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What are some examples of infinite sums, products or definite integrals that have conjectural closed form representations that were confirmed by numerical calculations up to whatever maximum precision anybody tried but still remain unproved?

I am also interested in values of special functions at certain points that have conjectural representations in terms of simpler functions (e.g. special values of hypergeometric functions, Meijer G-function or Fox H-function that representable in terms of elementary functions and well-known constants like $\pi$, $e$, Catalan, Euler–Mascheroni, Glaisher–Kinkelin or Khinchin).

To give some examples:

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    $\begingroup$ The book Experimental Mathematics in Action has a few conjectures of a similar flavor. It's a nice read! $\endgroup$
    – 77474
    May 21, 2013 at 16:34
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    $\begingroup$ Since you seem very interested in integrals, maybe you should have a look at Victor Moll's homepage. He has written a fun book called Irresistible Integrals. $\endgroup$
    – 77474
    May 23, 2013 at 17:22
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    $\begingroup$ Simon Plouffe's website also has many conjectures. $\endgroup$
    – Potato
    May 24, 2013 at 18:14
  • $\begingroup$ What about products ?? $\endgroup$
    – mick
    Oct 12, 2013 at 0:16
  • $\begingroup$ I believe the last one of the OP is actually proven somewhere if I recall correctly. $\endgroup$
    – mick
    Nov 22, 2023 at 12:23

3 Answers 3

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Some conjectural formulas for $\pi$ are given in:

For example (I expressed an infinite sum given in the paper in terms of hypergeometric functions): $$224 \, _5F_4\left(\frac{1}{3},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{2}{3 };1,1,1,1;8235 \sqrt{5}-18414\right)\\-100 \sqrt{5} \, _5F_4\left(\frac{1}{3},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{2}{3 };1,1,1,1;8235 \sqrt{5}-18414\right)\\-1655540 \, _5F_4\left(\frac{4}{3},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{5}{3 };1,2,2,2;\frac{27}{8} \left(5 \sqrt{5}-11\right)^3\right)\\+740380 \sqrt{5} \, _5F_4\left(\frac{4}{3},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{5}{3 };1,2,2,2;\frac{27}{8} \left(5 \sqrt{5}-11\right)^3\right)\\-1237563 \, _5F_4\left(\frac{4}{3},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{5}{3 };2,2,2,2;\frac{27}{8} \left(5 \sqrt{5}-11\right)^3\right)\\+553455 \sqrt{5} \, _5F_4\left(\frac{4}{3},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{5}{3 };2,2,2,2;\frac{27}{8} \left(5 \sqrt{5}-11\right)^3\right)\stackrel{?}{=}\frac{4}{\pi^2}$$

You can also look at:


And one more from J. Guillera homepage: $$\sum_{n=0}^\infty\frac{(5418n^2+693n+29)(6n)!}{(-23887872000)^n n!^6}\stackrel?=\frac{128\sqrt5}{\pi^2}.$$

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Many such conjectures can be seen in Borwein-Crandall's recent text, Closed Forms: What They Are and Why We Care. One cited case is the identity $$F(3,5)= \frac{15}{\pi^2} \sum_{n=0}^\infty \binom{2n}{n}^2 \frac{(1/16)^{2n+1}}{2n+1},$$ in which $F(x,y)$ denotes the $4$-dimensional lattice sum: $$F(x,y)=(1+x)(1+y)\sum_{a,b,c,d} \frac{(-1)^{a+b+c+d}}{\left((6a+1)^2+x(6b+1)^2+y(6c+1)^2+xy(6d+1)^2\right)^2}.$$ Both expressions have been numerically estimated; each is approximately $0.0628326$.

A second example of historical interest is the well-known Wallis product: $$\frac{\pi}{2} =\frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5}\cdot \frac{6}{7}\cdot\frac{8}{7}\cdots \frac{2n}{2n-1}\cdot \frac{2n}{2n+1}\cdots$$ Wallis conjectured the given value $\pi/2$ after lengthy calculation in 1655. Methods did not exist to derive such a result at least until the days of Euler (about 80 years later).

Wallis' product is notable in its relation to Stirling's approximation (i.e. completing the transition between the weak and strong Stirling approximations), and for its role in determining $$\zeta'(0)=-\log(2\pi)/2.$$

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You might want to have a look at the work of Zhi-Wei Sun and especially his 181 Conjectural Series for $\pi$ and Other Constants. Some examples:

$$\sum_{k=1}^\infty\frac{(10k-3)8^k}{k^3\binom{2k}{k}^2\binom{3k}{k}}=\frac{\pi^2}{2},$$

$$\sum_{k=1}^\infty\frac{(28k^2-18k+3)(-64)^k}{k^5\binom{2k}{k}^4\binom{3k}{k}}=-14\zeta(3),$$

$$\sum_{n=1}^\infty\frac{16n+5}{12^n}\sum_{k=1}^\infty\binom{2k}{k}\binom{2(n-k)}{n-k}\binom{-1/4}{k}\binom{-3/4}{n-k}=\frac{8}{\pi}.$$

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  • $\begingroup$ You seem to have a typo; the last identity's first summation should have $ n $ as the index, not $ k $. $\endgroup$
    – Jon Claus
    May 23, 2013 at 4:54
  • $\begingroup$ @JonClaus: thanks. $\endgroup$
    – 77474
    May 23, 2013 at 5:03

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