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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?

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See my answer here: math.stackexchange.com/questions/107329/… –  Byron Schmuland Aug 30 '12 at 18:52
    
To simplify the computations, one could include every $k\geqslant0$ in the sum in the LHS since the additional terms do not contribute to the coefficient of $x^n$ and note that the result is $1/(1-z)$ with $z=x+\cdots+x^6$. –  Did Aug 30 '12 at 18:52
    
@ByronSchmuland It seems that here the number of throws is not specified. Unlike in the answer you indicate, or am I missing something? –  Did Aug 30 '12 at 18:54
    
@did You are quite right, I simply missed that point. I'll leave my comment up anyways. Perhaps it will prove useful. –  Byron Schmuland Aug 30 '12 at 18:56

1 Answer 1

up vote 1 down vote accepted

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).

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