Each $ k_i$ is of the form $ k_i = 3^{u_1} + \ldots + 3^{u_s}$. This corresponds to the base $ 3$ representation of $ k_i$, which is unique; hence the monomial $ x^{k_i}$ is obtained from a single combination of powers of $ x$ in the product. This implies $ a_i = u_1\cdot u_2\cdots u_{s - 1}\cdot u_s$. Now each $ k_i$ has only digits $ 0$ and $ 1$ in its base $ 3$ representation and $ k_i$ is always a multiple of $ 3$. That's why $ k_{1998}$ is obtained by counting in base $ 3$ the number $ 1998\cdot2$ written in base $ 2$. Since $ (2\cdot1998)_{10} = (111110011100)_2$ we have $ k_{1998} = 3^2 + 3^3 + 3^4 + 3^7 + 3^8 + 3^9 + 3^{10} + 3^{11}$. Then $ a_{1998} = 2\cdot3\cdot4\cdot7\cdot8\cdot9\cdot10\cdot11 = 1330560$. Hope the argument is correct, and if it is, that I didn't do computational mistakes. ____ Edited after Rust's post below: sorry, the whole time I was reading $ a_{1998}$ instead of $ a_{1996}$!

But needed $ a_{1996}$, because $ 2*1996=2^{12}-2^7+24=2^{11}+2^{10}+2^9+2^8+2^7+2^4+2^3$ we get $ a_{1996}=3*4*7*8*9*10*11=665280.$

Quote: hat's why $ k_{1998}$ is obtained by counting in base $ 3$ the number $ 1998\cdot2$ written in base $ 2$. I don't understand why to find $k_i$ I have to write $2i$ in base 2 and counting in base 3.