<s>[INST] Solutions to equation 
Let $\alpha(n)$ be the number of pairs $(x, y)$ of integers such that $x+y = n, 0 \le y \le x$, and let $\beta(n)$ be the number of triples $(x, y, z)$ such that $x + y + z = n$ and $0 \le z \le y \le x.$ Find a simple relation between $\alpha(n)$ and the integer part of the number $\frac{n+2}{2}$ and the relation among $\beta(n), \beta(n -3)$ and $\alpha(n).$

Attempt:
We see that $\alpha(n) = \dfrac{\binom{n+1}{1}}{2} = \dfrac{n+1}{2}$ if $n$ is odd and $\alpha(n) = \dfrac{\binom{n+1}{1}-1}{2}+1 = \dfrac{n+2}{2}$ if $n$ is even. Thus, $\alpha(n) = \left[\dfrac{n+2}{2}\right]$. 
How do I relate $\beta(n), \beta(n -3)$ and $\alpha(n)$?
 [/INST] The triples that satisfy your condition are of two types. Type 1 are the triples with smallest element $0$, and Type 2 are the ones with smallest element $\ge 1$.
There are just as many triples of Type 1 with sum $n$ as there are pairs $(y,x)$  with $0\le y\le x$ such that $y+x=n$.
There are just as many triples of Type 2 with sum $n$ as there are triples $0\le z'\le y'\le x'$ with sum $n-3$. For let $z'=z-1$, $y'=y-1$, and $x'=x-1$. That gives a bijection between the Type 2 triples with sum $n$ and all good triples with sum $n-3$.
Thus we obtain the recurrence
$$\beta(n)=\alpha(n)+\beta(n-3).$$
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