The classical rational Ramanujan-type series for $1/\pi$ have the following four forms: \begin{align}\sum_{k=0}^\infty(ak+b)\frac{\binom{2k}k^3}{m^k}&=\frac{c}{\pi},\label{1}\tag{1} \\\sum_{k=0}^\infty(ak+b)\frac{\binom{2k}k^2\binom{3k}k}{m^k}&=\frac{c}{\pi},\label{2}\tag{2} \\\sum_{k=0}^\infty(ak+b)\frac{\binom{2k}k^2\binom{4k}{2k}}{m^k}&=\frac{c}{\pi},\label{3}\tag{3} \\\sum_{k=0}^\infty(ak+b)\frac{\binom{2k}k\binom{3k}k\binom{6k}{3k}}{m^k}&=\frac{c}{\pi},\label{4}\tag{4} \end{align} where $a,b,m\in\mathbb Z$, $am\not=0$ and $c^2\in\mathbb Q$. It is known that there are totally 36 such series, see, e.g., Chapter 14 of S. Cooper's book Ramanujan's Theta Functions (Springer, 2017).

For a positive integer $m$, can we find similar series for $(\log m)/\pi$? Motivated by Ramanujan-type series of the forms \eqref{1}-\eqref{4}, I have formulated the following general conjecture which involves the so-called harmonic numbers $$H_n=\sum_{0<k\le n}\frac1k\ \ \ \ (n=0,1,2,\ldots).$$

**Conjecture.** (i) If we have an identity \eqref{1} with $a,b,m\in\mathbb Z$, $am\not=0$ and $c^2\in\mathbb Q$, then
$$\sum_{k=0}^\infty\frac{\binom{2k}k^3}{m^k}(6(ak+b)(H_{2k}-H_k)+a)=c\frac{\log|m|}{\pi}.\label{I}\tag{I}$$

(ii) If we have an identity \eqref{2} with $a,b,m\in\mathbb Z$, $am\not=0$ and $c^2\in\mathbb Q$, then $$\sum_{k=0}^\infty\frac{\binom{2k}k^2\binom{3k}k}{m^k}((ak+b)(3H_{3k}+2H_{2k}-5H_k)+a)=c\frac{\log|m|}{\pi}.\label{II}\tag{II}$$

(iii) If we have an identity \eqref{3} with $a,b,m\in\mathbb Z$, $am\not=0$ and $c^2\in\mathbb Q$, then $$\sum_{k=0}^\infty\frac{\binom{2k}k^2\binom{4k}{2k}}{m^k}(4(ak+b)(H_{4k}-H_k)+a)=c\frac{\log|m|}{\pi}.\label{III}\tag{III}$$

(iv) If we have an identity \eqref{4} with $a,b,m\in\mathbb Z$, $am\not=0$ and $c^2\in\mathbb Q$, then $$\sum_{k=0}^\infty\frac{\binom{2k}k\binom{3k}{k}\binom{6k}{3k}}{m^k}(3(ak+b)(2H_{6k}-H_{3k}-H_k)+a)=c\frac{\log|m|}{\pi}.\label{IV}\tag{IV}$$

**Examples**. (i) Ramanujan [Quart. J. Math. 45(1914), 350-372] found that
$$\sum_{k=0}^\infty\frac{\binom{2k}k^3}{(-512)^k}=\frac{2\sqrt2}{\pi},$$
in view of this and part (i) of the above conjecture we should have
$$\sum_{k=0}^\infty\frac{\binom{2k}k^3}{(-512)^k}((6k+1)(H_{2k}-H_k)+1)=\frac{2\sqrt2}{\pi}\times\frac{\log512}6=3\sqrt2\frac{\log2}{\pi}.$$

(ii) Ramanujan [Quart. J. Math. 45(1914), 350-372] found that $$\sum_{k=0}^\infty(51k+7)\frac{\binom{2k}k^2\binom{3k}k}{(-12)^{3k}}=\frac{12\sqrt3}{\pi},$$ in view of this and part (ii) of our general conjecture we should have $$\sum_{k=0}^\infty\frac{\binom{2k}k^2\binom{3k}k}{(-12)^{3k}}((51k+7)(3H_{3k}+2H_{2k}-5H_k)+51) =\frac{12\sqrt3}{\pi}\times\log 12^3=36\sqrt3\frac{\log 12}{\pi}.$$

**QUESTION.** Can one modify the known appraoches to classical Ramanujan-type series for $1/\pi$ to prove the above general conjecture?

It seems that our conjecture also applies to irrational Ramanujan-type series for $1/\pi$. For example, Ramanujan [Quart. J. Math. 45(1914), 350-372] found that $$\sum_{k=0}^\infty\left(k+\frac{31}{270+48\sqrt5}\right)\frac{\binom{2k}k^3}{(2^{20}/(\sqrt5-1)^8)^k}=\frac{16}{(15+21\sqrt5)\pi},$$ motivated by this and part (i) of our general conjecture we guess that \begin{align}&\sum_{k=0}^\infty\frac{\binom{2k}k^3}{(2^{20}/(\sqrt5-1)^8)^k}\left(6\left(k+\frac{31}{270+48\sqrt5}\right)(H_{2k}-H_k)+1\right) \\=&\ \frac{16}{(15+21\sqrt5)\pi}\times\log\frac{2^{20}}{(\sqrt5-1)^8}, \end{align} which can be easily checked via Mathematica.

We also have conjectures on $p$-adic congruences corresponding to identities of the forms \eqref{I}-\eqref{IV}.

Your comments are welcome!