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A344845 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)).

%I A344845 #52 Feb 16 2022 11:56:46
%S A344845 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 A344845 52,55,58,56
%N A344845 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 A344845 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 A344845 It has been shown that A344710(n) can be described and computed as min{a(k) | n <= A001481(k+1) < 5n/4}.
%C A344845 For the following comments, let m := A001481(n+1).
%C A344845 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 A344845 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 A344845 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 A344845 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 A344845 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 A344845 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 A344845 For [b(n)], it is conjectured that 1/x(n) is always an integer, making the ceiling function omittable for the definition.
%H A344845 Jonathan F. Waldmann, <a href="https://docs.google.com/spreadsheets/d/19x9Qj8HcmQXVD855IMFH_3MZt70RBwckLld2zYk53L0/edit?usp=sharing">A more nuanced upright triangle sequence</a>
%H A344845 Jonathan F. Waldmann, <a href="https://docs.google.com/spreadsheets/d/19ylTMfMzm9XVvOOYr0xYvFAMOHt0wbQn3x-Q1ObSAIs">An algorithm for the upright triangle sequence</a>
%H A344845 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 A344845 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 A344845 Let m := A001481(n+1).
%F A344845    a(n) = min{m, 2*min{area(ABC) | A, B, C in Z^2;
%F A344845    A = (0,0);
%F A344845    B = (a,b) with a >= b >= 0;
%F A344845    |AB| = sqrt(m);
%F A344845    C = (c,d) with c >= 0 or d >= c;
%F A344845    sqrt(m) <= |AC| < sqrt(5m/4);
%F A344845    sqrt(m) <= |BC| < sqrt(2m) } } (proved in "A more nuanced upright triangle sequence").
%F A344845 a(n) <= m.
%F A344845 a(n) > sqrt(3/4)*m (conjectured).
%e A344845 [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 A344845 [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 A344845    XXXXXX   OXOXOX   OXOXOX
%e A344845    XXXXXX   XOXOXO   OOOOOO
%e A344845    XXXXXX   OXOXOX   OXOXOX
%e A344845    XXXXXX   XOXOXO   OOOOOO
%Y A344845 Cf. A001481, A344710.
%K A344845 nonn,more
%O A344845 1,2
%A A344845 _Jonathan F. Waldmann_, May 29 2021