11
$\begingroup$

The following sequence appears to be always an integer, experimentally.

QUESTION. Let $n\in\mathbb{Z}^{+}$. Are these indeed integers? $$\sum_{k=1}^n\frac{(4k - 1)4^{2k - 1}\binom{2n}n^2}{k^2\binom{2k}k^2}.$$

POSTSCRIPT. After Carlo's cute response and several useful comments, I like to ask this: is there a combinatorial proof?

$\endgroup$
16
$\begingroup$

$$\sum_{k=1}^n\frac{(4k - 1)4^{2k - 1}\binom{2n}n^2}{k^2\binom{2k}k^2}=16^n \left(1-\frac{\Gamma \left(n+\frac{1}{2}\right)^2}{\pi \Gamma (n+1)^2}\right)$$ $$\qquad=2^{4n}-c_n^2,\;\;\text{with}\;\;c_n=2^n\frac{(2n-1)!!}{n!}={{2n}\choose n}. \qquad\qquad\text{[thanks, Pietro Majer]}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How did you get the first equality? $\endgroup$ – Iosif Pinelis Oct 19 at 20:32
  • 1
    $\begingroup$ that's what Mathematica tells me. $\endgroup$ – Carlo Beenakker Oct 19 at 20:33
  • $\begingroup$ OK. :-) I mistyped the expression, and then Mathematica did not tell me anything good. $\endgroup$ – Iosif Pinelis Oct 19 at 20:35
  • 6
    $\begingroup$ after you have the answer, it is straightforward by induction (after division by $\binom{2n}n^2$) $\endgroup$ – Fedor Petrov Oct 19 at 20:36
5
$\begingroup$

Actually it is easy to give some other similar identities. For example, $$\sum_{k=1}^n\frac{(9k-2)27^{k-1}\binom{2n}n\binom{3n}n}{k^2\binom{2k}k\binom{3k}k}=\frac{27^n}3-\binom{2n}n\binom{3n-1}{n-1}\in\mathbb Z.$$ Also, $$\sum_{k=1}^n\frac{(16k-3)64^{k-1}\binom{4n}{2n}\binom{2n}n}{k^2\binom{4k}{2k}\binom{2k}k}=\frac{64^n-\binom{4n}{2n}\binom{2n}n}4\in\mathbb Z$$ and $$\sum_{k=1}^n\frac{(36k-5)432^{k-1}\binom{6n}{3n}\binom{3n}n}{k^2\binom{6k}{3k}\binom{3k}k}=\frac{432^n-\binom{6n}{3n}\binom{3n}n}{12}\in\mathbb Z.$$

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

There is a way to prove Zhi-Wei Sun's identity as well as Carlo Beenakker's identity. Of course, both can be treated in accord with Fedor Petrov's induction. Let's focus on Sun's identity. Divide through by $\binom{2n}n\binom{3n}n$ to write $$A_n:=\sum_{k=1}^n\frac{(9k-2)27^{k-1}}{k^2\binom{2k}k\binom{3k}k}=\frac{27^n}{3\binom{2n}n\binom{3n}n}-\frac13. \tag1$$ so that $$A_n-A_{n-1}=\frac{(9n-2)27^{n-1}}{n^2\binom{2n}n\binom{3n}n}.$$ Let $a_n=\binom{2n}n\binom{3n}nA_n$ (which is exactly Sun's LHS) to get the recursive equation $$n^2a_n-3(3n-1)(3n-2)a_{n-1}=(9n-2)27^{n-1}.\tag2$$ First, we find a solution to the homogeneous equation $n^2a_n-3(3n-1)(3n-2)a_{n-1}=0$ as follows $$a_n^{(h)}=\binom{2n}n\binom{3n}n. \tag4$$ A particular solution to the non-homogeneous equation (2) can be determined by mimicking the RHS as $a_n^{(p)}=(bn+c)27^n$. Now, plug this back in (2) to solve for $b$ and $c$: \begin{align*} n^2(bn+c)27^n-3(3n-1)(3n-2)(bn-b+c)27^{n-1}&=(9n-2)27^{n-1} \\ \iff 27n^2(bn+c)-3(3n-1)(3n-2)(bn-b+c)&=9n-2 \\ \iff \qquad b=0 \qquad \text{and} \qquad c=\frac13. \end{align*} Therefore, the general solution takes the form $$a_n=a_n^{(p)}+\beta\,a_n^{(h)}=\frac{27^n}3+\alpha\binom{2n}n\binom{3n}n.$$ Since $a_0=A_0=0$, we compute $\beta=-\frac13$ and hence $$a_n=\frac{27^n}3-\frac13\binom{2n}n\binom{3n}n=\frac{27^n}3-\binom{2n}n\binom{3n-1}{n-1}. \qquad \square$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.