$s\_2(n)$ denotes the sum of the standard base-2 digits of $n$.


For a fixed odd number $k>1$, can we construct $n\in \mathbb{Z}^+$, to make $s\_2(nk)<s\_2(n)$?


To clarify, that's not $s\_2(nk) \lt s\_2(k)$. For example, if $k=7$, we can take $n=23=10111\_2$ and $nk=161=10100001\_2$.



Construction of $n$ makes $s_2(nk)<s_2(n)$
Yes, basically because $1/k$ has infinitely many $1$s in its binary expansion. 


Let $n=\lceil 2^t/k \rceil$ for $t$ large enough, larger than the length of the period of $1/k$ times (the number of digits of $k$ plus $2$). Then $nk$ is just over $2^t$, $nk \lt 2^t+k$, so $s\_2(nk)$ is at most $1$ more than the number of digits of $k$, but $s\_2(n)$ is greater than the number of digits plus $2$. 


