When is $\binom{2n}{n}\cdot \frac{1}{2n}$ an integer?

Question

In a recent question here, asking about the number of necklaces of $n$ white and $n$ black beads (reworded in terms of apples and oranges), one of the naive and incorrect answers was that as there are $\binom{2n}{n}$ ways to arrange the beads in a straight line, dividing by $2n$ to account for the symmetry of the circle would correct the count.

In the case that $p$ is a prime not equal to two, it is clear to see that $\dfrac{(2p)!}{p!p!2p}$ has a factor of $p$ exactly twice in the numerator yet three times in the denominator and is thus not even an integer.

My question:

For what values of $n$ will $\binom{2n}{n}\frac{1}{2n}$ be an integer?

A similar question was asked here for the more general case of when $\frac{1}{n}\binom{n}{k}$ is an integer, generating some very nice graphics, and some special cases were mentioned such as when $\gcd(n,k)=1$, however that will never be the case in the special case I ask about (since $\gcd(2n,n)=n\neq 1$ for all $n>1$). Indeed, when looking at the lovely image made by @BrunoJoyal, one notices a black line down the center of the image, which would correspond to those positions I am interested in.

Wolfram gives us the beginnings of a sequence $1,6,15,28,42,45,66,77,91,\dots$ With the exception of $42$ and $77$ these numbers are the first hexagonal numbers, but that pattern breaks with $120$ not being in the sequence.

A longer list from the comments by @BanachTarski: There are 89 such numbers in the first 1000 natural numbers 1, 6, 15, 28, 42, 45, 66, 77, 91, 110, 126, 140, 153, 156, 170, 187, 190, 204, 209, 210, 220, 228, 231, 238, 266, 276, 299, 308, 312, 315, 322, 325, 330, 345, 378, 414, 420, 429, 435, 440, 442, 450, 459, 460, 468, 476, 483, 493, 496, 510, 527, 551, 558, 561, 570, 580, 589, 600, 609, 620, 651, 665, 682, 684, 696, 703, 740, 744, 748, 770, 777, 806, 812, 814, 851, 861, 868, 888, 902, 920, 924, 936, 943, 946, 950, 962, 966, 988, 989

Is there anything special we can say about those $n$ for which this is the case? (By imposing the extra condition on $k$ and $n$ in the generalized question, more patterns will hopefully emerge)

Answer 1

Suppose that $p$ is a prime dividing $2n$. Then introduce notation $2n=2c_pp^{k_p}$ where $c_p$ is the coprime coefficient of $p^{k_p}$ in the factorization of $n$.

The power of $p$ dividing $(2n)!$ is $$\left\lfloor\frac{2n}{p}\right\rfloor +\left\lfloor\frac{2n}{p^2}\right\rfloor +\left\lfloor\frac{2n}{p^3}\right\rfloor +\cdots =2c_pp^{k_p-1} +\left\lfloor2c_pp^{k_p-2}\right\rfloor +\left\lfloor2c_pp^{k_p-3}\right\rfloor +\cdots$$ and similarly the power of $p$ dividing $(n!)^2$ is $$2\left(c_pp^{k_p-1} +\left\lfloor c_pp^{k_p-2}\right\rfloor +\left\lfloor c_pp^{k_p-3}\right\rfloor +\cdots\right) =2c_pp^{k_p-1} +2\left\lfloor c_pp^{k_p-2}\right\rfloor +2\left\lfloor c_pp^{k_p-3}\right\rfloor +\cdots$$ So the power of $p$ dividing $\binom{2n}{n}$ is the difference: $$\left(\left\lfloor2c_pp^{k_p-2}\right\rfloor-2\left\lfloor c_pp^{k_p-2}\right\rfloor\right) +\left(\left\lfloor2c_pp^{k_p-3}\right\rfloor-2\left\lfloor c_pp^{k_p-3}\right\rfloor\right) +\cdots$$

Note that for $k_p$ large enough that $k_p-i$ is nonnegative, terms within parentheses are equal integers and contribute net zero. So this power is $$\left(\left\lfloor\frac{2c_p}{p}\right\rfloor-2\left\lfloor\frac{c_p}{p}\right\rfloor\right) +\left(\left\lfloor\frac{2c_p}{p^2}\right\rfloor-2\left\lfloor\frac{c_p}{p^2}\right\rfloor\right) +\cdots$$

These two-term expressions inside the parentheses are each either $0$ or $1$. They are $1$ precisely when $2c_p$ is between an odd multiple of $p^i$ and the next (even) multiple.

Now for $2n$ to divide this, we must have $$k_p\leq\left(\left\lfloor\frac{2c_p}{p}\right\rfloor-2\left\lfloor\frac{c_p}{p}\right\rfloor\right) +\left(\left\lfloor\frac{2c_p}{p^2}\right\rfloor-2\left\lfloor\frac{c_p}{p^2}\right\rfloor\right) +\cdots$$ when $p\neq2$ and $$\begin{align} k_2+1&\leq\left(\left\lfloor\frac{2c_2}{2}\right\rfloor-2\left\lfloor\frac{c_2}{2}\right\rfloor\right) +\left(\left\lfloor\frac{2c_2}{2^2}\right\rfloor-2\left\lfloor\frac{c_2}{2^2}\right\rfloor\right) +\cdots\\ \implies k_2&\leq\left(\left\lfloor\frac{2c_2}{2^2}\right\rfloor-2\left\lfloor\frac{c_2}{2^2}\right\rfloor\right) +\left(\left\lfloor\frac{2c_2}{2^3}\right\rfloor-2\left\lfloor\frac{c_2}{2^3}\right\rfloor\right) +\cdots\end{align}$$

Conversely if these inequalities hold for all $p$ dividing $n$ then $2n$ divides $\binom{2n}{n}$. So this rephrases the question into a question about what the relationship is between the powers of primes dividing $n$ and the coprime coefficients of those primes:

So $2n$ divides $\binom{2n}{n}$ if and only if

$k_2\leq\left(\left\lfloor\frac{2c_2}{2^2}\right\rfloor-2\left\lfloor\frac{c_2}{2^2}\right\rfloor\right) +\left(\left\lfloor\frac{2c_2}{2^3}\right\rfloor-2\left\lfloor\frac{c_2}{2^3}\right\rfloor\right) +\cdots$ and for all $p\neq2$, $k_p\leq\left(\left\lfloor\frac{2c_p}{p}\right\rfloor-2\left\lfloor\frac{c_p}{p}\right\rfloor\right) +\left(\left\lfloor\frac{2c_p}{p^2}\right\rfloor-2\left\lfloor\frac{c_p}{p^2}\right\rfloor\right) +\cdots$.

Some consequences:

• if a prime $p$ dividing $n$ is larger than twice its coprime coefficient, then $2n$ will not divide $\binom{2n}{n}$, because the right-hand side of the inequality will be $0$. This excludes numbers like $n=10,44,303$, etc.
• with the small exception of $3\cdot5$, a product of twin primes will never appear in this sequence, because $1=k_p \not\leq\left(\left\lfloor\frac{2(p+2)}{p}\right\rfloor-2\left\lfloor\frac{p+2}{p}\right\rfloor\right)=(2-2)=0$.
• more generally, $n=pq$ is in the sequence if and only if $2p$ is just above an odd multiple of $q$ and $2q$ is just above an odd multiple of $p$. This includes $n$ like $7\cdot11$, but excludes $n$ like $13\cdot19$.
• all even perfect numbers are in the sequence (note these are a special type of hexagonal number, discussed below).

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You have observed many hexagonal numbers in the sequence. That is not exactly a surprise given these inequalities. The $m$th hexagonal number is $n=m(2m-1)$. Note that $m$ and $(2m-1)$ are coprime. For small $m$, $2m-1$ is prime power more than half the time, and is just shy of an even multiple of its coprime coefficient. This gets many early hexagonal numbers past the hurdle of satisfying the inequality for its largest prime power factor.

For such $m$ (where $2m-1$ is a prime power) for smaller $p$ dividing $m$, their coprime coefficient is $\frac{m}{p^{k_p}}(2m-1)=2p^{k_p}\left(\frac{m}{p^{k_p}}\right)^2-\frac{m}{p^{k_p}}$ and so will be just below an even multiple of $p^i$ as long as $\frac{m}{p^{k_p}}<p^i$. With small $m$, and $p^{k_p}$ the largest prime power factor of $m$, this last conditions is automatic, clearing another hurdle for membership in the sequence.

For smaller prime divisors of $m$ (if there are any), it comes down to whether or not $\frac{m}{p^{k_p}}$ is just north of an even multiple of $p^i$ or an odd multiple. There's at least a 50/50 shot, but remember that many early hexagonal numbers don't even have a third prime factor. So these are all arguments why many early hexagonal numbers will be in the sequence.

Nothing said so far addresses hexagonal numbers where $2m-1$ is not prime. Some hexagonal numbers still fail the inequality, as you noted with $120$. With that number: $$1=k_3\not\leq\left(\left\lfloor\frac{80}{3}\right\rfloor-2\left\lfloor \frac{40}{3}\right\rfloor\right)+\left(\left\lfloor\frac{80}{9}\right\rfloor-2\left\lfloor\frac{40}{9}\right\rfloor\right)+\left(\left\lfloor\frac{80}{27}\right\rfloor-2\left\lfloor\frac{40}{27}\right\rfloor\right)=(26-26)+(8-8)+(2-2)=0$$ just barely fails. Here the issue boils down to $81$ (a power of $3$) being ever so slightly larger than $80$ (twice $3$'s coefficient. So the left term in the parentheses pairs never gets a chance to beat out the right term.

Here is a table investigating early hexagonal numbers. The above paragraphs "guarantee" that $n=m(2m-1)$ is part of this sequence when $2m-1$ and $m$ are prime powers.

m 2m-1 n 2m-1 prime power? factorization n in sequence 2 3 6 yes pq guaranteed 3 5 15 yes pq guaranteed 4 7 28 yes p^2q guaranteed 5 9 45 yes p^2q guaranteed 6 11 66 yes pqr yes 7 13 91 yes pq guaranteed 8 15 120 no p^3qr no 9 17 153 yes p^2q guaranteed 10 19 190 yes pqr yes 11 21 231 no pqr yes 12 23 276 yes p^2qr yes 13 25 325 yes p^2q guaranteed 14 27 378 yes pq^3r yes 15 29 435 yes pqr yes 16 31 496 yes p^4q guaranteed 17 33 561 no pqr yes 18 35 630 no pq^2rs no 19 37 703 yes pq guaranteed

Note that the two failures have high exponent ($5$) and that there are just as many successes which also have exponent $5$.

Summary of thoughts on hexagonals: the factorization of hexagonal numbers gives them a bump over general integers to be part of this sequence. I'm not sure that bump will survive into larger hexagonal numbers, since they will have a more broad array of prime power factors.

Answer 2

Just for the purpose of answering "for what values" part, let's note $$K_n=\binom{2n}{n}\cdot \frac{1}{2n}$$ Then

$$K_n=\frac{1}{2n} \cdot \frac{2n \cdot (2n-1)\cdot ... \cdot (n+1)}{1 \cdot 2 \cdot ... \cdot n}=$$ $$\frac{2n-1}{n^2} \cdot \frac{(2n-2)\cdot ... \cdot (n+1) \cdot n}{1 \cdot 2 \cdot ... \cdot (n-1)}=\frac{2n-1}{n^2} \cdot \binom{2(n-1)}{n-1}$$

From this perspective: $$K_n \in \mathbb{N} \Leftrightarrow n^2 \mid \binom{2(n-1)}{n-1}$$ because $\gcd(n^2, 2n-1)=1$

This leads to another observation: $$2n-2 \equiv -2 \pmod{n}$$ $$2n-3 \equiv -3 \pmod{n}$$ $$...$$ $$n+1=2n-(n-1) \equiv -(n-1) \pmod{n}$$ So that: $$(2n-2)\cdot ... \cdot (n+1) \equiv (-1)^{n-2} (n-1)! \pmod{n}$$

Or $$(2n-2)\cdot ... \cdot (n+1)=n \cdot Q + (-1)^{n-2} (n-1)!$$ And $$K_n=\frac{2n-1}{n^2} \cdot \binom{2(n-1)}{n-1}=\frac{2n-1}{n}\cdot \left ( \frac{n \cdot Q}{(n-1)!} +(-1)^{n-2} \right )$$

The only time this is not true is, when: $$K_n \in \mathbb{N} \Leftrightarrow (n-1)! \equiv 0 \pmod{n}$$

E.g. $K_6 \in \mathbb{N}$ because $5! \equiv 0 \pmod{6}$

$K_{15} \in \mathbb{N}$ because $14! \equiv 0 \pmod{15}$

...

It is obvious that $n$ can not be prime, because, according to Wilson's theorem we would have $$(n-1)! \equiv -1 \pmod{n}$$

Answer 3

A little long for a comment, but not quite an answer...

A necessary, but woefully insufficient, condition is that the number of "$1$"s in the binary representation of $n$ must exceed the power of $2$ in $2n$. See the OEIS: 1's-counting sequence ("a(n) is also the largest integer such that 2^a(n) divides binomial(2n, n)").

So, for $n=4$, the number of $1$s in its binary representation is $1$, but $2\cdot4 = 2^3$, so $2n$ does not divide $\binom{2n}{n}$. With small numerical experiments, this seems to kill about half the integers.

Sadly, I don't know a similar characterization for any other primes.