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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.)

Gottfried Helms
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Tito Piezas III
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  • What's the huge number? Is it possible that it's twice a square? – vadim123 Sep 29 '13 at 00:54
  • It's 34041759472536138536782994687493766710446015122061244605489282359202. (And it's square-free.) – Tito Piezas III Sep 29 '13 at 00:55
  • Alas, it is not square-free; in fact it is twice a square: http://www.wolframalpha.com/input/?i=factor+34041759472536138536782994687493766710446015122061244605489282359202 – vadim123 Sep 29 '13 at 00:57
  • The real mystery is how the hell Wolfram Alpha factors it so quickly. – vadim123 Sep 29 '13 at 00:59
  • No, the mystery is my old Mathematica (Ver. 4) does not show sqrt{34041759472536138536782994687493766710446015122061244605489282359202} as twice a square. Another bug! :( – Tito Piezas III Sep 29 '13 at 01:01
  • A001653 can be calculated by (((3 + 2sqrt(2))n - (3 - 2sqrt(2))n) - ((3 + 2*sqrt(2))(n-1) - (3 - 2sqrt(2))(n-1))) / (4sqrt(2)). Thus, the sqrt(2) is right in it. Line 92 is an error of Mathematica (your version at least). – Stefan Gruenwald Mar 13 '15 at 02:21
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    @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.) – Tito Piezas III Mar 13 '15 at 13:43

1 Answers1

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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.

vadim123
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  • 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.) – Tito Piezas III Sep 29 '13 at 01:08