Over a decade ago, I developed a formula that generates the subset $S$ of Pythagorean triples where $GCD(A,B,C)$ is an odd square. (Note: This includes all primitive where $GCD(A,B,C)=1.)\quad$ This also includes all triples where $GCD(A,B,C)\in\{9,25,49,\cdots\}$ but this is $\frac{1}{3}$ of the non-primitives produced by Euclid's formula (which include doubles and even square multiples of primitives). The number of non-primitives should not matter because, for example, with $(6,8,10)$, $A>B$ just like $(3,4,5)$.
The formula $F(n,k)$ produces distinct sets of triples where $n$ is the set number and where $k$ is the "member" in the set. These sets reveal patterns. The formula is
\begin{equation} A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2 \end{equation} and it produces the following sample of subset $S$. \begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 & k=6 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 & 85,132,157 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline Set_{6} &43,24,145 &165,52,173 &187,84,205 &209,120,241 &231,160,281 & 275,252,373 \\ \hline \end{array}
Notice that, to a depth $(k=n)$ in any set, there are $n$ members where $A>B.\quad$ A count of all qualifying triples up to $Set_n$ is a sum of integers so
$$P=\frac{\frac{n(n+1)}{2}}{n^2}\land n=6\implies P=\frac{21}{36}=\frac{7}{12}= 58.33\%$$
My feeling is that $A>B$ approaches $50\%$ for large numbers Pythagorean triples. Is there proof of this? What is
$$\lim_{n\rightarrow\infty} \frac{n(n+1)}{2n^2}=\space ?$$