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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.)

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  • $\begingroup$ What's the huge number? Is it possible that it's twice a square? $\endgroup$ – vadim123 Sep 29 '13 at 0:54
  • $\begingroup$ It's 34041759472536138536782994687493766710446015122061244605489282359202. (And it's square-free.) $\endgroup$ – Tito Piezas III Sep 29 '13 at 0:55
  • $\begingroup$ Alas, it is not square-free; in fact it is twice a square: wolframalpha.com/input/… $\endgroup$ – vadim123 Sep 29 '13 at 0:57
  • $\begingroup$ The real mystery is how the hell Wolfram Alpha factors it so quickly. $\endgroup$ – vadim123 Sep 29 '13 at 0:59
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    $\begingroup$ @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.) $\endgroup$ – Tito Piezas III Mar 13 '15 at 13:43
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The likelihood of a new sequence agreeing with a known sequence for 45 terms, then never again, is very small. The likelihood of a sequence agreeing with a known sequence for (apparently) infinitely many terms, but disagreeing for some scattered subset, is almost nil. This is how I suspected that the disagreement was illusory.

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  • $\begingroup$ Thanks. It agreed with the first 45 terms, then sporadically some more within the search radius I used. (Mathematica ver 4's \Sqrt[n] function leaves much to be desired.) $\endgroup$ – Tito Piezas III Sep 29 '13 at 1:08

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