Ralph, I am not sure if you know this trick or not, but if you start with
interger (n) whose binary representation contains exactly k ones then
{ int t,b;
t=n^(n&(n-1));
b=t+n;
n=b|(((b^n)/t)>>2);
}
generates next larger number with exactly k ones. As long as you stay within
int size (in your case with 28 elements you do) it is a decent method.
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