(0,1) matrices with distinct rows and distinct columns

[Max]

In symmetric form the formula looks like

a(n) = Sum(i=0..n) Sum(j=0..n) Stirling1(n,i) * Stirling1(n,j) * 2^(i*j)

BTW, for (0,1) matrices of size n x m, there are exactly
Sum(i=0..n) Sum(j=0..m) Stirling1(n,i) * Stirling1(m,j) * 2^(i*j)
ones with pairwise distinct rows and columns.

Regards,
Max

Vladeta wrote:

>I got that
>
>a(n) = n!*Sum_{k=0..n} Stirling1(n,k)*binomial(2^k,n).
>
>Vladeta J.
>
>___________
>
>
>----- Original Message -----
>From: "Yuval Dekel" <dekelyuval at hotmail.com>
>To: <seqfan at ext.jussieu.fr>
>Sent: Friday, November 07, 2003 5:57 PM
>Subject: (0,1) matrices with distinct rows and distinct columns
>
>
>  
>
>>Hi,
>>The number of n X n (0,1) matrices with distinct rows is in sequence
>>    
>>
>A088229
>  
>
>>and this is trivial .
>>
>>Let me ask if there is a formula for a(n) where:
>>a(n) = number of n X n (0,1) matrices with distinct rows and distinct
>>columns .
>>
>>(or perhaps someone can compute a(n) ) .
>>
>>Thanks,
>>Yuval
>>
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>>    
>>
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