Special rational numbers that appear as answers to natural questions 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:


*

*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.

*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.)
 A: 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.  
A: Not a huge denominator, but I still think it's amazing that $$\prod_{\text{$p$ prime}} \frac{p^2-1}{p^2+1} = \frac25.$$
A: 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 
A: 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 $\frac{27812926939574093625}{7042788441228875157149}$.
A: 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.
A: Many delicate estimates in analytic number theory lead to bounds involving unusual rational numbers. Two examples I am aware of are:


*

*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.

*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.
A: 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.
A: The maximum density of a letter in an infinite
ternary squarefree word in $255/653$. See P. Ochem, Letter frequency in infinite
repetition-free words, Theoretical Computer                                     Science 380 (2007), 388--392. The minimum density of a letter in an infinite
ternary squarefree word is $883/3215$. See A. Khalyavin, The minimal density of a letter in an infinite ternary square-free word is 883/3215, J. Integer                        Sequences 10(2) (2007), Article 07.6.5.
A: 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:


*

*Henze, N. "Random Triangles in Convex Regions." J. Appl. Prob. 20, 111-125, 1983 http://www.jstor.org/stable/3213725

*http://mathworld.wolfram.com/SquareTrianglePicking.html
)
A: 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 $\pi$.
A: 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
$$
b_n = \begin{cases}
\displaystyle\frac{b_{n-1}b_{n-3}-b_{n-2}^2}{b_{n-4}} & \textrm{if }n \not\equiv 2 \ \ (\textrm{mod }3),\\
\displaystyle\frac{b_{n-1}b_{n-3}-3b_{n-2}^2}{b_{n-4}} & \textrm{if }n \equiv 2 \ \ 
(\textrm{mod }3).\\
\end{cases}
$$
A: 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).  
A: [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
$$\sin\pi/39\sin9\pi/26\sin14\pi/39\sin11\pi/26=1/16$$
A: 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.
A: $5\over 8$ths of (propositional) classical tautologies are intuitionistically valid.
This is surprising to me on multiple levels. First, it's fairly robust: a couple different reasonable approaches yield the same answer. Second, of course the fact that it's rational strikes me as deeply weird, I'd have definitely expected something quite odd. Finally, I'm quite surprised that it's greater than $1\over 2$ - my naive take would definitely be that a randomly-picked classical tautology probably isn't intuitionistically valid.
A: 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. 
A: 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.
A: 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 $\beta(\pi)$. 

A: 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.
A: 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 $\frac{16}{9}$!
A: $$\pi^{-4}\sum_1^{\infty}n^{-4}{2n\choose n}^{-1}=2\pi^{-4}\int_0^{\pi/3}x\biggl(\log\bigl(2\sin(x/2)\bigr)\biggr)^2\,dx={17\over3240}$$ 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." 
A: 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."
A: 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.
A: 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*}
A: 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.
A: For any sufficiently large even value of $n$, the Bernoulli number $B_n$ is probably a good example.
A: 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.
A: 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
A: 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
