The following Pythagorean hypotenuses have many possibilities of triangles. $125$ has three triangles $$35, 120, 125$$ $$44, 117, 125$$ $$75, 100, 125$$ the $365$ has $4$ triangles , $85$ has $4$ triangles , $1125$ has $3$ triangles ,$845$ has $7$ traingles, $1885$ has $13$ traingles , $2859545$ has $141$ triangles , and $2859547$ has only one.
We notice that the number of Pythagorean triangle not depend on a clear relation.So, I want to know if there is a specific Pythagorean hypotenuse which gives a maximum number of triangles?
You got the most efficient case from Robert in your recent question. Take the primes $$ 5,13,17,29,37,41, \ldots $$ that are the $$ p_i \equiv 1 \pmod 4. $$ Then take hypoteneuse $$ H = 5 \cdot 13 \cdot 17 \cdots p_k. $$ If you allow nonprimitive triangles, the total count of nonzero, ordered, positive triples, giving triangles, is $$ (3^k - 1)/ 2. $$ If you restrict to primitive triangles, it is just $$ 2^{k-1} $$