Duplicates in "n × hamming weight of n" sequence Background
The sequence in the title is A245788 &quot;n times the number of 1's in the binary expansion of n&quot; (&quot;times&quot; here means multiplication), which starts like this:
1,   2,   6,   4,   10,  12,  21,  8,   18,  20,
33,  24,  39,  42,  60,  16,  34,  36,  57,  40,
63,  66,  92,  48,  75,  78,  108, 84,  116, 120,
155, 32,  66,  68,  105, 72,  111, 114, 156, 80,
123, 126, 172, 132, 180, 184, 235, 96,  147, 150,
204, 156, 212, 216, 275, 168, 228, 232, 295, 240, ...

Though it is not immediately obvious, this sequence has some duplicates (same number appearing at least twice). The first duplicate appears at \$a(22) = a(33) = 66\$. The next ones are \$a(39) = a(52) = 156\$, and \$a(44) = a(66) = 132\$. Note that the duplicates do not appear in ascending order.
It is easy to prove that there exist infinitely many duplicates in this sequence, and I suspect that there exist infinitely many \$k\$-plicates (the numbers that appear at least \$k\$ times in the sequence) as well. For the record, the first triplicate entry is \$a(1236) = a(1545) = a(2060) = 6180\$.
Challenge
Produce the sequence of unique duplicates in A245788, in ascending order. For example, 156 should appear after 132, and 6180 should appear only once in the sequence. The following is the list of all 124 terms under 10000, generated using this Python program:
66,   132,  156,  228,  264,  300,  312,  420,  435,  456,
528,  588,  600,  624,  804,  840,  870,  912,  1056, 1100,
1164, 1176, 1200, 1220, 1248, 1545, 1572, 1608, 1635, 1680,
1740, 1824, 2050, 2100, 2112, 2196, 2200, 2244, 2316, 2328,
2340, 2352, 2400, 2440, 2496, 2580, 2628, 2820, 3090, 3108,
3140, 3144, 3216, 3270, 3325, 3360, 3480, 3535, 3648, 3690,
4025, 4098, 4100, 4200, 4224, 4260, 4392, 4400, 4488, 4620,
4632, 4656, 4680, 4704, 4800, 4880, 4914, 4992, 5160, 5256,
5640, 5880, 6125, 6180, 6195, 6216, 6280, 6288, 6432, 6510,
6540, 6650, 6660, 6720, 6960, 7070, 7296, 7380, 7608, 8040,
8050, 8196, 8200, 8220, 8292, 8400, 8448, 8490, 8520, 8580,
8784, 8800, 8976, 9228, 9240, 9264, 9312, 9360, 9408, 9600,
9732, 9760, 9828, 9984

Default I/O methods for sequence are allowed:

Take no input, and output the terms infinitely.
Take \$n\$ as input, and output the \$n\$-th term (0- or 1-indexed).
Take \$n\$ as input, and output the first \$n\$ terms.

Standard code-golf rules apply. The shortest code in bytes wins.
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)

Try it online!
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.