What is the coefficient of the $x^{302}$

Question

What's the coefficient of the $x^{302}$ in the following expression:

$$\dfrac{1}{(1-x^2)(1-x^3)(1-x^6)}$$

I don't know where to start. Also x are in the denominator how this expression can have x to power of a positive term?

Answer 1

Let $[x^a](...)$ be the coefficient of $x^a$ in $(...)$.

Notice in the expansion of $\frac{1}{(1-x^2)(1-x^3)(1-x^6)}$, the monomials from $2^{nd}$ and $3^{rd}$ factors in denominator have exponents divisible by $3$ and $302 \equiv 2 \pmod 3$. One must take a $x^2$ from $1^{st}$ factor to get $302$. As a result,

$$ K \stackrel{def}{=} [x^{302}]\frac{1}{(1-x^2)(1-x^3)(1-x^6)} = [x^{300}]\frac{1}{(1-x^2)(1-x^3)(1-x^6)} $$ We can rewrite RHS as $$[x^{300}]\frac{1 + x^2 + x^4}{(1-x^3)(1-x^6)^2}$$ Notice the $x^2$ and $x^4$ in numerator cannot contribute to $x^{300}$ because the exponent in denominator are all divisible by $3$ and $3$ dividies $300$. We have

$$K = [x^{300}]\frac{1}{(1-x^3)(1-x^6)^2} = [y^{100}]\frac{1}{(1-y)(1-y^2)^2} = [y^{100}]\frac{1+y}{(1-y^2)^3} $$ Once again, the $y$ in numerator cannot contribute to $y^{100}$ because the exponents in denominator are all divisible by $2$ and $2$ divides $100$. This implies

$$K = [y^{100}]\frac1{(1-y^2)^3} = [z^{50}]\frac{1}{(1-z)^3}$$

Recall the power series expansion for $|z| < 1$. $$\frac{1}{(1-z)^m} = \sum_{k=0}^\infty \binom{m-1+k}{k} z^k$$

This means the coefficient $K$ we seek equals to $\binom{50+2}{2} = 1326$.

Answer 2

If $|x|<1$, then the sum of a geometric series tells us that $$\displaystyle\frac{1}{1-x}=\displaystyle\sum_{k=0}^{\infty} x^k$$

As $|x|<1$, $|x^2|,|x^3|, |x^6|<1$ and so we may substitute $x$ for $x^2, x^3$ or $x^6$ in the previous expression, obtaining $$\dfrac{1}{(1-x^2)(1-x^3)(1-x^6)}=\dfrac{1}{1-x^2}\cdot\dfrac{1}{1-x^3}\dfrac{1}{1-x^6}=(\displaystyle\sum_{k=0}^{\infty} x^{2k})\cdot(\displaystyle\sum_{k=0}^{\infty} x^{3k})\cdot(\displaystyle\sum_{k=0}^{\infty} x^{6k})$$

The coefficient of $x^{302}$ can be found then either multiplying those series or as the number of non-negative integer solutions $x_1,x_2,x_3$ of $$2x_1+3x_2+6x_3=302$$