrt 6}$ and $x_{n-2}>\frac{n-2}{\sqrt 6}$ Then $x_{n-1}x_{n-2}+\frac n2>\frac{n-2-3n+2}6+\frac n2=\frac{n^2+2}6>\frac {n^2}6$ And so $x_n>\frac n{\sqrt 6}$ 2) If $x_{n-1}<\frac{n-1}{\sqrt 6}+1$ and $x_{n-2}<\frac{n-2}{\sqrt 6}+1$ Then $x_{n-1}x_{n-2}+\frac n2<\left(\frac{n-1}{\sqrt 6}+1\right)\left(\frac{n-2}{\sqrt 6}+1\right)+\frac n2$ Which is $x_n^2<\frac{n^2}6+\frac{2n}{\sqrt 6}+1+\frac 13-\frac 3{\sqrt 6}$ $<\frac{n^2}6+\frac{2n}{\sqrt 6}+1$ Which is $x_n^2<\left(\frac n{\sqrt 6}+1\right)^2$ And so $x_n<\frac n{\sqrt 6}+1$ Hence the claim