Ordinary Generating Functions 3 In football, a team scores points in the following ways: * two points (safety), * three points (field goal), * six points (touchdown only), * seven points (touchdown plus extra point), * and eight points (touchdown plus two-point conversion). Find a concise Ordinary Generating Function (OGF) of $\left\\{a_k\right\\}_{k\geq 0}$ where $a_k$ is the number of ways a team can score a total of $k$ points.

$a_k$ is the number of solutions of the diophantine equation : $a\times2+b\times3+c\times6+d\times7+e\times8=k$.
Thus its generating function is : $$\sum_{k=0}^\infty a_kt^k=\frac{1}{(1-t^2)(1-t^3)(1-t^6)(1-t^7)(1-t^8)}$$ Indeed : $$\frac{1}{(1-t^2)(1-t^3)(1-t^6)(1-t^7)(1-t^8)}=(\sum_{k=0}^\infty t^{2k})(\sum_{k=0}^\infty t^{3k})(\sum_{k=0}^\infty t^{6k})(\sum_{k=0}^\infty t^{7k})(\sum_{k=0}^\infty t^{8k})\\\=\sum_{k=0}^\infty(\sum_{a2+b3+c6+d7+e8=k\\\a,b,c,d,e\geq0}t^k)=\sum_{d=0}^\infty a_k t^k$$