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We can see that the generating function for the natural numbers $1,2,3,\cdots$ (which are also the 2-rough numbers) is then \begin{equation}  G_2(z)=\frac{z}{(z-1)^2}=z+2z^2+3z^3+\cdots\\  a_2(n)=n  \end{equation} Then for the 3-rough numbers we have the odd numbers, \begin{equation}  G_3(z)=\frac{z(1+z)}{(z-1)^2} \\  a_3(n) = 2n-1  \end{equation}  we can use the Mathematica expression \begin{equation}  \mathrm{Simplify[InverseZTransform[F[1/z],z,n],Element[n,Integers]&&n>0]}  \end{equation}  to extract the sequence function $a_3(n)$.