A344845 - OEIS

%I #52 Feb 16 2022 11:56:46

%S 1,2,4,5,8,9,10,12,16,15,18,20,23,24,28,32,30,36,34,38,39,42,49,45,48,

%T 52,55,58,56

%N a(n)/2 is the smallest possible area of a non-obtuse triangle with integer coordinates in the plane and with shortest side of length sqrt(A001481(n+1)).

%C 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.

%C It has been shown that A344710(n) can be described and computed as min{a(k) | n <= A001481(k+1) < 5n/4}.

%C For the following comments, let m := A001481(n+1).

%C 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.

%C 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.

%C 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.

%C 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.

%C 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.

%C 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.

%C For [b(n)], it is conjectured that 1/x(n) is always an integer, making the ceiling function omittable for the definition.

%H Jonathan F. Waldmann, <a href="https://docs.google.com/spreadsheets/d/19x9Qj8HcmQXVD855IMFH_3MZt70RBwckLld2zYk53L0/edit?usp=sharing">A more nuanced upright triangle sequence</a>

%H Jonathan F. Waldmann, <a href="https://docs.google.com/spreadsheets/d/19ylTMfMzm9XVvOOYr0xYvFAMOHt0wbQn3x-Q1ObSAIs">An algorithm for the upright triangle sequence</a>

%H Jonathan F. Waldmann, <a href="https://docs.google.com/spreadsheets/d/1cJo_cP683PiU3ZUKEtNWKvSSMfGMhcI4efwyVwFBb4M">Proofs for the first few terms in the discrete circle packing sequence</a>

%H Jonathan F. Waldmann, <a href="https://docs.google.com/spreadsheets/d/1bKbl6Qq_PsWtb7zBx_PPmbrlIeVmAMioJ7_gFJRUae0">More proofs for the discrete circle packing sequence</a>

%F Let m := A001481(n+1).

%F    a(n) = min{m, 2*min{area(ABC) | A, B, C in Z^2;

%F    A = (0,0);

%F    B = (a,b) with a >= b >= 0;

%F    |AB| = sqrt(m);

%F    C = (c,d) with c >= 0 or d >= c;

%F    sqrt(m) <= |AC| < sqrt(5m/4);

%F    sqrt(m) <= |BC| < sqrt(2m) } } (proved in "A more nuanced upright triangle sequence").

%F a(n) <= m.

%F a(n) > sqrt(3/4)*m (conjectured).

%e [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.

%e [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:

%e    XXXXXX   OXOXOX   OXOXOX

%e    XXXXXX   XOXOXO   OOOOOO

%e    XXXXXX   OXOXOX   OXOXOX

%e    XXXXXX   XOXOXO   OOOOOO

%Y Cf. A001481, A344710.

%K nonn,more

%O 1,2

%A _Jonathan F. Waldmann_, May 29 2021