Egyptian fraction series for $\frac{99}{70}-\sqrt{2}$

This question is motivated by series found while answering How is the sequence 1, 1.4, 1.41, 1.414 generated?

A fast series for $\sqrt{2}$ is given by the Egyptian fraction mentioned in the Wikipedia

$$\sqrt{2}=\frac{3}{2}-\frac{1}{2}\sum_{n=0}^\infty \frac{1}{a(2^n)}=\frac{3}{2}-\frac{1}{2}\left(\frac{1}{6}+\frac{1}{204}+\frac{1}{235416}+\dots \right)$$

with denominators defined by $2^n$th terms of a second order recurrence relation $$a(n)=34a(n-1)-a(n-2)$$ with $a(0)=0, a(1)=6$ (OEIS sequence A082405).

The corresponding closed form is $$\sqrt{2}=\frac{3}{2}-\sum_{k=0}^\infty \frac{2\sqrt{2}}{(17+12\sqrt{2})^{2^k}-(17-12\sqrt{2})^{2^k}}$$

as obtained in this answer.

A similar series seems to exist starting from the closer convergent $\dfrac{99}{70}$, because when applying the Babylonian method starting from $\dfrac{7}{5}$ the following sequence is obtained: $$\frac{7}{5}, \frac{99}{70}, \frac{19601}{13860}, \frac{768398401}{543339720}, \frac{1180872205318713601}{835002744095575440},...$$

and the difference between consecutive approximations has unit numerator, although the fractions are not consecutive convergents, so similarly we have

$$\sqrt{2}=\frac{99}{70}-\frac{1}{13860}-\frac{1}{543339720}-\frac{1}{835002744095575440}-...$$

Is there an underlying recurrence that can be sampled to obtain the denominators of these negative fractions?

After the answer

Formulas for $\dfrac{3}{2}-\sqrt{2}$ (question) and $\dfrac{10}{7}-\sqrt{2}$ (from the answer) can be written in terms of the silver ratio and the index starting from $0$.

$$\sqrt{2} = \frac{3}{2} - \sum_{k=0}^\infty \frac{2\sqrt{2}}{(1+\sqrt{2})^{2^{k+2}}-(1-\sqrt{2})^{2^{k+2}}}$$

$$\sqrt{2} = \frac{10}{7} - \sum_{k=0}^\infty \frac{2\sqrt{2}}{(1+\sqrt{2})^{3·2^{k+1}}-(1-\sqrt{2})^{3·2^{k+1}}}$$

Related questions

Numbers $p-\sqrt{q}$ having regular egyptian fraction expansions?

Newton's method for square roots 'jumps' through the continued fraction convergents.

We repeatedly apply the mapping $\frac{a}{b} \to \frac{1}{2}(\frac{a}{b} + \frac{2b}{a}) = \frac{a^2 + 2b^2}{2ba}$.

So $a(n+1) = a(n)^2 + 2b(n)^2$ and $b(n+1) = 2a(n)b(n)$.

There's a pattern in $b(n)$. Continuing the pattern: $b(n+2) = 2a(n+1)2a(n)b(n)$.

Generally, $\displaystyle b(n) = 5\cdot2^{n-1}\prod_{k=1}^{n-1}a(k)$.

As you've noticed, in $\frac{a^2 + 2b^2}{2ba} - \frac{a}{b} = \frac{2b^2 - a^2}{2ba}$ we have $2b^2 - a^2 = -1$.

$a(n+1) = a(n)^2 + 2b(n)^2 = a(n)^2 + a(n)^2 - 1 = 2a(n)^2 - 1$.

We have $a(n) = \cos(2^{n-2} \cos^{-1}(99))$ for $n > 1$.

However, I notice that all $b$ are Pell numbers, in fact $P(6), P(12), P(24), \dots P(6\cdot 2^n)$.

This gives us a formula for $b$:

$$b(n) = \frac{\sqrt2}{4}\left((7+5\sqrt 2)^{2^{n-1}} - (7-5\sqrt 2)^{2^{n-1}} \right)$$

I don't believe that the $7$ and $5$ constants in the formula are a coincidence.