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Recursionhard
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For the continued fraction [1;1,1][1; 1, 1][1;1,1] (meaning 1+11+111 + \cfrac{1}{1 + \cfrac{1}{1}}1+1+11​1​), the convergents pnqn\frac{p_n}{q_n}qn​pn​​ satisfy: pn=anpn−1+pn−2,qn=anqn−1+qn−2p_n = a_n p_{n-1} + p_{n-2}, \quad q_n = a_n q_{n-1} + q_{n-2}pn​=an​pn−1​+pn−2​,qn​=an​qn−1​+qn−2​ with p−1=1,p0=a0,q−1=0,q0=1p_{-1} = 1, p_0 = a_0, q_{-1} = 0, q_0 = 1p−1​=1,p0​=a0​,q−1​=0,q0​=1. What is the second convergent p2q2\frac{p_2}{q_2}q2​p2​​?