The "skip-pure" numbers
JavaScript (ES7),  59  31 bytes
Returns the \$n\$-th term, 0-indexed. This is a closed-form formula.

n=&gt;n*1.5^Math.log2(n)&amp;n%4==2^!n

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Formula
The only edge case is \$a(0)=1\$. For \$n\ge1\$:
$$a(n)=\cases{\left\lfloor\dfrac{3n}{2}\right\rfloor-1,&amp;\text{if $\lfloor log_2(n)\rfloor$ is odd and $n\equiv 2\pmod 4$} \\
\left\lfloor\dfrac{3n}{2}\right\rfloor,&amp;\text{otherwise}
}
$$