While writing an answer to an unrelated question, I stumbled onto a curious fact that I hadn't noticed before: the binary expansion of $x = k(2^n-1)$, for all non-negative integers $n$ and all integers $1 ≤ k ≤ 2^n$, has exactly $n$ bits set (i.e. the Hamming weight of $x$ written in binary is $n$; or, in other words, $x$ is the sum of exactly $n$ distinct non-negative integer powers of two).
For example, for $n = 4$, we have:
$$\begin{array}{rcrcrcl}
1 \times (2^4-1) &=& 15 &=& 0000\,1111_2 &=& 2^3 + 2^2 + 2^1 + 2^0 \\
2 \times (2^4-1) &=& 30 &=& 0001\,1110_2 &=& 2^4 + 2^3 + 2^2 + 2^1 \\
3 \times (2^4-1) &=& 45 &=& 0010\,1101_2 &=& 2^5 + 2^3 + 2^2 + 2^0 \\
4 \times (2^4-1) &=& 60 &=& 0011\,1100_2 &=& 2^5 + 2^4 + 2^3 + 2^2 \\
5 \times (2^4-1) &=& 75 &=& 0100\,1011_2 &=& 2^6 + 2^3 + 2^1 + 2^0 \\
6 \times (2^4-1) &=& 90 &=& 0101\,1010_2 &=& 2^6 + 2^4 + 2^3 + 2^1 \\
7 \times (2^4-1) &=& 105 &=& 0110\,1001_2 &=& 2^6 + 2^5 + 2^3 + 2^0 \\
8 \times (2^4-1) &=& 120 &=& 0111\,1000_2 &=& 2^6 + 2^5 + 2^4 + 2^3 \\
9 \times (2^4-1) &=& 135 &=& 1000\,0111_2 &=& 2^7 + 2^2 + 2^1 + 2^0 \\
10 \times (2^4-1) &=& 150 &=& 1001\,0110_2 &=& 2^7 + 2^4 + 2^2 + 2^1 \\
11 \times (2^4-1) &=& 165 &=& 1010\,0101_2 &=& 2^7 + 2^5 + 2^2 + 2^0 \\
12 \times (2^4-1) &=& 180 &=& 1011\,0100_2 &=& 2^7 + 2^5 + 2^4 + 2^2 \\
13 \times (2^4-1) &=& 195 &=& 1100\,0011_2 &=& 2^7 + 2^6 + 2^1 + 2^0 \\
14 \times (2^4-1) &=& 210 &=& 1101\,0010_2 &=& 2^7 + 2^6 + 2^4 + 2^1 \\
15 \times (2^4-1) &=& 225 &=& 1110\,0001_2 &=& 2^7 + 2^6 + 2^5 + 2^0 \\
16 \times (2^4-1) &=& 240 &=& 1111\,0000_2 &=& 2^7 + 2^6 + 2^5 + 2^4 \\
\end{array}$$
The pattern breaks at $k=17$, where $17 \times 15 = 255 = 1111\,1111_2$. (This is not surprising; indeed, it's easy to see that the binary expansion of $(2^n+1)(2^n-1) = 2^{2n}-1$ will always have $2n > n$ bits set.) But it's remarkable that it does seem to hold for all $1 ≤ k ≤ 2^n$ and for all $n ≥ 0$.
But how to prove that it does?
I tried to look for an existing proof on math.SE, but all I found (perhaps because I was having trouble finding good search keywords) was this Q&A about showing that the binary expansion of $k(2^n-1)$ has at least $n$ bits set for all $k$, so I figured I'd ask a new question (and try to answer it myself).
Best Answer
Note that $k(2^n-1) = k2^n - k = (k - 1)2^n + (2^n - k)$.
Since $1 ≤ k ≤ 2^n$, it follows that $0 ≤ 2^n - k < 2^n$, so the binary expansion of $2^n - k$ is at most $n$ bits long. Meanwhile, the binary expansion of $(k - 1)2^n$ is the same as that of $k-1$, but with $n$ zero bits appended to the right.
Thus, when $k(2^n-1)$ is written out as a $2n$ bit binary number, the leftmost $n$ bits of it will be identical to those of $k - 1$, while the rightmost $n$ bits are identical to those of $2^n - k$.
It remains to be shown that the binary expansions of $k - 1$ and $2^n - k$ together always have exactly $n$ bits set. But that's indeed always the case, since $2^n - k = (2^n - 1) - (k - 1)$ is the $n$-bit ones' complement of $k - 1$, and thus has exactly those bits set in its binary expansion that are not set in the binary expansion of $k - 1$.