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Motivation:

Many interesting irrational numbers (or numbers believed to be irrational) appear as answers to natural questions in mathematics. Famous examples are $e$, $\pi$, $\log 2$, $\zeta(3)$ etc. Many more such numbers are described for example in the wonderful book "Mathematical Constants" by Steven R. Finch.

The question:

I am interested in theorems where a "special" rational number makes a surprising appearance as an answer to a natural question. By a special rational number I mean one with a large denominator (and preferably also a large numerator, to rule out numbers which are simply the reciprocals of large integers, but I'll consider exceptions to this rule). Please provide examples.

For illustration, here are a couple of nice examples I'm aware of:

  1. The average geodesic distance between two random points of the Sierpinski gasket of unit side lengths is equal to $\frac{466}{885}$. This is also equivalent to a natural discrete math fact about the analysis of algorithms, namely that the average number of moves in the Tower of Hanoi game with $n$ disks connecting a randomly chosen initial state to a randomly chosen terminal state with a shortest number of moves, is asymptotically equal to $\frac{466}{885}\times 2^n$. See here and here for more information.

  2. The answer to the title question of the recent paper ""The density of primes dividing a term in the Somos-5 sequence" by Davis, Kotsonis and Rouse is $\frac{5087}{10752}$.

Rules:

1) I won't try to define how large the denominator and numerator need to be to for the rational number to qualify as "special". A good answer will maximize the ratio of the number's information theoretic content to the information theoretic content of the statement of the question it answers. (E.g., a number like 34/57 may qualify if the question it answers is simple enough.) Really simple fractions like $3/4$, $22/7$ obviously do not qualify.

2) The question the number answers needs to be natural. Again, it's impossible to define what this means, but avoid answers in the style of "what is the rational number with smallest denominator solving the Diophantine equation [some arbitrary-sounding, unmotivated equation]".

Edit: a lot of great answers so far, thanks everyone. To clarify my question a bit, while all the answers posted so far represent very beautiful mathematics and some (like Richard Stanley's and Max Alekseyev's answers) are truly astonishing, my favorite type of answers involve questions that are conceptual in nature (e.g., longest increasing subsequences, tower of Hanoi, Markov spectrum, critical exponents in percolation) rather than purely computational (e.g., compute some integral or infinite series) and to which the answer is an exotic rational number. (Note that someone edited my original question changing "exotic" to "special"; that is fine, but "exotic" does a better job of signaling that numbers like 1/4 and 2 are not really what I had in mind. That is, 2 is indeed quite a special number, but I doubt anyone would consider it exotic.)

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    $\begingroup$ If one is allowed to cheat and use quadratic irrationalities, then there is Freiman's constant: the last gap in the Markov spectrum, namely, $\frac{2221564096 + 283748\sqrt{462}}{491993569}$. $\endgroup$ – Richard Stanley Sep 9 '15 at 17:12
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    $\begingroup$ Richard, that is indeed a beautiful example and very much in the spirit of what I had in mind. $\endgroup$ – Dan Romik Sep 9 '15 at 18:04
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    $\begingroup$ There are two kinds of "famous" rational constants. First kind is when it is clear a priori that it is rational (like Bernoulli numbers). Second kind where this comes as a suprise. Like the examples in my answer. $\endgroup$ – Alexandre Eremenko Sep 9 '15 at 22:01
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    $\begingroup$ I don't want to type it in, but there's the simplest rational right triangle with area 157, which Koblitz tells me was computed by Zagier. $\endgroup$ – Kimball Sep 12 '15 at 17:11

28 Answers 28

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We have $$\int\limits_0^\infty {\frac{{\sin x}}{x}dx} = \int\limits_0^\infty {\frac{{\sin x}}{x}\frac{{\sin \left( {{x/3}} \right)}}{{{x/3}}}dx} = \ldots = \int\limits_0^\infty {\frac{{\sin x}}{x}\cdots\frac{{\sin \left( {{x/{13}}} \right)}}{{{x/{13}}}}dx} = \frac{\pi }{2}$$

But $$\int\limits_0^\infty {\frac{{\sin x}}{x}\cdots\frac{{\sin \left( {{x/{15}}} \right)}}{{{x/{15}}}}dx} = \frac{{467807924713440738696537864469}}{{935615849440640907310521750000}} \cdot \pi$$

See http://link.springer.com/article/10.1023%2FA%3A1011497229317

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    $\begingroup$ I thought of this too but couldn't find the reference. $\endgroup$ – Noam D. Elkies Sep 9 '15 at 20:35
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    $\begingroup$ This is essentially a duplicate of that answer. $\endgroup$ – Stefan Kohl Sep 9 '15 at 23:13
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Gobel's sequence is defined by the recurrence relation: $$\begin{cases} a_0=a_1=1,\\ a_{n+1} = \frac{a_0^2 + a_1^2 +\dots + a_n^2}{n} & \text{for}\ n\geq 1. \end{cases}$$ It turns out that $a_k$ for all $k\leq 43$ are integer, but $a_{44}$ and on are not. Namely, $$a_{44} = \frac{m}{43},$$ where $m$ is a huge integer not divisible by 43.

See http://oeis.org/A003504

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    $\begingroup$ Is there any explanation of this phenomenon? $\endgroup$ – Fedor Petrov Sep 10 '15 at 11:47
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    $\begingroup$ @BlueRaja about 400 billion digits according to the asymptotic estimate given in the OEIS entry (I originally figured it grew as 2^n, but it actually grows as k^(2^n) where k is slightly larger than 1). $\endgroup$ – hobbs Sep 12 '15 at 4:52
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    $\begingroup$ @FedorPetrov: Another recurrent formula: $a_{n+1} = a_n\cdot (a_n+n-1) / n$ may justify why $a_n$ remains integer for quite a while. It shows that $a_n$ accumulates most of the factors of the previous terms and some others. The "$/n$" cancellations happen to be minor and hit the existing factors up until $n=43$. $\endgroup$ – Max Alekseyev Sep 14 '15 at 18:01
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    $\begingroup$ But if we replace $a_n+n-1$ to something like $a_n+n+1$ or $a_n+2n-1$, or add additional multiple $a_n$ in the recurrent formula for $a_{n+1}$, does similar thing happen? $\endgroup$ – Fedor Petrov Sep 14 '15 at 18:33
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    $\begingroup$ @FedorPetrov: For some other coefficients, it does. E.g., $a_{n+1} = a_n(a_n+n+5)/n$ with $a_2 = 2$ remains integer for up to $a_{59}$. $\endgroup$ – Max Alekseyev Sep 14 '15 at 19:00
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Perhaps rational numbers appearing in geometric probability qualify, e.g. Sylvester's four point problem:

What is the probability that four randomly chosen points (independently with uniform probability measure) in a planar region give a convex quadrilateral? In my opinion this is a "natural" question.

For a square the probability is $\frac{25}{36}$, for a hexagon it's $\frac{683}{972}$.

(Source: http://mathworld.wolfram.com/SylvestersFour-PointProblem.html)

EDIT (Sept,15; another example for a “special” rational number appearing as answer of a “natural” geometric probability question):

I was mentioning this MO question and my answer today to Prof Henze (KIT) and he pointed me to the fact that the expectation value of the (unsigned) area of a triangle the vertices of which are randomly chosen in the unit square (independently with uniform probability measure) is $\frac{11}{144}$.

(Sources:

)

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Another example from probability theory: in critical bond percolation on the square lattice $\mathbb{Z}^2$, one can define a certain family of "connectivity events," which encode information about certain points being connected to certain other points, and whose probabilities turn out to be \begin{align*} \mathbb{P}(1\leftrightarrow 2) &= \frac{3}{8}, \\[5pt] \mathbb{P}(1\leftrightarrow 2, 3\leftrightarrow 4) &= \frac{97}{512}, \\[5pt] \mathbb{P}(1\leftrightarrow 2, 4\leftrightarrow 5) &= \frac{135}{1024}, \\[5pt] \mathbb{P}(1\leftrightarrow 4, 2\leftrightarrow 3) &= \frac{59}{1024}, \\[5pt] \mathbb{P}(1\leftrightarrow 2, 3\leftrightarrow 4, 5\leftrightarrow6) &= \frac{214093}{2^{21}}, \\[5pt] \mathbb{P}(1\leftrightarrow 2, 3\leftrightarrow 6, 4\leftrightarrow5) &= \frac{69693}{2^{21}}, \ldots. \end{align*} There is no known simple explanation for why these probabilities are rational (let alone dyadic rational) numbers. Only the top three of these identities have been proved, using extremely complicated methods relating to the combinatorics of alternating sign matrices, totally symmetric self complementary plane partitions, and something called the quantum Knizhnik-Zamolodchikov equation. The remaining identities are conjectural, but the evidence that they are correct is very strong. This paper reduces the proof of these conjectures to the purely algebraic problem of proving a certain family of conjectural constant term identities.

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Many percolation critical exponents are mysterious rationals, prescribed by physicists and in rare cases proved mathematically.

Example. Given $p\in (1/2,1)$, open each hexagon in the honeycomb lattice with probability $p$. Then probability that given hexagon lies in an infinite cluster behaves as $(p-1/2)^{5/36+o(1)}$ when $p$ goes to $1/2$. Why $5/36$?

This is taken from the paper "Critical exponents for two-dimensional percolation" by Stanislav Smirnov and Wendelin Werner.

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    $\begingroup$ I don't understand the terms (and, of course, have not read the paper) but could it just be $1/6(1-1/6)$? $\endgroup$ – Mark Hurd Sep 10 '15 at 11:46
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    $\begingroup$ It definitely equals $1/6(1-1/6)$, but is it so typical for the exponent? $\endgroup$ – Fedor Petrov Sep 10 '15 at 12:04
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    $\begingroup$ Wouldn't one expect $\frac{1}{6}$ to show up with connectedness in a hexagonal grid? $\endgroup$ – user21820 Sep 12 '15 at 15:14
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    $\begingroup$ @user21820, the same constant $5/36$ is believed to still be the correct one for percolation on other lattices (square, triangular etc.), when the question is modified appropriately ($1/2$ should be replaced with the critical percolation threshold). So the $1/6$ is unrelated to the hexagonal strcture of the lattice. Here is a page with more details and additional interesting critical exponents such as $43/18$, $187/91$, ... $\endgroup$ – Dan Romik Sep 13 '15 at 4:30
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    $\begingroup$ @DanRomik: Ah very interesting thanks! $\endgroup$ – user21820 Sep 15 '15 at 1:40
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I wasn't thinking of mentioning this, but some of the other answers reminded me of the elegant formulas \begin{align} \pi^{-6} \sum_{n,m=1}^\infty \frac{1}{(n m(n+m))^2} &= \frac{1}{2835}, \\ \pi^{-12} \sum_{n,m=1}^\infty \frac{1}{(n m(n+m))^4} &= \frac{19}{273648375}, \end{align} due essentially to Mordell (1958).

While I requested rational numbers that arise as answers to conceptual questions and these seem more of the "purely computational" variety, these formulas actually arise in connection with some deeper mathematics (the so-called Witten zeta function of $sl(3,\mathbb{C})$; volumes of moduli spaces of vector bundles of curves), some of which came up recently in connection with my own work (see section 1.2 of my recent paper and for additional background section 7 of this paper by Zagier). Anyway, they are pretty identities so I thought I'd mention it even if it's not completely in the spirit of my question.

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    $\begingroup$ What's the Mordell reference? Did he do it by integrating cubes of Bernoulli polynomials $B_2$ and $B_4$? $\endgroup$ – Noam D. Elkies Sep 11 '15 at 4:28
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    $\begingroup$ Yes, and the same technique will work with the Bernoulli polynomial $B_{2k}$ to show that $\sum_{n,m\ge 1} (n m (n+m))^{-2k}$ is a rational multiple of $\pi^{6k}$. The reference is: L. J. Mordell, On the evaluation of some multiple series. J. London Math. Soc. 33 (1958), 368--371. $\endgroup$ – Dan Romik Sep 11 '15 at 4:36
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    $\begingroup$ Thank you! I rediscovered this trick of integrating $B_{2k}(x)^3$ around 1980, and never knew a source, though it was clearly unlikely that I was the first to notice it. $\endgroup$ – Noam D. Elkies Sep 11 '15 at 4:50
  • $\begingroup$ Great! This problem has an interesting history, and the fact I mentioned about $\sum_{n,m} (nm(n+m))^{-2k}$ was rediscovered several times. There are also better formulas for the rational coefficient of $\pi^{6k}$ than the one you get by integrating $B_{2k}(x)^3$. See the other references I mentioned. $\endgroup$ – Dan Romik Sep 11 '15 at 4:55
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$$\pi^{-4}\sum_1^{\infty}n^{-4}{2n\choose n}^{-1}=2\pi^{-4}\int_0^{\pi/3}x\biggl(\log\bigl(2\sin(x/2)\bigr)\biggr)^2\,dx={17\over3240}$$ The part with the sum is an exercise on page 89 of Comtet, Advanced Combinatorics; the part with the integral was a remark of Lewin; there is some discussion in Alf van der Poorten's paper on Apéry's proof of the irrationality of $\zeta(3)$, "A proof that Euler missed."

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An adventitious quadrangle is a (convex) quadrilateral with the property that if its two diagonals are drawn in, all angles formed are rational multiples of $\pi$. A classification of all such quadrilaterals was published by Bol, whose method was essentially correct but whose list contained some errors. A correct classification was published by Poonen and Rubinstein. It turns out that there are 65 "sporadic" solutions. The rational numbers arising in these sporadic solutions could be considered exotic, although the largest denominator that arises is "only" 210. One such sporadic adventitious quadrilateral has interior angles (in cyclic order) $139\pi/210$, $5\pi/14$, $6\pi/7$, and $13\pi/105$.

Note that adventitious angles have shown up in the popular press occasionally, e.g., the Washington Post in 1995 and in a United Airlines magazine in 2004. It's typically phrased as an innocent-looking "find the angle in the diagram" problem that looks easy but is actually very difficult, at least if trigonometry is not allowed.

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    $\begingroup$ Fantastic! These are just the sort of exotic numbers and results I was hoping to learn about. $\endgroup$ – Dan Romik Sep 10 '15 at 17:59
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    $\begingroup$ In my paper, Rational products of sines of rational angles, Aeq. Math. 45 (1993) 70-82, I found all the solutions of $$\sin\pi x\sin\pi y\sin\pi z\sin\pi w=r$$ in rational $x,y,z,w,r$. These included $$\sin\pi/42\sin5\pi/14\sin8\pi/21\sin10\pi/21=1/16$$ and $$\sin\pi/39\sin9\pi/26\sin14\pi/39\sin11\pi/26=1/16$$ $\endgroup$ – Gerry Myerson Sep 10 '15 at 23:28
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    $\begingroup$ Gerry, that's awesome. Why not post it as an answer? $\endgroup$ – Dan Romik Sep 11 '15 at 4:46
  • $\begingroup$ Done (didn't notice your suggestion at the time). $\endgroup$ – Gerry Myerson Oct 4 '17 at 23:14
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There is a nice "Birthday" probability value (not sure though if it fulfills the OP's criteria fully).

Suppose you have to get in line for a movie. The theatre announces that the first person who has a birthday matching any of the people in front of her or him will get a free ticket. Assuming uniform distribution of birthdays across the year, the aim is to find at which position you should get in line to maximize the probability of your win.

Some calculation shows that you should try to be the 20th person, and that \begin{equation*} \text{Pr}(20) = \frac{249547857325289514001589444555797521991729152} {7721192983187403134097091636121110137939453125} \end{equation*}

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    $\begingroup$ This answer fails on the "surprising" part; the large values are merely calculations. but the original puzzle is nice, so I included it here. $\endgroup$ – Suvrit Sep 10 '15 at 1:34
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    $\begingroup$ @PerAlexandersson "Assuming uniform distribution of birthdays across the year" (exact quote from this answer), no, I don't think that's right. Uniform distribution of birthdays across the year means that given a large enough sample size, for each day of the year (including Feb 29), the number of people with that day as their birthday is the same. What you're thinking of is when for each day, the number of people with that day as their birthday is the same. It's a subtle difference and your interpretation is less unrealistic, I'll admit that, but I don't think it matches this answer. $\endgroup$ – hvd Sep 10 '15 at 13:01
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    $\begingroup$ The year? Which year? This year, 2015, has 365 days for the cinema-goers to be uniformly distributed over. Next year, 2016, has 366 days for them. I don't know whether this answer assumed a 365 or a 366 day year, but either way I expect you'll get some big numbers as numerator and denominator. What would be surprising is if one of them simplified nicely, thereby disqualifying itself from being an answer to this question! $\endgroup$ – Steve Jessop Sep 11 '15 at 8:55
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    $\begingroup$ Related: MO 103892: Solving a modified birthday problem at a glance. $\endgroup$ – Benjamin Dickman Sep 11 '15 at 11:06
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    $\begingroup$ This doesn't seem like a particularly 'nice' number or question. The number 365 is arbitrary, and presumably any combinatorial question like these gives as result which is a fraction with very large numerator and denominator. $\endgroup$ – jwg Sep 15 '15 at 22:00
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A somewhat silly example, which is a "fun" (i.e. hard) multivariable calculus exercise: Take a sphere of unit radius, then remove two cylinders of radius 1/2 that are tangent to each other along the $z$-axis (and so also tangent to the sphere on the boundary). What is the volume of the remaining sphere-with-holes?

Punchline: it's $\frac{16}{9}$!

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Some answers to this question might provide examples for this question also:

For example, $32/27$, in this answer, about a bound on real zeros of chromatic polynomials for graphs.

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Rational number 1/4 occurs as a universal constant in several problems of Analysis. The most famous is the "Koebe 1/4 Theorem". Let $f(z)=z+\ldots$ be an injective holomorphic function in the unit disk. Then the image contains a disk $|z|\geq c$ where $c$ is an absolute constant. This was discovered by Koebe, and for some time the optimal value was even called the "Koebe constant". The optimal (greatest for which this is true) constant was later proved to be $1/4$.

An apparently unrelated subject. The family $\phi_n(x)=e^{inx}$ is a Riesz basis in the space $L^2(-\pi,\pi)$. For a sequence of real numbers $\Lambda=\lambda_n$, consider the family $e^{i\lambda_nx}$. If $|\lambda_n-n|<c$, and $c$ is sufficiently small, this is also a Riesz basis. (This is due to Golovin, I guess). This was known for some time before M. Kadets proved that the optimal (greatest) $c$ for which this is true is $1/4$. This remarkable result is called the "Kadets 1/4 theorem".

A priori reasons why these two numbers should be rational are not clear. Both numbers are defined as solutions of sophisticated extremal problems, so it was totally unclear why they should be rational until these extremal problems were solved.

EDIT. There are also conjectured extremal problems with a rational answer. One is mentioned in the comment of Noam Elkies below. Another is a conjecture of Carleson and Jones, where (surprise!) the extremal constant is also 1/4, or 3/4, depending on notation, but I was never able to understand the reasons why they made such a conjecture: MR1162188. They say it is suggested by numerical computation but the details of this computation were never published.

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    $\begingroup$ There's also Selberg's $1/4$ conjecture en.wikipedia.org/wiki/Selberg%27s_1/4_conjecture , but $1/4$ is surely too ordinary a rational number to qualify. (Selberg proved $3/16$, and the current bound seems to be $975/4096$ mathoverflow.net/questions/145637 which would be sufficiently exotic except that it's a bound that's not expected to be the actual minimum.) $\endgroup$ – Noam D. Elkies Sep 9 '15 at 21:03
  • $\begingroup$ Yes, sure. I just forgot it:-( You may enter this as another answer or to edit my answer. $\endgroup$ – Alexandre Eremenko Sep 9 '15 at 21:05
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    $\begingroup$ I won't answer with a rational number as simple as $1/4$; if you want to edit your answer go ahead. $\endgroup$ – Noam D. Elkies Sep 9 '15 at 21:09
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Let $v(n)$ be the supremum, over all symplectic embeddings of the disjoint union of $n$ equal balls into $B(1)$, of the fraction of the volume of $B(1)$ that is filled. As of 1997, these $v(n)$ are known for all $n$: $$ \begin{array}{c|ccccccccc} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & n \geq 9 \\ \hline v(n) & 1 & \frac12 & \frac34 & 1 & \frac45 & \frac{24}{25} & \frac{63}{64} & \frac{288}{289} & 1 \end{array} $$ At least the cases $n=7$ and $n=8$ should be legitimate examples given that $v(n)=1$ for all larger $n$ (and that it is not obvious a priori that $v(n)$ is rational for any $n > 1$). The cases $n\leq 9$ are due to McDuff and Polterovich (1994); $n \geq 9$, to Biran (1997). This is according to M.Hutchings, Recent progress on symplectic embedding problems in four dimensions, PNAS 108 #20 (2011), 8093-8099. See OEIS sequences A125846 and A125847, where it is reported that "Explicit constructions for $n = 8$ and $n = 9$ are still unknown."

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For any sufficiently large even value of $n$, the Bernoulli number $B_n$ is probably a good example.

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    $\begingroup$ He probably had $B_{12}$ in mind. The numerators of $B_n$ for $n \leq 14$ are all $0$ or $\pm 1$ except for $n=12$ when it's $691$ (and $B_{12}$ itself is $-691/2730$). Hence the appearance of $691$ in Ramanujan's congruence for $\tau(n)$, "etc.". $\endgroup$ – Noam D. Elkies Sep 9 '15 at 20:37
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    $\begingroup$ @DanRomik: For any sufficiently large even value of $n$, the Bernoulli number $B_n$ is a particular rational number, not a sequence. $\endgroup$ – Steven Landsburg Sep 9 '15 at 21:06
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    $\begingroup$ $$\pi^{-12}\sum_1^{\infty}n^{-12}={691\over638512875}$$ $\endgroup$ – Gerry Myerson Sep 9 '15 at 23:16
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    $\begingroup$ It's also a good vitamin. $\endgroup$ – Gerry Myerson Sep 10 '15 at 5:05
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    $\begingroup$ Well there's the poison formerly known as Vitamin B$_{17}$: en.wikipedia.org/wiki/Amygdalin $\endgroup$ – Noam D. Elkies Sep 11 '15 at 4:51
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Randomly choose a subset $\{a,b,c,d\}$ of size exactly 4 uniformly from among all $N \choose 4$ size 4 subsets of $\{1,2,3,...,N\}$.

Form the polynomial $p(z) = 1 + z^a + z^b + z^c + z^d$.

What's the probability that $p(z)$ has at least one root on the unit circle? The probability depends on $N$, but how does the probability behave when $N$ approaches infinity?

Some plausible conjectures imply that the answer is $909/9464$. See paper A6 here: http://math.colgate.edu/~integers/vol12.html

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  • $\begingroup$ In the paper (by Idris Mercer) linked, it is seen that if the size of 4 is substituted with a size of 3 or 2, this solution becomes "simpler" rational numbers, $\frac37$ respectively $\frac14$. $\endgroup$ – Jeppe Stig Nielsen Apr 14 '18 at 13:25
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For a prime number $p$, the number of nonisomorphic groups of order $p^n$ is $p^{(2/27)n^3 + O(n^{8/3})}$. I was surprised when I first saw this formula with leading coefficient $2/27$ in the exponent.

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If $f(\pi)$ is a length of the longest increasing subsequence of a permutation $\pi$, than $f(\pi)$ concentrates when $\pi$ runs over $S_n$, and expected (or median, or whatever) value behaves like $c\sqrt{n}$. There are no obvious reasons why $c$ must be good number, but it appears to be equal to 2. This was known as Ulam's problem, solved by Vershik and Kerov and by Logan and Shepp. Of course, 2 is not very original rational number, but I believe that this example is something that Dan is asking about.

And let me popularize here an intriguing problem by Ambrus and Bárány: if we consider $n$ random independent points in a triangle $ABC$ and look for a convex chain $AP_1\dots P_kB$, $P_i$ are chosen from our points and $k$ has to be maximized, then $k$ is concentrated near $c\sqrt[3]{n}$ (not a surprise), and numerical computations suggest that $c=3$. My opinion is that it would be great if this is true and we really have $2\sqrt{n}$ for the longest monotone chain and $3\sqrt[3]{n}$ for the longest convex chain.

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I can hardly ever be more surprised than with this expression (by Ramanujan):

$$ \frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{ (4k)! (1103+26390k) }{ (k!)^4 396^{4k} } = \frac1{\pi}. $$

In particular, it's intriguing that such "special" rational numbers (in the sense defined by the OP) appear in a series related to good old $\pi$.

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    $\begingroup$ That isn't rational. $\endgroup$ – PyRulez Sep 12 '15 at 1:12
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    $\begingroup$ @PyRulez, the LHS has some cool fractions, which are rational except for the $\sqrt{2}$ factor. Regardless, it is never really wrong to cite Ramanujan... $\endgroup$ – Dan Romik Sep 12 '15 at 7:36
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    $\begingroup$ To me, the "rules of the game" of the actual question are not so clear if PyRulez's objection is dismissed. $\endgroup$ – Todd Trimble Sep 13 '15 at 5:24
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    $\begingroup$ @ToddTrimble You could rephrase it as: $$ \frac{4}{9801} \sum_{k=0}^\infty \frac{ (4k)! (1103+26390k) }{ (k!)^4 396^{4k} } = \frac{\sqrt 2}{\pi}. $$ and then the fact that those "special" rational numbers appear on the LHS is still surprising (what do they have to do with $\sqrt 2/\pi$?) $\endgroup$ – Luis Mendo Sep 13 '15 at 15:01
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    $\begingroup$ @LuisMendo I can't see how this is responsive to my comment. Is the question now allowing not just single rational values, but sequences of rational values that converge to... whatever? Couldn't you then just stick a pin in Ramanujan's notebooks and wherever you land, there you have it? $\endgroup$ – Todd Trimble Sep 13 '15 at 15:32
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In the dissection of the unit square into the least number of square pieces of different sizes (Duijvestijn's dissection), the area of the largest piece is $\frac{625}{3136}$ and the harmonic mean of all areas is $\frac{27812926939574093625}{7042788441228875157149}$.

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    $\begingroup$ This is very nice, but may I ask why the harmonic mean is interesting to compute (as opposed to the arithmetic mean, or any other function of the areas)? $\endgroup$ – Dan Romik Sep 11 '15 at 22:30
  • $\begingroup$ Just because it has a simple name and gives a scarier rational :) $\endgroup$ – Vladimir Reshetnikov Sep 12 '15 at 2:34
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    $\begingroup$ Even using area instead of side is artificially inflating the relevant size: area 625/3136 is just side 25/56. $\endgroup$ – Noam D. Elkies Sep 12 '15 at 14:48
  • $\begingroup$ I feel that an area is a natural measure of size in this case, because areas must add up to $1$. $\endgroup$ – Vladimir Reshetnikov Sep 12 '15 at 16:53
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John H. Conway's prime producing machine (also known as PRIMEGAME) is a weird algorithm that produces prime numbers using the following bizarre ordered sequence of fourteen rational numbers: $$ \frac{17}{91},\ \ \frac{78}{85},\ \ \frac{19}{51},\ \ \frac{23}{38},\ \ \frac{29}{33},\ \ \frac{77}{29},\ \ \frac{95}{23},\ \ \frac{77}{19},\ \ \frac{1}{17},\ \ \frac{11}{13},\ \ \frac{13}{11},\ \ \frac{15}{2},\ \ \frac{1}{7},\ \ 55. $$ See here, here and (subscription required) here.

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The game of Bidding Chess (see Discrete Bidding Games by Mike Develin and Sam Payne) offers some rather tricky rational values.

Let's put the two kings on their original squares e1 and e8, and a white knight on g1 (admittedly not a natural position for any mathematical reason). As is explained in the paper, bidding games are related to random turn games, where before each move, a coin toss decides who gets to move. The objective of the game is to capture the opponent's king, so there are no concepts of check, checkmate or stalemate.

We can define two numbers $\alpha \leq \beta$ (representing the value of the position from White's perspective) such that $\alpha$ is White's maximum probability of winning, and $\beta$ is White's maximum probability of not losing, taken over all White's strategies (and minimizing over all Black's strategies). For Black, the maximum probability of winning is $1-\beta$ and the maximum probability of not losing is $1-\alpha$.

It is not true in general that $\alpha=\beta$, but in this particular position, numerical evidence (!) suggests that $\alpha = \beta$, and that they are equal to $$ {\tiny 140297187507809718787571560535429924880742491099805023919271172220178831973930689386662219848285322420407060840990534312850343}\\ \big /\\ {\tiny 194465010677647568859234587998029627946213150285790861705846175978932032134223961344086399663594245257797038922571722208051200}$$

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  • $\begingroup$ This is under some assumption on how many chips each player has at the outset? $\endgroup$ – Gerry Myerson Sep 16 '15 at 0:02
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    $\begingroup$ Gerry, just assuming one can bid any real number. Or equivalently, taking the "infinitely many chips"-limit. So the given number is the limit proportion of chips that Black needs in order to win/survive. $\endgroup$ – Johan Wästlund Sep 16 '15 at 0:13
  • $\begingroup$ I feel like the huge number here isn't too surprising - presumably this is related to some particular Markov chains on the $(64\times63\times62)$-node position graph... $\endgroup$ – Steven Stadnicki Jan 20 '18 at 20:54
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Many delicate estimates in analytic number theory lead to bounds involving unusual rational numbers. Two examples I am aware of are:

  1. The Lindelof hypothesis asks about the rate of growth of $|\zeta(1/2+it)|$ as $|t|\to\infty$, that is the infimum $\lambda$ of all numbers $c>0$ such that $|\zeta(1/2+it)|=O(|t|^c)$. The conjecture is that $\lambda=0$. This is known to follow from the Riemann hypothesis but does not imply it. A series of papers by many authors over the last 100+ years, starting with Lindelof in 1908, proved that $\lambda$ is bounded by $$ \frac14, \frac16, \frac{163}{988}, \frac{27}{164}, \frac{229}{1392}, \frac{19}{116}, \frac{15}{92}, \frac{6}{37}, \frac{173}{1067}, \frac{35}{216}, \frac{139}{858}, \frac{32}{205}, \frac{53}{342} $$ (it's not obvious at a glance, but this is a decreasing sequence!). The last and currently best known bound of $\frac{53}{342}\approx 0.1549$ is due to Bourgain (2014). See Wikipedia for more details.

  2. There is a similar story about the Dirichlet divisor problem, which is the problem of estimating the average order of the number of divisors $\sigma(n)$. That is, denote $$ \sigma(n) = \sum_{d|n} 1, \qquad D(x) = \sum_{n\le x} d(n). $$ Dirichlet proved that $$ D(x) = x \log x + x(2\gamma-1) + \Delta(x) $$ where $\gamma$ is the Euler-Mascheroni constant and $\Delta(x) = O(x^{1/2})$. The Dirichlet divisor problem asks for the infimum $\theta$ of numbers $c$ such that $\Delta(x)=O(x^c)$. In this case a lower bound due to Hardy says that $\theta\ge\frac14$. As with the Lindelof hypothesis, Dirichlet's upper bound $\theta\le \frac12$ has been improved over the years by many people who published the successive bounds $$ \frac12, \frac13, \frac{33}{100}, \frac{27}{82}, \frac{15}{46}, \frac{12}{37}, \frac{346}{1067}, \frac{35}{108}, \frac{7}{22}, \frac{131}{416}. $$ The most recent bound $\frac{131}{416}\approx 0.3149$ is due to Huxley (2003). See Wikipedia. (Paraphrasing a joke Cris Moore made recently on the Domino Forum, it would be nice to look up this sequence on OEIS and see what it converges to...)

Of course all these numbers are only non-sharp bounds, so they are sociological rather than mathematical constants and therefore a bit lame as an answer to the question. I still think the fact that such numbers appear repeatedly in analytic number theory says something interesting (and potentially deep) about the known methods we have for estimating number theoretic analytic functions. I'm not an expert on this, but would be happy to hear from someone who can add insight on this phenomenon.

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    $\begingroup$ I, too, am no expert on these matters, but I wonder whether we are dealing with results relying on "exponent pairs". E.g., we find in Sid Graham's paper, An algorithm for computing optimal exponent pairs, jlms.oxfordjournals.org/content/s2-33/2/203.full.pdf that the infimum of $k+\ell$ over all exponent pairs $(k,\ell)$ is at least $73604982/88785387$. $\endgroup$ – Gerry Myerson Sep 16 '15 at 2:39
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Not a huge denominator, but I still think it's amazing that $$\prod_{\text{$p$ prime}} \frac{p^2-1}{p^2+1} = \frac25.$$

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  • $\begingroup$ True, but noted already in a previous answer by Geoff Robinson (in the equivalent $5/2$ form) that he then deleted because indeed the fraction is so simple. To repeat my comment on that question: $\endgroup$ – Noam D. Elkies Sep 18 '15 at 14:19
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    $\begingroup$ Thus $2/5$ is the probability that $\gcd(a,b)=\gcd(c,d)$ for "random" $a,b,c,d$. math.harvard.edu/~elkies/Misc/sol1.html But $2/5$ is surely not "special" enough for the OP. Maybe the probability that $\gcd(a,b,c,d,e,f)=\gcd(i,j,k,l,m,n)$? $\endgroup$ – Noam D. Elkies Sep 18 '15 at 14:20
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    $\begingroup$ (which is $\zeta(12)/\zeta(6)^2 = 691/715$) $\endgroup$ – Noam D. Elkies Sep 18 '15 at 14:21
  • $\begingroup$ Ok, but then we are back at the discussion of Bernoulli numbers... $\endgroup$ – Johan Wästlund Sep 18 '15 at 14:25
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    $\begingroup$ The question has been designated as community wiki, which (in my opinion) means my opinion as OP should be irrelevant. For what it's worth, these are both cute examples, and yes, $691/715$ is a lovely fraction. $\endgroup$ – Dan Romik Sep 18 '15 at 16:37
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In the recent paper The density of primes dividing a particular non-linear recurrence sequence by Gorman et al, it is proved that the density of the primes dividing a certain sequence $(b_n)$ is $\frac{179}{336}$, where $(b_n)$ is defined by $b_0=1$, $b_1=2$, $b_2=1$, $b_3=-3$, and the recurrence $$ b_n = \begin{cases} \displaystyle\frac{b_{n-1}b_{n-3}-b_{n-2}^2}{b_{n-4}} & \textrm{if }n \not\equiv 2 \ \ (\textrm{mod }3),\\ \displaystyle\frac{b_{n-1}b_{n-3}-3b_{n-2}^2}{b_{n-4}} & \textrm{if }n \equiv 2 \ \ (\textrm{mod }3).\\ \end{cases} $$

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    $\begingroup$ That's a good example but it's basically the same as the "Somos sequence" example in the original question (the $b_n$ form a "Somos-4" sequence). $\endgroup$ – Noam D. Elkies Sep 14 '15 at 21:00
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    $\begingroup$ Thanks, I love these sorts of results. I edited your answer to add more detail. Note that this example is a bit similar to example 2 in my question (and there is a nonempty intersection between the authors of both papers). I'm curious if there are many other density results along these lines. $\endgroup$ – Dan Romik Sep 14 '15 at 21:00
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    $\begingroup$ Very similar, yes . . . It must be known that every (weighted) Somos sequence related with a theta function has this kind of behavior (including all "Somos-$k$" for $k=4,5,6,7$). $\endgroup$ – Noam D. Elkies Sep 14 '15 at 21:02
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This may not exactly qualify, because it's still only the conjectured answer to a question.

The question: Consider all $2$-colorings of $\{1,\dots,n\}$. What is the minimum number of monochromatic $3$-term arithmetic progressions (solutions to $x+z=2y$ where all of $x,y,z$ are the same color) formed, and what coloring achieves it?

The conjectured answer (supported by a fair amount of computational evidence): The minimum is asymptotic to $\frac{117}{2192} n^4$. The optimal coloring is to divide your interval into alternating blocks of size proportional to $$\frac{28}{548}, \frac{6}{548}, \frac{28}{548}, \frac{37}{548}, \frac{59}{548}, \frac{116}{548}, \frac{116}{548}, \frac{59}{548}, \frac{37}{548}, \frac{28}{548}, \frac{6}{548}, \frac{28}{548}.$$

The best lower bound currently known is $\frac{1675}{32768}(1+o(1)) n^4$, achieved by Parrilo, Robertson, and Saracino using semidefinite programming (this paper also contains the conjecture described here).

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    $\begingroup$ This is fascinating - is there any sense for where that factor of 137 in the denominator comes from? $\endgroup$ – Steven Stadnicki Jan 20 '18 at 20:48
  • $\begingroup$ I don't really have a sense if it has any meaning other than coincidence. The problem ends up being equivalent to minimizing $x^T A x$, where $x$ is a $\pm 1$ vector, and $a_{ij}$ is the number of progressions containing both $i$ and $j$. The conjectured answer comes from assuming (after running some experiments) a $12$ block solution with roughly these block sizes, and writing down the local optimality condition that perturbing the block sizes can't reduce the number of solutions. The original function was quadratic, so this turns into a system of linear equations in the block sizes. $\endgroup$ – Kevin P. Costello Jan 23 '18 at 20:11
  • $\begingroup$ (c'td) You can see the system on page 7 here math.ucr.edu/~costello/research/constellations.pdf . The coefficients of the system are all small integers, and the $137$ just shows up along the way when you solve it. $\endgroup$ – Kevin P. Costello Jan 23 '18 at 20:13
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[Elevated from a comment to an answer at suggestion of OP]

In my paper, Rational products of sines of rational angles, Aeq. Math. 45 (1993) 70-82, I found all the solutions of $$\sin\pi x\sin\pi y\sin\pi z\sin\pi w=r$$ in rational $x,y,z,w,r$. These included $$\sin\pi/42\sin5\pi/14\sin8\pi/21\sin10\pi/21=1/16$$ and $$\sin\pi/39\sin9\pi/26\sin14\pi/39\sin11\pi/26=1/16$$

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The best current lower bound on the packing density of regular tetrahedra is $4000/4671$, which is the density of an explicit periodic packing by Chen, Engel, and Glotzer.

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Interesting rational constant concerning base-10 normal numbers.

Let $a$ be a real number with a base-10 decimal representation $a_1a_2\ldots a_n \ldots$ Denote the number of ways to write $a_n$ as the sum of positive integers as $p(a_n)$ - also called the partition of $a_n$. I write $A(n)=\sum_{k=1}^n p(a_k)$ and $B(n)= \sum_{k=1}^n a_k$ and set $$\beta(a)=\lim_{n\to \infty}{A(n)\above 1.5 pt B(n)}$$ If $a$ is a base-10 normal number then $\beta(a)={97 \above 1.5 pt 45}$ .

Note the converse of the above statement is false!

Numerically ${97\above 1.5pt 45}$ can be written $2.15\ldots$. If we believe $\pi$ is normal then we would expect $\beta(\pi)={97 \above 1.5 pt 45}$. Up to the $500\text{ }000$-th digit $\beta(\pi)=2.153781\ldots$ See the plot of $\beta(\pi)$ below. Note the blue line is the constant value ${97\above 1.5 pt 45}$ and the orange "graph" are the values of $\beta(\pi)$.

enter image description here

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    $\begingroup$ As written the sum is infinite. If you fix it (to what's written in the stack exchange post) I would argue that this is unsurprising. You could replace $p$ with any rational-valued function and it would still be true with constant $(p(0) + p(1) + \cdots + p(9))/(0 + 1 + \cdots + 9)$. $\endgroup$ – Tim Carson Aug 22 '17 at 19:05
  • $\begingroup$ @TimCarson I finally understand what you mean by "any rational-valued function. " I understand now. $\endgroup$ – Antonio Hernández Maquívar Jun 26 '18 at 19:22
  • $\begingroup$ Actually now I don't remember why I said rational-valued. It seems like any function $p : \{0, 1, ..., 9\} \to \mathbb{R}$ would do. $\endgroup$ – Tim Carson Jul 13 '18 at 14:48

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