

A344710


a(n)/2 is the smallest possible area of a nonobtuse triangle with coordinates in Z^2 and no side shorter than sqrt(n).


1



1, 2, 4, 4, 5, 8, 8, 8, 9, 10, 12, 12, 12, 15, 15, 15, 15, 18, 20, 20, 23, 23, 23, 23, 23, 24, 28, 28, 28, 30, 30, 30, 30, 30, 34, 34, 34, 38, 38, 38, 39, 42, 42, 42, 42, 45, 45, 45, 45
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OFFSET

1,2


COMMENTS

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.
If n is not in A001481, a(n) = a(n+1) because a distance of sqrt(n) can never occur in Z^2.
The smallest area of a nonobtuse triangle with an exact rather than a minimum shortest sidelength is documented in A344845. For A344845, only the terms corresponding to A001481/{0} can exist, as only those are norms in Z^2.
For n in A001481, it was proved that a valid triangle with a sidelength of sqrt(2n) or larger has at least area n/2, and that for n in A001481 a valid triangle with area n/2 always exists.
For n in A001481, all valid triangles with an area smaller than n/2 can be found by checking for triangles with no sidelength of sqrt(2n) or longer, and at most one sidelength of sqrt(5n/4) or longer.
These criteria can be used to set A to (0,0), and look for B and C in the set of points X with sqrt(n) <= AX < sqrt(5n/4). Furthermore, B can be assumed to be in the first octant, and C in a different octant but at most 2 octants away. Lastly, AB <= AC can be assumed. It has been shown that for n in A001481, this suffices to find congruent versions of all valid triangles with an area below n/2.
Up to at least n=17, the following sequence [b(n)] has been proved to have the same terms: 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 pairwise minimum distance sqrt(n).
If it is not identical for larger n, [a(n)] has been shown to at least be an upper bound on [b(n)].
It has been shown that an alternative interpretation of the problem described in b(n) is the packing of circles with diameter sqrt(n) with centers in the discrete plane Z^2.
For [b(n)], it is conjectured that 1/x(n) is always an integer, so the ceiling function can be omitted.
Conjectured next terms (from the conjectured inequality a(n) > sqrt(3/4)*n) are a(50,...,64) ?= 45, 48, 48, 52, 55, 55, 55, 55, 55, 56, 56, 56, 56, 56, 56.


LINKS

Table of n, a(n) for n=1..49.
Jonathan F. Waldmann, An algorithm for the upright triangle sequence
Jonathan F. Waldmann, A more nuanced 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

a(n) = min{A344845(k)  n <= A001481(k+1) < 5n/4}.
a(n+1) >= a(n).
a(n) <= n if n is in A001481.
a(n) > sqrt(3/4)*n (conjectured).


EXAMPLE

[a(n)]: For n=4, a triangle with 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).
[b(n)]: For n=1, n=2, and n=3, 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

Cf. A001481, A344845.
Sequence in context: A257174 A327625 A084824 * A184615 A151969 A261393
Adjacent sequences: A344707 A344708 A344709 * A344711 A344712 A344713


KEYWORD

nonn,more


AUTHOR

Jonathan F. Waldmann, May 26 2021


STATUS

approved



