OFFSET
1,2
COMMENTS
Since sqrt(m) is a distance between two points in Z^2 iff m is the sum of two squares, the "shortest side length" requirement can only be met for m in A001481\{0}. Thus this sequence cannot be extended to more arguments.
It has been shown that A344710(n) can be described and computed as min{a(k) | n <= A001481(k+1) < 5n/4}.
For the following comments, let m := A001481(n+1).
The validity of a triangle can be checked via four Diophantine inequalities: three Euclidean norms for the distances, and one derived from the law of cosines 2*max{|AB|,|BC|,|CA|}<=|AB|+|BC|+|CA| for the angles.
It was proved that a valid triangle with a sidelength of sqrt(2m) or larger has at least area m/2, and that a valid triangle with area m/2 always exists.
All valid triangles with an area smaller than m/2 can be found by checking for triangles with no sidelength of sqrt(2m) or longer, and at most one sidelength of sqrt(5m/4) or longer.
These criteria can be used to set A to (0,0), and look for B in the set of points X with |AX| = sqrt(m), and C in the set of points X with sqrt(m) <= |AX| < sqrt(5m/4). Furthermore, B can be assumed to be in the first octant, and C in a different octant but at most 2 octants away. It has been shown that this suffices to find congruent versions of all valid triangles with an area below m/2.
The sequence c(n) := min{a(k) | k >= n} is an upper bound on the sequence b(n) := ceiling(1/x(n)), where x(n) is the supremum on the density of marked points ("dots") in the discrete plane Z^2 with increasing pairwise minimum distance. Up to n=10, the two sequences have been shown to contain the same terms.
It has been shown that an alternative interpretation of the problem described in b(n) is the packing of circles with increasing diameter with centers in the discrete plane Z^2.
For [b(n)], it is conjectured that 1/x(n) is always an integer, making the ceiling function omittable for the definition.
LINKS
Jonathan F. Waldmann, A more nuanced upright triangle sequence
Jonathan F. Waldmann, An algorithm for the upright triangle sequence
Jonathan F. Waldmann, Proofs for the first few terms in the discrete circle packing sequence
Jonathan F. Waldmann, More proofs for the discrete circle packing sequence
FORMULA
Let m := A001481(n+1).
a(n) = min{m, 2*min{area(ABC) | A, B, C in Z^2;
A = (0,0);
B = (a,b) with a >= b >= 0;
|AB| = sqrt(m);
C = (c,d) with c >= 0 or d >= c;
sqrt(m) <= |AC| < sqrt(5m/4);
sqrt(m) <= |BC| < sqrt(2m) } } (proved in "A more nuanced upright triangle sequence").
a(n) <= m.
a(n) > sqrt(3/4)*m (conjectured).
EXAMPLE
[a(n)]: For n = 3, i.e. A001481(n+1) = 4, a triangle with a shortest sidelength of sqrt(4) and the minimal area of 4/2 = 2 can be placed at A=(0,0), B=(2,0), and C=(0,2). Alternatively, C can be placed at (1,2) or (2,2). -> a(3) = 4.
[b(n)]: For the smallest 3 minimum distances sqrt(1), sqrt(2), and sqrt(4), the following repeating patterns (X for dots, O for empty spaces) achieve the highest possible densities of 1, 1/2, and 1/4 respectively:
XXXXXX OXOXOX OXOXOX
XXXXXX XOXOXO OOOOOO
XXXXXX OXOXOX OXOXOX
XXXXXX XOXOXO OOOOOO
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Jonathan F. Waldmann, May 29 2021
STATUS
approved