Recently the following identity was conjectured by The Ramanujan Machine:

$$ \frac{8}{7\zeta(3)}=1-\frac{u_1}{v_1-\frac{u_2}{v_2-\frac{u_3}{v_3-\ddots}}}, $$ where $u_n=n^6$ and $v_n=(2n+1)(3n^2+3n+1)$.

It has not yet been proven (authors are not aware of a proof), however this is somewhat similar to known cfrac in this answer, the main difference there being $v_n=(2n+1)(17n^2+17n+5)$.

I couldn't prove it (to be expected), but I've found a following observation. The polynomial $(2n+1)(17n^2+17n+5)$ above is in various related articles used in certain recurrence which is satisfied by an integer sequence (see for example section 3 in Apéry’s theorem. Thirty years after): $$ (n+1)^3r_{n+1}-(2n+1)(17n^2+17n+5)r_n+n^3r_n=0. $$ But in the paper of A Proof that Euler Missed (footnote page 202) we can find the conjectured polynomial $(2n+1)(3n^2+3n+1)$ to be in similar recurrence as well (after shift on $n$):

$$ (n+1)^3r_{n+1}-2(2n+1)(3n^2+3n+1)r_n-(4n+1)4n(4n-1)r_{n-1}=0. $$ This recurrence is mentioned to be satisfied by integer sequence $r_n=\sum_{k=0}^n\binom{n}{k}^4$ (apparently with $r_0=1$ and $r_1=2$).

Can this observation be any helpful in proving the conjecture, or is this just coincidence? Maybe do you see another way to prove it? (the problem has not been open for that long, so who knows...)

EDIT: One more instance of the same polynomial, we have $$ \sum_{k=0}^{\infty} \frac{1}{(3k+1)^3}-\sum_{k=0}^{\infty} \frac{1}{(3k+2)^3}=\frac{4\pi^3\sqrt{3}}{243}=\frac{1}{P(1)+\frac{3\cdot 1^6}{P(2)+\frac{3\cdot 2^6}{P(3)+\dots}{}}} $$ where $P(n)=6n^3-9n^2+5n-1=(2n-1)(3n^2-3n+1)$ (the first equality is well known, the second equality is found in https://lfant.math.u-bordeaux.fr/seminar/slides/2019-11-05T10:00--Henri_Cohen.pdf ). Also notice $$ \frac{1}{9}((3n+1)^3+(3n+2)^3)=(2n+1)(3n^2+3n+1). $$

  • 1
    $\begingroup$ Damn, this Ramanujan project is neat. $\endgroup$
    – Evariste
    May 5, 2021 at 23:29

1 Answer 1


Finally! Found this result buried in the literature... In a French article Sur l’accélération de la convergence de certaines fractions continues by Christian Batut and Michel Olivier from 1980, a method for accelerating continued fractions for series of form $\sum_{n=1}^{\infty}\frac{1}{f(n)}$ is described. In examples we then find accelerated version of well-known $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^3}=\frac{7}{8} \zeta(3)$ which gives the desired continued fraction (the accelaration process and meaning of sequences $a_{n,k},b_{n,k},r_{n,k},R_{n,k},d_{n,k}$ is described earlier in the article):

3.2.1 - Pour la série $S=\sum_{n=1}^{\infty}\frac{1}{f(n)}$, on obtient: \begin{align} a_{n+1,k}&=f(n)+f(n+1)+2ak(k+1)(2n+1)\\ b_{n+1,k}&=-f(n)^2\\ r_{n,k}&=-an^3+2a(k+1)n^2-[2a(k+1)^2+b]n+(k+1)[(k+1)^2a+b]\\ R_{n,k}&=an^3+2a(k+1)n^2+[2a(k+1)^2+b]n+(k+1)[(k+1)^2a+b]\\ d_{n,k}&=-(k+1)^2[(k+1)^2a+b]^2 \end{align} Notons que: $r_{n+1,k-1}=-r_{k,n}$, d'oú il résulte que le tableau des $(\frac{p_{n,k}}{q_{n,k}})_{n \geq 0, k \geq 0}$ est symétrique.

3.2.2 - Cas particuliers

En remplacant dans le cas précédent, $n$ par $n + \frac{b}{a}$ et en multipliant par $a^3$ on obtient les suites $(r_{n,k}),(a_{n,k}),(b_{n,k})$ et $(R_{n,k})$ correspondant au polynome $f(n)=(an+b)^3$.


3.2.4 - $S=\sum_{n=1}^{\infty}\frac{1}{(2n-1)^3}$

Dans (3.2.2) on choisit $a=2, b=-1$. On construit ainsi la fraction continue: $$ \frac{7}{8} \zeta(3)=\frac{1}{p(0)-\frac{1^6}{p(1)-\frac{2^6}{p(2)-\ddots}}} $$ ou $p(n)=6n^3+9n^2+5n+1$.

Here $6n^3+9n^2+5n+1=(2n+1)(3n^2+3n+1)$.


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