The number of different lattice triangles

Question

Two convex lattice polygons are equivalent if there is a lattice-preserving affine transformation mapping one of them to the other. Equivalent polygons have the same area. Let $H(A)$ denote the number of equivalence classes of such polygons having area $A$. In the paper "On the number of convex lattice polygons" Barany and Pach proved that $\log H(A)\asymp A^{1/3}$. (In 1980 Arnolʹd proved that $c_1A^{1/3}<\log H(A)<c_2A^{1/3}\log A$.)

Let $H_3(A)$ denote the number of equivalence classes of lattice triangles having area $A$. What is known about behaviour of $H_3(A)$?

Answer 1

Let $A$ be double area of triangle. I calculate number of equivalence classes of triangles $OPQ$ with labelled vertices. Your problem is slightly different, but mine has nicer answer:) Without loss of generality two vertices are $(0,0)$ and $(a,0)$, where $a$ is some divisor of $A=ab$, and the third vertex has coordinates $(u,b)$. Unimodular transforms which fixes vertices $(0,0)$ and $(a,0)$ (and also upper half-plane) is $(x,y)\rightarrow (x+cy,y)$. That is, third vertex $(u,b)$ is equivalent to $(u+cb,b)$. Therefore there are $b$ different classes of equivalence. Sum up by pairs $(a,b)$ such that $ab=A$, we get that number of equivalence classes equals $\sigma(A)=$sumof divisors of $A$.