No construction from 2, 3 or 5 equilateral triangles exists. The first difference to the Padovan numbers occurs for a(15)=39, where the corresponding term A000931(19)=37. a(16)=A000931(20)=49. a(n)>=A000931(n+3). From the growth behavior of A290697 it is conjectured that a(k)>A000931(k+3) for all k>20.
a(19) is at least 130. This compares with A000931(23) = 114. It hints of growth behavior similar to sqrt(A014529) or sqrt(A001590). Ceiling(sqrt(A001590(n)) matches a(n) to n=14, then runs 38, 52, 70, 95, 128, ... . - Peter Munn, Mar 10 2018
From Peter Munn, Mar 14 2018 re monotonicity: (Start)
For n >= 6, a(n+1) > a(n).
Sketch of proof (inductive step) expressed in terms of tiling:
Given a triangle of side a(n) tiled with n equilateral triangular tiles. Let X, Y and Z be the tiles incident on its vertices, with X being not smaller than Y or Z.
Case 1: Y and Z have no vertices coincident. Remove Y and Z, thereby reducing the tiled area to a pentagon that has edges A and C that were previously internal to the area, and an edge B between A and C. Fit a new tile T against edge B, thereby extending edges A and C. Make the tiled area triangular by fitting a new tile against each of the extended edges.
Case 2: X, Y and Z have pairwise coincident vertices. It follows that these tiles are the same size. Remove Y and Z, thereby reducing the tiled area to a rhombus. Remove the tile at the rhombus vertex opposite X. The remaining area is a pentagon, since n >= 6. Extend the area by resiting Y against X, and Z against Y so that X and Z have external edges aligned. Make the area trapezoidal by fitting a new tile against the area's edge that includes an edge of Y. Fit another tile T against the smaller of the trapezoid's parallel edges.
In each case, we now have n+1 tiles, tiling an equilateral triangle with side length a(n) plus the side of T. As the sides of new and removed tiles can be calculated by adding sides of tiles that stayed in place, the gcd of the sides is unchanged.