A possible candidate for OEIS

[Dan Dima]

I think the answer for your question is the following sequence.
In fact I found n x n matrices with M(n) 1's but i did not entirely prove
there are no n x n matrices with more 1's, so this sequence is so far like a
lower bound.
For matrices m^2 x m^2 I found M(m^2) = m^3 as m^2 m x m matrices with one
row of m of 1's and (m-1) rows of m of 0's.
M(m^2) = m^3
...
M(m^2+k) = m^3 + km, 0 <= k <= m.
...
M(m(m+1)) = (m+1)m^2
...
M(m(m+1)+k) = (m+1)m^2 + k(2m+1), 0 <= k <= m+1.
...
M((m+1)^2) = (m+1)^3
1,2,5,8,10,12,17,22,27,30,33,36,43,50,57,64,68,72,76,80,89,98,107,116,125,...
M(n) = n^{3/2} + o(n)



On 8/21/06, Brendan McKay <bdm at cs.anu.edu.au> wrote:
A question: what is the maximum number of 1s in such a matrix?

Brendan.
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