<s>[INST] Calculating $ \sum_{n \ge 3|m|}^{} (-1)^n q^{(2n + 1)^2 - 32m^2} $. $ \sum_{n \ge 3|m|}^{} (-1)^n q^{(2n + 1)^2 - 32m^2} = η(8t)η(16t)$.
(Kac-Peterson)
I try to confirm the formula, however I don't know the left.
(The right is $q -q^9 -2q^{17} + q^{25} + 2q^{41} + \cdots $)
Define $a(k)$ as the following.

$ \sum_{n \ge 3|m|}^{} (-1)^n q^{(2n + 1)^2 - 32m^2} = \sum a(k) q^{k}.$

For example,
$(2n + 1)^2 - 32m^2 = 1$ and $n \ge 3|m|⇔ (n, m) = (0, 0)$. So $a(1) = (-1)^0 = 1.$
How do we calculate $a(9), a(17), \cdots$?
 [/INST] $(2n + 1)^2 - 32m^2 = 9$ and $n \ge 3|m|⇔ (n, m) = (1, 0)$.
So $a(9) = (-1)^1 = -1.$
$(2n + 1)^2 - 32m^2 = 17$ and $n \ge 3|m|⇔ (n, m) = (3, -1), (3, 1)$.
So $a(17) = (-1)^3 + (-1)^3 = -2.$
The other $a(k)$ is calculated by the same way.
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