Prove that the coefficients of F(x) are positive integers

Question

Consider the formal power series $$F(x) = \sum_{k\ge0}(x+x^2-x^3)^k$$ How can I show that the coefficients of F(x) are positive integers?

I wrote the F as $F(x) = 1/(1-x-x^2 +x^3) = 1/((x-1)^2(x+1))$

But I don't know what to do next.

Answer 1

From $$ \sum_{n=0}^{\infty} a_{n} \, t^n = \frac{1}{1-t-t^2+t^3}$$ the difference equation $$ a_{n} = a_{n-1} + a_{n-2} - a_{n-3}$$ is obtained with the initial conditions $a_{0} = a_{1} = 1, a_{2} = 2$. Making use of this leads to one way of showing all the coefficients are positive.

By factoring the generating function and then using it in the form \begin{align} \frac{1}{1-t-t^2+t^3} &= \frac{1}{(1+t)(1-t)^2} = \frac{1}{4} \, \left( \frac{1}{1+t} + \frac{1}{1-t} + \frac{2}{(1-t)^2} \right) \\ &= \frac{1}{4} \, \sum_{n=0}^{\infty}\left( (-1)^n + 1 + 2 \, (n+1)\right)t^n\\ &= \sum_{n=0}^{\infty}\frac{2 n + 3 + (-1)^n}4\,t^n \end{align} then it can be seen that $$a_{n} = \frac{2 n + 3 + (-1)^n}{4} $$ which shows all the coefficients are positive.

Another way is to show that $$ a_{n} =\left\lfloor{ \frac{n+2}{2}}\right\rfloor. $$

Answer 2

Following great hint of @Mike Earnest for $a\ge b \ge 0 $ we have the expression with a positive expansion

$$\frac{1}{(1-x)^a (1+x)^b} = \frac{1}{(1-x)^{a-b}\cdot (1- x^2)^b}$$