Why does this pattern fail (sometimes) for the continued fraction convergents of $\sqrt{2}$?

Remark: this question is based on a wrong numerical computation and persists only as an example for spurious imprecision in software (see comments)

This is connected to my post on the continued fraction convergents of pi. Motivated by Calvin Lin's comment whether a similar pattern exists for other constants, I checked $$\sqrt{2}$$. Its convergents are,

$$p_n = \frac{1}{1}, \frac{3}{2}, \frac{7}{5}, \frac{17}{12}, \frac{41}{29}, \frac{99}{70}, \frac{239}{169},\dots$$

Define the analogous $$a,b,c$$,

$$a_n,\,b_n,\,c_n = p_{n-2}-1,\;\; p_{n-1}-1,\;\; p_n-1$$

$$v_n=\text{Numerator}\,(a_n)\,\text{Numerator}(b_n)$$

and the same function in the other post,

$$F(n) = \sqrt{\frac{a_n c_n}{a_n-c_n}-v_n}$$

then for even $$n>2$$, we have,

$$\begin{array}{cc} n&F(n) \\ 4& \sqrt{2} \\ 6&5\sqrt{2} \\ 8&29\sqrt{2} \\ 10&169\sqrt{2} \\ 12&985\sqrt{2} \\ 14&5741\sqrt{2}\\ 16&33461\sqrt{2} \\ \vdots \\ 92&\sqrt{\text{huge number}} \\ 94&\text{integer}\sqrt{2} \\ \vdots \\ \end{array}$$

The sequence $$1,5, 29, 169,985,\dots$$ is A001653.

Question: Why does it fail at $$n = 92$$ (and other n as well) but, when it is $$N\sqrt{2}$$ again for some integer N, then N resumes being the correct kth term of the OEIS sequence?

Edit: As vadim123 pointed out, the case $$n=94$$ does in fact yield twice a square (and was just a bug in my old Mathematica V 4.)

• What's the huge number? Is it possible that it's twice a square? Commented Sep 29, 2013 at 0:54
• It's 34041759472536138536782994687493766710446015122061244605489282359202. (And it's square-free.) Commented Sep 29, 2013 at 0:55
• Alas, it is not square-free; in fact it is twice a square: wolframalpha.com/input/… Commented Sep 29, 2013 at 0:57
• The real mystery is how the hell Wolfram Alpha factors it so quickly. Commented Sep 29, 2013 at 0:59
• @StefanGruenwald: I've been tempted to delete this post, but decided to keep it as an example of a failed experiment. (In this case, using Mathematica Ver 4's Sqrt[] command on very large integers.) Commented Mar 13, 2015 at 13:43