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I searched numbers $N$, such that the continued fraction of $N^{1/3}$ has very large entries. I only searched for a single large entry, but I was surprised that two continued fractions contained not only one large entry, but multiple large entries. Here the two amazing expansions :

$$102175 [46, 1, 2, 1, 8741, 2, 186, 2, 13112, 1, 6, 1, 8, 2, 9, 2, 623, 1, 33, 1 , 9, 1, 2, 2, 17484, 14, 2, 2, 1, 4, 19021, 2, 1, 1, 1, 1, 1, 1, 3437888, 2, 2, 6, 21510, 2, 1, 2, 55063048, 1, 1, 1, 1, 1, 2, 8, 44, 2, 1, 4, 1, 4, 61, 2, 1666 1, 2, 1, 3, 1, 1, 23, 1, 4, 2, 2, 8, 3, 3, 1, 1, 2, 6, 3, 1, 1, 3, 5, 17, 21, 17 , 3, 168, 3, 1, 1, 17, 1, 3, 2, 3, 4, 3] 55063048$$ $$267090 [64, 2, 2, 31104, 1, 4, 64, 4, 1, 46657, 1288, 55545, 1127, 62210, 2, 2, 40, 1, 1, 2, 1, 1, 4, 559, 8, 1, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1, 1, 101091, 3, 1, 1, 8, 6, 10, 3, 1, 2, 2, 1, 1, 2, 17, 2, 1, 1, 2, 1, 4897902700, 1, 54, 1, 288, 1, 1, 1, 20, 1, 1, 5, 31360929, 1, 15, 9, 1, 1, 30, 1, 5, 6, 2, 7, 16, 2, 3, 1, 2, 3, 9935, 1, 3, 2, 1, 5, 4, 4, 2, 1, 28, 1, 27] 4897902700$$

Explanation : First, the number $N$ is displayed, then the first $100$ terms of the continued fraction of $N^{1/3}$ and finally the maximum of the entries.

The cubic roots of the numbers $102175$ and $267090$ seem to have a very special continued fraction.

In the second continued fraction, we have even $5$ consecutive large entries, and in both continued fractions we have an entry larger than $10^6$ besides the maximum entry.

This is not at all what I expected, in particular because almost every real number has a continued fraction expansion that follows a special distribution mentioned here :

Typicality of boundedness of entries of continued fraction representations

The continued fraction expansions above are far away from this distribution (even if we do not consider the maximum entry).

How can this phenomen be explained ?

Community
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Peter
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  • You might note that after a while the expansion does seem to "settle down": e.g. in $102175^{1/3}$, after term #47 = $55063048$ you get term #63 = $16661$ and no more terms $> 2000$ from #64 to #1000. So this behaviour seems not to be inconsistent with the continued fraction having the right limiting distribution; it's just that there are a few unusually good rational approximations. – Robert Israel Jan 25 '17 at 00:36
  • @RobertIsrael True that this trend need not continue, but in particular the $5$ consecutive entries larger than $1000$ are very surprising. – Peter Jan 25 '17 at 00:48
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    Would "coincidence" be an answer? I know it's not very satisfying… How many numbers have you tested before finding these two samples? – ThomasR Jan 26 '17 at 01:29

1 Answers1

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This does not answer the question, but it may give some ideas as to where to look for an answer.

In 1965, Brillhart found that the real root of $x^3-8x-10$ has several very large partial quotients in its continued fraction expansion, e.g.,
$a_{17}=22986$
$a_{33}=1501790$
$a_{59}=35657$
$a_{81}=49405$
$a_{103}=53460$
$a_{121}=16467250$
$a_{139}=48120$
$a_{161}=325927$
After this, it "calms down". There is an explanation for this behavior, involving some fairly advanced stuff (modular forms and the like). See, for example,
Churchhouse and Muir, Continued fractions, algebraic numbers and modular invariants, J. Inst. Maths Applics 5 (1969) 318-328.
Stark, An explanation of some exotic continued fractions found by Brillhart, in Atkin and Birch, eds., Computers in Number Theory, Academic Press 1971, 21-35.

https://oeis.org/A002937 is also worth a look, and possibly R.P. Brent, Alfred J. van der Poorten, Herman J.J. te Riele: "A comparative study of algorithms for computing continued fractions of algebraic numbers" Algorithmic Number Theory (Talence, 1996), pages 35-47, Lecture Notes in Computer Science, 1122, Springer, Berlin, 1996.

Gerry Myerson
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    Given the cubic $x^3-8x-10=0$, if we do the translation $x=y-2$, then $$y^3-6y^2+4y-2=0$$ the real root of which is $$y=e^{\pi,i/24}\frac{\eta(\tau)}{\eta(2\tau)} = 5.318628\dots$$ with Dedekind eta function $\eta(\tau)$ and $\tau = \frac{1+\sqrt{-163}}2$. Another nice thing about this root is, $$e^{\pi\sqrt{163}} = y^{24}-24.00000000000000105\dots$$ I also noticed the OP's $N=102175 = \color{blue}{67}\times1525$ and $N=267090= \color{blue}{58}\times4605$. I wouldn't be surprised if there is an $N = \color{blue}{163},m$ for some appropriate $m$. – Tito Piezas III Aug 24 '17 at 09:22
  • @Tito, yes, the translation $x=y-2$ is noted and discussed in the Stark paper at the OEIS page. Well done spotting the 67 and the 58. – Gerry Myerson Aug 24 '17 at 09:30
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    I searched $N = 163,m$. Didn't find anything terribly impressive, but did find the rather unusual cfrac expansion for $N = 163\times 6135$, $$N; [100, 6000, 100, 9000, 80, 10714, 3, 2, 1, 588000, 1,\dots]$$ it being the near-cube $N = 100^3+6$. Hmm. – Tito Piezas III Aug 24 '17 at 10:45
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    I meant the near-cube $N = 163\times 6135 = 100^3+5$. Jeez, this typo went unnoticed this long? – Tito Piezas III Jul 19 '23 at 15:37