We are throwing the die (original cube for the board games). How many are ways to get the sum of the points equal to $n$ ?
I've heard this problem today in the morning and still can't deal with it, which is tiring. The only way I see it, is that I am looking for the number of solutions of equations: $\sum_{i=1}^k x_i = n$ for all possible $k$, where $1\le x_i\le 6$ for all $1\le i\le k$. So if I find the coefficient before $x^n$ in expansion to series this sum: $$\sum_{k=1}^n (x+x^2+x^3+x^4+x^5+x^6)^k=\sum_{k=1}^n\left(\frac{1-x^7}{1-x}\right)^k$$ it will be over. But I completely don't know how to do that. Or maybe there is a simpler solution for this problem?
If the number of ways is $a(n)$ then $$a(n) = a(n-1)+ a(n-2) +a(n-3)+a(n-4)+a(n-5)+a(n-6)$$ starting with $a(0)=1$, and $a(n)=0$ for $-5 \le n \le -1$.
So the generating function is $$\frac{1}{1-x-x^2-x^3-x^4-x^5-x^6}$$ and you want the coefficient of $x^n$.
With an offset this is OEIS A001592 (Hexanacci numbers).