How to get the coefficients of a power series without multiplying

Question

Power series are considered:

$$\begin{align*} A(x) &= 1 + x^{2} + x^{4} + x^{6} + x^{8} + ...\\ B(x) &= 1 + x^{3} + x^{6} + x^{9} + x^{12} + ...\\ C(x) &= 1 + x^{4} + x^{8} + x^{12} + x^{16} + ... \end{align*}$$

Determine the values of $q_{1}$, $q_{2}$, $q_{3}$, $q_{4}$, $q_{5}$, $q_{6}$ in

$$A(x) \cdot B(x) \cdot C(x) = 1 + q_{1}x^{} + q_{2}x^{2} + q_{3}x^{3} + q_{4}x^{4} + ...$$

without doing the product of $A(x)$, $B(x)$, $C(x)$

Answer 1

I have finally realized that the solution of each $ q_n $ is the number of forms that $n$ can be represented by the sum of $2$ or $3$ or $4$.

• For $ q_1 $ is 0 because you can not add one with those numbers.
• For $ q_2 $ is 1. ($2$)
• For $ q_3 $ is 1. ($3$)
• For $ q_4 $ is 2. ($2+2$ or $4$)
• For $ q_5 $ is 1. ($2+3$)
• For $ q_6 $ is 3. ($2+2+2$ or $3+3$ or $4+2$)

Answer 2

That depends on what you mean by not forming the product. Only needing terms up to $q_6$ I would just do a hand count of how many ways there are to add to each number $1$ through $6$ using $2,2,2,3,3,4$, which is not very many. That is equivalent to multiplying, but it looks different.

You could sum the series, saying $A(x)=\frac 1{1-x^2}$, multiply those, and find the expansion in $x$, but that is more work.