A possible candidate for OEIS

[Dan Dima]

And this new M(n) upper bounds A121231 as:
C(n^2,0)+C(n^2,1)+...+C(n^2,M(n))
M(n) = n^{3/2} + O(n)


On 8/24/06, Dan Dima <dimad72 at gmail.com> wrote:
>
>  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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