Continued fraction but no recurrence?

By using Euler's continued fraction formula , we change the fraction into another form by expanding or reducing fractions and the formula. $\quad\dfrac{1}{1-\dfrac{1}{5-\dfrac{16}{13-\dfrac{81}{25-\dfrac{256}{41-\dfrac{625}{61-\dfrac{1296}{85-\cdots}}}}}}}\\=\dfrac{1}{1-\dfrac{\frac{1}{4}}{\frac{5}{4}-\dfrac{\frac{4}{9}}{\frac{13}{9}-\dfrac{\frac{9}{16}}{\frac{25}{16}-\dfrac{\frac{16}{25}}{\frac{41}{25}-\dfrac{\frac{25}{36}}{\frac{61}{36}-\dfrac{\frac{36}{49}}{\frac{85}{49}-\cdots}}}}}}} \\= 1+1\times\dfrac{1}{4}+1\times\dfrac{1}{4}\times\dfrac{4}{9}+1\times\dfrac{1}{4}\times\dfrac{4}{9}\times\dfrac{9}{16}+1\times\dfrac{1}{4}\times\dfrac{4}{9}\times\dfrac{9}{16}\times\dfrac{16}{25}+1\times\dfrac{1}{4}\times\dfrac{4}{9}\times\dfrac{9}{16}\times\dfrac{16}{25}\times\dfrac{25}{36}+1\times\dfrac{1}{4}\times\dfrac{4}{9}\times\dfrac{9}{16}\times\dfrac{16}{25}\times\dfrac{25}{36}\times\dfrac{36}{49}+\cdots \\= 1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\cdots=\zeta(2)=\dfrac{\pi^2}{6}$

Therefore, $A=\dfrac{1}{6}, B=2, \dfrac{B}{A}=\boxed{12}$