MHB What is $a_{1996}$ in this series?

[anemone]

Here is this week's POTW:

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Let $$\prod_{n=1}^{1996} (1+nx^{3n})=1+a_1x^{k_1}+a_2x^{k_2}+\cdots+a_mx^{k_m}$$ where $a_1,\,a_2,\,\cdots a_m$ are non-zero and $k_1<k_2<\cdots<k_m$.

Find $a_{1996}$.

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[anemone]

No one answered last week's POTW.(Sadface) You can find the suggested solution below.

Spoiler

Note that $k_i$ is the number obtained by writing $i$ in base 2 and reading the result as a number in base 3, and $a_i$ is the sum of the exponents of the powers of 3 used. In particular,

$1996=2^{10}+2^9+2^8++2^7+2^6+2^3+2^2$

So $a_{1996}=10+9+8+7+6+3+2=45$