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Given $f(x) = x^4\left(\frac{1-x^6}{1-x}\right)^4 = (x+x^2+x^3+x^4+x^5+x^6)^4$. This is the generating function $f(x)$ of $a_n$, which is the number of ways to get $n$ as the sum of the upper faces of four thrown dice.

How do I calculate a coefficient from said generating function, for example $a_{17}$?

I could of course write it all out, but that would take a lot of time. Is there a quicker way to do it?

I've rewritten the formula as $x^4(1-x^6)^4\left(\frac1{1-x}\right)^4$, giving me to find the coefficient of $x^{17-4}=x^{13}$ of $(1-x^6)^4\left(\frac1{1-x}\right)^4$. I'm, however, stuck there.

RBS
  • 841
  • A friend just helped me. You can rewrite $(1-x^6)^4$ in terms of $1 - \binom41 x + ...$ and $\left(\frac{1}{1-x}\right)^4$ like $\sum\limits_{n=0}^\infty \binom{n+4-1}{n}$. This gives the answer $104$. – RBS Mar 31 '14 at 17:45

2 Answers2

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You want: \begin{align} [z^{17}] z^4 (1 - z^6)^4 (1 - z)^{-4} &= [z^{13}] (1 - 4 z^6 + 6 z^{12} - 4 z^{18} + z^{24}) \cdot \sum_{k \ge 0} \binom{-4}{k} (-1)^k z^k \\ &= [z^{13}] (1 - 4 z^6 + 6 z^{12}) \cdot \sum_{k \ge 0} \binom{k + 3}{3} z^k \\ &= \binom{16}{3} - 4 \cdot \binom{10}{3} + 6 \cdot \binom{4}{3} \\ &= 104 \end{align}

vonbrand
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$$f(x) = x^4\left(\frac{1-x^6}{1-x}\right)^4 = (x+x^2+x^3+x^4+x^5+x^6)^4=x^4(1+x+^2+x^3+x^4+x^5)^4=\\ =x^4\left(\sum_{k=0}^{5}x^k\right)^4=x^4\left(\frac{1-x^6}{1-x}\right)^4=x^4(1-x^6)^4(1-x)^{-4}=\\=x^4\sum_{j=0}^{4}(-1)^j\binom{4}{j}x^{6j}\sum_{h=0}^{\infty}(-1)^h\binom{-4}{h}x^h$$

Note that $(-n)!=(-n)(-n-1)(-n-2)\cdots=\infty$, and $\binom{-n}{k}=\frac{(-n)!}{k!(-n-k)!}=(-1)^k\frac{n^{(k)}}{k!}=(-1)^k\frac{(n+k-1)_k}{k!}=(-1)^k\binom{n+k-1}{k}$ so

$$f(x)=x^4\sum_{j=0}^{4}(-1)^j\binom{4}{j}x^{6j}\sum_{h=0}^{\infty}\binom{h+4-1}{h}x^h $$

Any exponent S of the polynomial will come from the sum $S=4+6j+h \rightarrow h=S-6j-4$ and it coefficient will be

$$c(S)=\sum_{j=0}^{4}(-1)^j\binom{4}{j}\binom{S-6j-1}{S-6j-4}$$

where$\binom{S-6j-1}{3}\neq 0$ iff $S-6j-1\geq 3$ i.e. $j\leq \frac{S-4}{6}$. So

$$c(S)=\sum_{j=0}^{\lfloor\frac{S-4}{6}\rfloor}(-1)^j\binom{4}{j}\binom{S-6j-1}{3};\ S\in [4,24]_{\Bbb N}$$

Masacroso
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