Consider a subset of $n$ points in an equilateral triangular lattice. Draw all the edges between nearest-neighbor points.


What is the maximum, over all such subsets, of the number of edges? This sequence appears to start 0, 1, 3, 5, 7, 9, 12, 14, 16...


What is the maximum number of triangular lattice cells? (Not the number of all triangles, just the number of smallest possible equilateral triangles in the lattice.) This sequence appears to start 0, 0, 1, 2, 3, 4, 6, 7, 8, 10, 11, 13...


<http://oeis.org/A047932> is related to the first sequence but I have no proof it's the same. (There might be some other way of arranging the pennies that yields a higher number of contacts. A047932 is a lower bound on my sequence.) I can't find any OEIS sequences relevant to the second one.



Maximal number of edges and triangular cells for n points in a triangular lattice
The following was conjectured by D. Reutter in problem 664A, Elemente der mathematik 27 and proved by H. Harborth in Solution to problem 664A, Elemente der mathematik 29, 14-15



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> The maximum number of times the minimum distance can occur among $n$ points in the plane is $\lfloor 3n-\sqrt{12n-3}\rfloor$.
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This is achieved by a hexagonal piece in the triangular lattice, i.e. from points forming a hexagonal spiral. In particular, this gives a formula for your first sequence.


