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?


3 Answers 3


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

  • $\begingroup$ How did you get the first equality? $\endgroup$ Oct 19, 2020 at 20:32
  • 1
    $\begingroup$ that's what Mathematica tells me. $\endgroup$ Oct 19, 2020 at 20:33
  • $\begingroup$ OK. :-) I mistyped the expression, and then Mathematica did not tell me anything good. $\endgroup$ Oct 19, 2020 at 20:35
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    $\begingroup$ after you have the answer, it is straightforward by induction (after division by $\binom{2n}n^2$) $\endgroup$ Oct 19, 2020 at 20:36

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 $$\sum_{k=1}^n\frac{(36k-5)432^{k-1}\binom{6n}{3n}\binom{3n}n}{k^2\binom{6k}{3k}\binom{3k}k}=\frac{432^n-\binom{6n}{3n}\binom{3n}n}{12}\in\mathbb Z.$$


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 $$a_n=\frac{27^n}3-\frac13\binom{2n}n\binom{3n}n=\frac{27^n}3-\binom{2n}n\binom{3n-1}{n-1}. \qquad \square$$


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