$x_2 + x_3 + x_4 + x_5 = n$ where $x_i$ is not divisible by $i$?

Question

How can I find the generating function for the number of solutions of non-negative integers for the equation $x_2$ + $x_3$ + $x_4$ + $x_5 = n$ where $x_i$ is not divisible by $i$?

Attempt:

I know $x_2$ is a variable for odd numbers and its generating function is $\frac{x}{1-x^2}$. But how do I find the other three?

Answer 1

You have to multiply the series

$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 2} } x^i \right)$$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 3} } x^i \right)$$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 4} } x^i \right)$$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 5} } x^i \right)$.

We have

$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 2} } x^i \right) = \left(\sum_{i \geq 0} x^i\right) - \left(\sum_{i \geq 0} x^{2i}\right) = \frac{1}{1-x} - \frac{1}{1-x^2} = \frac{x}{1-x^2}$.

In a similar way, we also have

$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 3} } x^i \right) = \left(\sum_{i \geq 0} x^i\right) - \left(\sum_{i \geq 0} x^{3i}\right) = \frac{1}{1-x} - \frac{1}{1-x^3} = \frac{x + x^2}{1-x^3}$,

$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 4} } x^i \right) = \left(\sum_{i \geq 0} x^i\right) - \left(\sum_{i \geq 0} x^{4i}\right) = \frac{1}{1-x} - \frac{1}{1-x^4} = \frac{x+x^2+x^3}{1-x^4}$,

and

$\left(\sum_{i \geq 0 \\ \text{is not} \\ \text{divisible} \\ \text{by 5} } x^i \right) = \left(\sum_{i \geq 0} x^i\right) - \left(\sum_{i \geq 0} x^{5i}\right) = \frac{1}{1-x} - \frac{1}{1-x^5} = \frac{x +x^2+x^3+x^4}{1-x^5}$.

Multiplying the results and having some simplification yields the desired answer.