Duplicates in "n × hamming weight of n" sequence
JavaScript (V8),  72 71 70  69 bytes
A full program that prints the sequence forever.

for(n=1;c=2;c||print(n))for(d=++n;d--;)c&gt;&gt;=n==(h=n=&gt;n&amp;&amp;h(n&amp;n-1)+d)(d)

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Commented
for(                   // infinite outer loop:
  n = 1;               //   start with n = 1
  c = 2;               //   before each iteration: initialize c to 2
  c || print(n)        //   after each iteration: print n if c = 0
)                      //
  for(                 //   inner loop:
    d = ++n;           //     increment n and start with d = n
    d--;               //     stop when d = 0 / decrement d
  )                    //
    c &gt;&gt;=              //     right-shift d by 1 position if ...
      n == (           //       ... n is equal to ...
        h = n =&gt;       //       ... the result of the recursive function h
          n &amp;&amp;         //       which adds d as many times as there are
          h(n &amp; n - 1) //       bits set in d, using n &amp; n - 1 to clear
          + d          //       the least significant bit at each iteration
      )(d)             //

Note
We decrement \$d\$ right away in the inner loop, which means that we never consider the case where the divisor of \$n\$ is \$n\$ itself. But that would only work if there's a single 1 in the binary representation of \$n\$, which means that \$n\$ is a power of \$2\$ and all its divisors have a single 1 in their binary representation as well. Therefore, \$n=1\times n\$ is the only possible solution and \$n\$ does not belong to the sequence.