It is easily verifiable that $$\sum_{k\geq0}\binom{2k}k\frac1{2^{3k}}=\sqrt{2}.$$ It is not that difficult to get $$\sum_{k\geq0}\binom{4k}{2k}\frac1{2^{5k}}=\frac{\sqrt{2-\sqrt2}+\sqrt{2+\sqrt2}}2.$$

Question. Is there something similarly "nice" in computing $$\sum_{k\geq0}\binom{8k}{4k}\frac1{2^{10k}}=?$$ Perhaps the same question about $$\sum_{k\geq0}\binom{16k}{8k}\frac1{2^{20k}}=?$$

NOTE. The powers of $2$ are selected with a hope (suspicion) for some pattern.

  • 11
    $\begingroup$ I don't see how $3$ fits the pattern in $3,5,10,20,...$. $\endgroup$
    – user21820
    Oct 8, 2018 at 18:18

4 Answers 4


It is moderately nice, I would say. We have $\sum \binom{2k}k x^k=(1-4x)^{-1/2}$ for $|x|<1/4$. If we need only terms with $k$ divisible by 4 and $x=2^{-5/2}$, $4x=2^{-1/2}$, we get $$\sum \binom{8k}{4k}2^{-10k}=\frac14\sum_{w^4=1}(1-w/\sqrt{2})^{-1/2}$$ and so on.


More generally, it seems that $$ \sum_{k\ge0}\binom{2^{j+1}k}{2^jk}2^{-a_{j+2} k}=2^{-j}\sum_{w^{2^j}=1}(1-w/\sqrt2)^{-1/2} $$ where $a_1=2,a_2=3,a_{j+1}=a_j+a_{j-1}+\dots+a_1$ (cf. A257113).


Let $$f(x):=\sum_{k\ge 0}\binom{2k}{k}x^k=\frac{1}{\sqrt{1-4x}}$$ with $|x|<1/4$. We have for $N,j\in\mathbb{Z}_+$, \begin{align} \frac{1}{N}\sum_{r=1}^Ne(-{jr}/{N})f\left(xe({r}/{N})\right)&=\frac{1}{N}\sum_{r=1}^Ne(-{jr}/{N})\sum_{k\ge 0}\binom{2k}{k}x^ke({kr}/{N})\\ &=\sum_{k\ge 0}\binom{2k}{k}x^k\frac{1}{N}\sum_{r=1}^Ne({(k-j)r}/{N}), \end{align} where $e(x):=e^{2\pi{\rm i}x}$. Hence clearly, $$\sum_{\substack{k\ge 0\\ k\equiv j\pmod N}}\binom{2k}{k}x^k=\frac{1}{N}\sum_{r=1}^N\frac{e(-{jr}/{N})}{\sqrt{1-4xe(r/N)}},$$ holds for all $N,j\in\mathbb{Z}_+$, which is a more general result.

  • $\begingroup$ Well I would say under the "and so on" at the end of another answer is most probably understood (more or less an equivalent of) this more general formula. $\endgroup$ Jan 24, 2019 at 18:14
  • $\begingroup$ @მამუკაჯიბლაძე Yes, you are right. The above is just a very usual and basic technique. $\endgroup$
    – Zhou
    Jan 25, 2019 at 1:15

Mathematica says, for the first:

$$\, _4F_3\left(\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8};\frac{1}{4},\frac{1}{2},\frac{ 3}{4};\frac{1}{4}\right)$$


$$\, _8F_7\left(\frac{1}{16},\frac{3}{16},\frac{5}{16},\frac{7}{16},\frac{9}{16},\frac{11}{16} ,\frac{13}{16},\frac{15}{16};\frac{1}{8},\frac{1}{4},\frac{3}{8},\frac{1}{2},\frac{5}{8}, \frac{3}{4},\frac{7}{8};\frac{1}{16}\right) $$ for the second.

  • 2
    $\begingroup$ What's with the downvotes? $\endgroup$
    – Igor Rivin
    Oct 9, 2018 at 0:36
  • $\begingroup$ Is the answer wrong? $\endgroup$
    – Igor Rivin
    Oct 9, 2018 at 3:25
  • 7
    $\begingroup$ It's not wrong, but I think this is obvious from the definition of the generalized hypergeometric function ${}_pF_q$, and doesn't seem to be a helpful step toward a closed form. (I didn't downvote, fwiw.) $\endgroup$ Oct 9, 2018 at 5:58
  • $\begingroup$ @DavidZhang Interestingly, for the first two cases, mathematica gives non-hypereometric forms. $\endgroup$
    – Igor Rivin
    Oct 9, 2018 at 16:25

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