find a group of lowest N numbers so that no 2 pairs have the same bitwise or I am trying to find the lowest group of N numbers (i.e. N=1000)
so that no 2 pairs from the group have the same bit-wise or.
more specific need to find a group
$A = \{a_1,a_2,a_3,..,a_N\} $
such that  for any $ 1\le i,j,k,m \le n $ 
$ i \ne j , k \ne m $
$OR(a_i,a_j) \ne OR(a_k,a_m)$
$a_1 .. a_n$ lowest as possable
example for N=3 
A = {0,1,2}
as Or(0,1) = 1 , Or(0,2) = 2 , Or(1,2) = 3
 A: The question "Can we find $1000$ numbers in the range $[0,2^{n}-1]$?" can be reworded as "Can we find $1000$ subsets of $\{1,\dots, n\}$ such that 


*

*There do not exist distinct $A,B,C,D$ with $A \cup B=C \cup D$

*There do not exist distinct $A,B,C$ with $A \cup B=A \cup C$


In extremal set theory, such a collection of subsets is called a strongly union free family.  Coppersmith and Shearer have some asymptotic constructions of fairly large (roughly $2^{0.31n}$ subsets) collections satisfying this property.  I'm not too familiar with the details of the constructions, but it might be possible to extract something in the non-asymptotic regime from them.  
A: A rewritten start:  If you continue greedily, you will get the set $2^k$.  Given that you have $0$ and $1$, you cannot have any number $n$ with a $1$ in the $2^0$ position because $n \text{ or } 0 = n \text{ or } 1$.  Similarly, you cannot have a number with a $1$ in the $2^1$ position.  $3$ violates this, but $4$ is acceptable.  Now you can't have any other number with a $1$ in the $2^2$ position.  
It would be better (get you a smaller top number) to have each number have more bits.  Given $N$ numbers, you have $\frac 12N(N-1)$ pairs that should have different ORs.  You could let each have $ b$ one bits and $c$ zero bits.  You need at least ${b+c \choose b}=N$.  Then if all the ORs are different you are home.  The greedy strategy above is $b=1,c=N$  There must be a nice pattern of numbers with $b$ bits that guarantees no collisions, but I haven't found it. I practice, I think I would choose $b$ and $c$, let all the $N$ numbers have $b$ one bits and choose them at random from the ${b+c \choose b}$ and try.  If it fails too many times, increase $c$ and try again.
