Circle packing in an equilateral triangle

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Circle packing in an equilateral triangle is a packing problem in discrete mathematics where the objective is to pack Template:Mvar unit circles into the smallest possible equilateral triangle. Optimal solutions have been proved for Template:Math, and for any triangular number of circles, and conjectures are available for Template:Math.^[1]^[2]^[3]^[4]

A conjecture of Paul Erdős and Norman Oler states that, if Template:Mvar is a triangular number, then the optimal packings of Template:Math and of Template:Mvar circles have the same side length: that is, according to the conjecture, an optimal packing for Template:Math circles can be found by removing any single circle from the optimal hexagonal packing of Template:Mvar circles.^[5] This conjecture is now known to be true for Template:Math.^[6] In a paper by Graham and Lubachevsky concerning solutions for Template:Math they also conjectured seven infinite families of optimal solutions in addition to the one by Erdős and Oler. These families give conjectured solutions for many more numbers, including Template:Math.^[3]

Minimum solutions for the side length of the triangle:^[1]

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Number of circles	Triangle number	Length	Area
1	Yes	$2 \sqrt{3}$ = 3.464...	5.196...
2		$2 + 2 \sqrt{3}$ = 5.464...	12.928...
3	Yes	$2 + 2 \sqrt{3}$ = 5.464...	12.928...
4		$4 \sqrt{3}$ = 6.928...	20.784...
5		$4 + 2 \sqrt{3}$ = 7.464...	24.124...
6	Yes	$4 + 2 \sqrt{3}$ = 7.464...	24.124...
7		$2 + 4 \sqrt{3}$ = 8.928...	34.516...
8		$2 + 2 \sqrt{3} + \frac{2}{3} \sqrt{33}$ = 9.293...	37.401...
9		$6 + 2 \sqrt{3}$ = 9.464...	38.784...
10	Yes	$6 + 2 \sqrt{3}$ = 9.464...	38.784...
11		$4 + 2 \sqrt{3} + \frac{4}{3} \sqrt{6}$ = 10.730...	49.854...
12		$4 + 4 \sqrt{3}$ = 10.928...	51.712...
13		$4 + \frac{10}{3} \sqrt{3} + \frac{2}{3} \sqrt{6}$ = 11.406...	56.338...
14		$8 + 2 \sqrt{3}$ = 11.464...	56.908...
15	Yes	$8 + 2 \sqrt{3}$ = 11.464...	56.908...