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Proof
From the recursive definition of continued fractions, we have:

Let:

In other words, $a_{3n+1} = 2n$ and $a_{3n+0} = a_{3n+2} = 1$

Then $p_i$ and $q_i$ are as follows:


 * $\begin{array}{r|cccccccccc}

\displaystyle i & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9  \\ \hline

p_i & 1 & 1 & 2 & 3 & 8 & 11 & 19 & 87 & 106 & 193 \\

q_i & 1 & 0 & 1 & 1 & 3 & 4 & 7 & 32 & 39 & 71 \\ \hline

\end{array}$

Then $p_i$ and $q_i$ satisfy the following $6$ recurrence relations:

Our ultimate aim is to prove that:

Let us define the integrals:

Lemma 1 - separate page


 * For $n \in \Z_{>0}$,

We begin by demonstrating the relationships hold for the base case, $n = 0$:

Next, we assert the following three recurrence relations hold:

To prove the first relation, we note that the derivative of the integrand of $A_n$ is equal to the sum of the integrand of $A_n$ with the integrand of $B_{n-1}$ and the integrand of $C_{n-1}$

By integrating both sides of the equation, we verify the first recurrence relation:

To prove the second relation, we note that the derivative of the integrand of $A_n$ is equal to the sum of the integrand of $A_n$ with the integrand of $B_{n-1}$ and the integrand of $C_{n-1}$

By integrating both sides of the equation, we verify the first recurrence relation: