OFFSET
1,2
COMMENTS
Diagonal of A181019.
Three or more "1"s may be adjacent in an L-shape or step shape (cf. bottom of first example) or 2 X 2 square (top right of 2nd example) or similar. One possible (not always optimal) solution is therefore to fill the square with 2 X 2 squares of "1"s, separated by rows of "0"s: this yields the lower bound (n - floor(n/3))^2 = ceiling(2n/3)^2 given in FORMULA. I conjecture that this is optimal for n = 2 (mod 3) and that a(n) ~ (2n/3)^2. For n = 3k, the array can be filled with 2k(2k+1) "1"s by repeating the optimal solution for n = 3 on the diagonal, and filling the rest with 2 X 2 blocks separated by rows of "0"s, cf. the 4th example for 6 X 6. - M. F. Hasler, Jul 17 2015 [Conjecture proved to be wrong, see below. - M. F. Hasler, Jan 19 2016]
74 <= a(12) <= 77. - Manfred Scheucher, Jul 23 2015
You can repeat a 4 X 2 block [1100; 0011] infinitely in both directions and then crop the needed square. That gives ceiling(n^2/2). It eventually surpasses the solutions we've found so far: at 17*17 the pattern above gives 12*12=144 but this one ceiling(17*17/2)=145. The credit for finding this goes to Jaakko Himberg. - Juhani Heino, Aug 11 2015
LINKS
Manfred Scheucher, Python Script
Peter J. Taylor, Java program to compute terms
FORMULA
a(n) >= ceiling(2n/3)^2; a(3k) >= A002943(k) = 2k(2k+1). - M. F. Hasler, Jul 17 2015; revised by Juhani Heino, Aug 11 2015
a(n) >= ceiling(n^2/2). - Juhani Heino, Aug 11 2015
EXAMPLE
Some solutions for 6 X 6:
0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1
1 0 1 0 0 1 1 0 1 0 1 1 1 0 1 0 0 1 1 0 1 0 1 1
1 1 0 0 1 0 1 1 0 0 0 0 1 1 0 0 1 0 1 1 0 0 0 0
0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 1 1
1 0 1 1 0 1 1 0 1 1 0 1 1 1 0 1 0 1 1 1 0 1 0 1
1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0
A solution with 73 ones for 12 X 12 (I replaced "0" with "." for readability):
1 1 . 1 1 . 1 1 . 1 . 1
1 1 . . 1 1 . 1 1 . 1 1
. . . 1 . . . . . . 1 .
1 1 . 1 . 1 . 1 1 . . 1
. 1 1 . . 1 1 . . 1 1 .
1 . . . 1 . 1 . 1 . . 1
1 1 . . 1 1 . . 1 . 1 .
. 1 . 1 . 1 . 1 . . 1 1
1 . . 1 1 . . 1 1 . . 1
. 1 . . . . 1 . 1 . 1 .
1 1 . 1 1 . 1 1 . . 1 1
1 . 1 . 1 1 . 1 . 1 . 1
- Manfred Scheucher, Jul 23 2015
An optimal solution with 74 ones (denoted by O) for 12 X 12 (also symmetric):
O . O . O . O O . O O .
O O . O O . . . O O . O
. O . O . O O . . . O O
O . . . O O . O O . O .
. O O . . . O . . . . O
O O . O O . O . O O . .
. . O O . O . O O . O O
O . . . . O . . . O O .
. O . O O . O O . . . O
O O . . . O O . O . O .
O . O O . . . O O . O O
. O O . O O . O . O . O - Giovanni Resta, Jul 29 2015
PROG
(Java) See Taylor link
(MATLAB with CPLEX)
function v = A181018(n)
%
Grid = [1:n]' * ones(1, n) + n*ones(n, 1)*[0:n-1];
f = -ones(n^2, 1);
A = sparse(4*(n-2)*(n-1), n^2);
count = 0;
for i =1:n
for j = 1:n-2
count = count+1;
A(count, [Grid(i, j), Grid(i, j+1), Grid(i, j+2)]) = 1;
end
end
for i = 1:n-2
for j = 1:n
count = count+1;
A(count, [Grid(i, j), Grid(i+1, j), Grid(i+2, j)]) = 1;
end
end
for i = 1:n-2
for j = 1:n-2
count = count+2;
A(count-1, [Grid(i, j+2), Grid(i+1, j+1), Grid(i+2, j)]) = 1;
A(count, [Grid(i, j), Grid(i+1, j+1), Grid(i+2, j+2)]) = 1;
end
end
b = 2*ones(4*(n-2)*(n-1), 1);
[x, v, exitflag, output] = cplexbilp(f, A, b);
end;
for n = 1:11
A(n) = A181018(n);
end
A % Robert Israel, Jan 14 2016
CROSSREFS
KEYWORD
nonn,more,nice
AUTHOR
R. H. Hardin, Sep 30 2010
EXTENSIONS
a(11)-a(12) from M. F. Hasler, Jul 20 2015
a(12) deleted by Manfred Scheucher, Jul 23 2015
a(12) from Giovanni Resta, Jul 29 2015
PARI code (which implemented a conjectured formula shown to underestimate) deleted by Peter J. Taylor, Jan 06 2016
a(13)-a(15) from Peter J. Taylor, Jan 09 2016
a(16) from Peter J. Taylor, Jan 14 2016
STATUS
approved