OFFSET
0,3
COMMENTS
For every n, there are trivial solutions where an entire row is filled with 1's and an entire column is filled with 1's, and the column index is equal to the row index. This easily follows from the nature of matrix multiplication. Every matrix that has at least one of these row/column pairs along with any other 1's is also a solution because there are no negative numbers involved here. The number of trivial solutions is given by A307248.
LINKS
Wikipedia, Logical matrix.
EXAMPLE
For n = 2, the a(2) = 3 solutions are
1 1 0 1 1 1
1 0 1 1 1 1
MATHEMATICA
a[n_] := Module[{b, iter, cnt = 0}, iter = Sequence @@ Table[{b[k], 0, 1}, {k, 1, n^2}]; Do[If[FreeQ[MatrixPower[Partition[Array[b, n^2], n], 2], 0], cnt++], iter // Evaluate]; cnt]; a[0] = 1;
Do[Print[a[n]], {n, 0, 5}] (* Jean-François Alcover, Jun 23 2019 *)
PROG
(MATLAB)
%Exhaustively searches all matrices
%from n = 1 to 5
result = zeros(1, 5);
for n = 1:5
for m = 0:2^(n^2)-1
p = fliplr(dec2bin(m, n^2) - '0');
M = reshape(p, [n n]);
D = M^2;
if(isempty(find(D==0, 1)))
result(n) = result(n) + 1;
end
end
end
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Christopher Cormier, Mar 29 2019
EXTENSIONS
a(6) from Giovanni Resta, May 29 2019
STATUS
approved