We have discussed a proof of Chudnovsky series for $1/\pi$ in this post based on Ramanujan's ideas as presented in this post. However a serious look at one of the pages from his lost notebook suggests that Ramanujan used a slightly different approach to obtain Chudnovsky type series and he also performed all the desired calculations needed to get the series in explicit form. This is what we intend to discuss in the current post.
Bruce C. Berndt and his collaborators tried to make some guesses and discovered correctly the functions $Q,R$ as well the cryptic expression involved and its relation to the integers listed at the beginning of each row in the table. Based on their research let us demystify the above page and to that end we assume $n$ to be a positive integer and define
\begin{align}
P(q)&= 1-24\sum_{j=1}^{\infty}\frac{jq^j}{1-q^j}\tag{1}\\
Q(q)&= 1+240\sum_{j=1}^{\infty}\frac{j^3q^j}{1-q^j}\tag{2}\\
R(q)&= 1-504\sum_{j=1}^{\infty}\frac{j^5q^j}{1-q^j}\tag{3}
\end{align}
and $$P_n=P(-e^{-\pi\sqrt{n}}),Q_n=Q(-e^{-\pi\sqrt{n}}),R_n=R(-e^{-\pi\sqrt{n}}).\tag{4}$$ The table by Ramanujan above then consists of relations between $Q_n,R_n$ for different values of $n$ listed in first column. The cryptic expression in last column is $(\sqrt{n}P_n -(6/\pi))/\sqrt{Q_n}$. Ramanujan barely writes this expression with just first term $1$ for $P_n$ and two terms for $Q_n$. Let us then rewrite the table by Ramanujan as
\begin{array}{|c|c|c|}
\hline
n& \text{Relation between } Q_n\text{ and } R_n&\frac{\sqrt{n} P_n-(6/\pi)}{\sqrt{Q_n}}\\
\hline
11& (8^3+27)Q_n^3-8^3R_n^2=0&\sqrt{2}\\
\hline
19&(8^3+1)Q_n^3-8^3R_n^2=0&\sqrt{6}\\
\hline
27&(40^3+9)Q_n^3-40^3R_n^2=0&3\sqrt{\frac{6}{5}}\\
\hline
43&(80^3+1)Q_n^3-80^3R_n^2=0&6\sqrt{\frac{3}{5}}\\
\hline
67&(440^3+1)Q_n^3-440^3R_n^2=0&19\sqrt{\frac{6}{55}}\\
\hline
163&(53360^3+1)Q_n^3-53360^3R_n^2=0&362\sqrt{\frac{3}{3335}}\\
\hline
35&((60+28\sqrt{5})^3+27)Q_n^3-(60+28\sqrt{5})^3R_n^2=0&(2+\sqrt{5})\sqrt{\frac{2}{\sqrt{5}}} \\
\hline
\end{array}
The table thus deals with values of Ramanujan's functions $P(-q),Q(-q),R(-q)$ with nome $q=e^{-\pi\sqrt{n}}$ for certain integer values of $n$. These functions can be evaluated in terms of elliptic moduli and integrals associated with nome $q$ as follows
\begin{align}
P(-q)&=\left(\frac{2K(k)}{\pi}\right)^2\left(\frac{6E(k)}{K(k)}+4k^2-5\right)\tag{5a}\\
Q(-q)&=\left(\frac{2K(k)}{\pi}\right)^4\left(1-4G^{-24}\right)\tag{5b}\\
R(-q)&=\left(\frac{2K(k)}{\pi}\right)^6\left(1-2k^2\right)\left(1+8G^{-24}\right)\tag{5c}\\
\end{align}
where $G=(2kk')^{-1/12}$ is one of Ramanujan's class invariants, $k\in(0,1)$ is the elliptic modulus, $k'=\sqrt{1-k^2}$ is the complementary modulus and elliptic integrals $K,E$ are given by $$K(k)=\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}},E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2x}\,dx\tag{6}$$ and nome $q$ is related to elliptic integrals via $$q=\exp\left(-\pi\frac{K(k')}{K(k)}\right).$$
The function $P(q)$ is crucial in obtaining series for $1/\pi$ as decribed by Ramanujan in his famous paper "Modular equations and approximations to $\pi$" and I have discussed his approach in detail in my blog posts (see posts starting from here). The approach described in these posts deals with values of $P(q^2)$ and its connection with Dedekind's eta function and elliptic integral $K=K(k)$
\begin{align}
\eta(q)&=q^{1/24}\prod_{j=1}^{\infty}(1-q^j)\tag{7a}\\
P(q^2)&=12q\frac{d}{dq}\log\eta(q^2)\tag{7b}\\
P(q^2)&= \left(\frac{2K}{\pi}\right)^2(1-2k^2)+\frac{3kk'^2}{2}\frac{d}{dk} \left(\frac{2K}{\pi}\right)^2\tag{7c}
\end{align}
We also have the relation $$2P(q^2)-P(-q)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)\tag{8}$$ and thus the values of $P_n=P(-q)$ given in lost notebook can be used to calculate the values of $P(q^2)$ and one can follow the procedure mentioned in Ramanujan's paper to obtain series for $1/\pi$ by choosing a suitable hypergeometric series for $(2K/\pi)^2$. However it is better to deal with the function $P(-q)$ directly and while doing so we will also see the usefulness of the relationship between $Q_n,R_n$ given in table above.
Let us now take another elliptic modulus $l$ which is of degree $n$ over $k$ so that its nome is $q^n$ and the corresponding elliptic integral is denoted by $L$ and $L'/L=nK'/K$. Then we have $$\eta^{24}(-q^n)=-2^{-4}(2L/\pi)^{12}(ll')^{2}$$ Next we have $$nP(-q^n) =q\frac{d} {dq} \log\eta^{24}(-q^n)$$ so that \begin{align} nP(-q^n) - P(-q) &=q\frac{d} {dq} \log\frac{\eta^{24}(-q^n)}{\eta^{24}(-q)} \notag\\ &=q\frac{dk} {dq} \frac{d} {dk} \log\left\{\left(\frac{L} {K} \right) ^{12}\left(\frac{ll'} {kk'} \right) ^2\right\} \notag\\ &=\left(\frac{2K}{\pi}\right) ^2kk'^2\frac{d} {dk} \log\left\{\left(\frac{L} {K} \right) ^{6}\frac{ll'} {kk'} \right\} \notag\\ &=\frac{4KL}{\pi^2}mkk'^2\frac{d} {dk} \log\frac{ll'} {m^6kk'}\notag \end{align} where $m=K/L$ is the multiplier. Since $m$ as well as $l$ are algebraic functions of $k$ it follows that $$nP(-q^n) - P(-q) =\frac{4KL} {\pi^2}B_n(l, k)\tag{9} $$ where $B_n(l, k) $ is an algebraic function of $l, k$. If we put $l=k'$ so that $L=K'$ and $K'/K=1/\sqrt{n}$ the $q=e^{-\pi/\sqrt {n}} $ and we get $$nP(-e^{-\pi\sqrt{n}} ) - P(-e^{-\pi/\sqrt{n}} ) =\left(\frac{2L}{\pi}\right)^2B_n\tag{10}$$ where $L$ corresponds to nome $e^{-\pi\sqrt{n}} $ and $B_n$ is some algebraic number dependent on $n$.
Next note that $$\frac{\eta^{24}(-e^{-\pi\sqrt{n}})} {\eta^{24}(-e^{-\pi/\sqrt{n}}) } =\left(\frac{K'} {K} \right) ^{12}=n^{-6}$$ and logarithmic differentiation with respect to $n$ gives us $$nP(-e^{-\pi\sqrt{n}}) +P(-e^{-\pi/\sqrt{n}}) =\frac{12\sqrt{n}}{\pi}\tag{11}$$ Using equations $(10),(11)$ we see that $$P(-e^{-\pi\sqrt{n}}) =\frac{6}{\pi\sqrt{n}}+\left(\frac{2K}{\pi} \right) ^2\frac{B_n}{2n}$$ where elliptic integral $K$ corresponds to nome $e^{-\pi\sqrt{n}} $. Ramanujan perhaps calculated certain values of this expression for different values of $n$ and figured out that multiplying the above expression by $\sqrt{n} $ and further dividing the result by $\sqrt{Q(-q)} $ to get rid of factor $(2K/\pi) ^2 $ leads to very simple algebraic numbers. And this probably led to the table of values of the expression $(\sqrt{n} P_n-(6/\pi)) /\sqrt{Q_n} $.
Ramanujan gave the value $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n}P_n-\frac{6}{\pi}\right)=\sqrt{2}$$ for $n=11$ and we show the calculations needed to obtain this value. We need two ingredients here with the first one being the modular equation of degree $11$ given by Ramanujan $$\sqrt{kl} +\sqrt{k'l'} +2(4klk'l')^{1/6}=1\tag{12}$$ The equation is symmetric in $k, l$ and hence works when modulus $l$ is of degree $11$ over modulus $k$ as well as when $k$ is of degree $11$ over $l$. Let us assume the latter here and let $k, l$ correspond to nomes $e^{-\pi\sqrt{11}},e^{-\pi/\sqrt{11}} $ respectively so that $l=k'$ and then the equation $(12)$ reduces to $$2\sqrt{kk'}+2(2kk')^{1/3}=1$$ Let us further observe that $G=G_{11}=(2kk')^{-1/12}$ and hence the above equation can be written in terms of $G$ as $$2(G^{-12}/2)^{1/2}+2G^{-4}=1$$ or $$\sqrt{2}+2G^{2}=G^6$$ so that $a=G^2=G_{11}^2$ is the root of $$f(x) = x^3-2x-\sqrt{2}\tag{13}$$ It can be checked that this equation has no roots in $\mathbb{Q} (\sqrt{2})$ and hence $a$ is of degree $3$ over $\mathbb{Q} (\sqrt{2})$ and hence of degree $6$ over $\mathbb{Q} $.
The minimal polynomial for $a$ over rationals is then $$(x^3-2x-\sqrt{2})(x^3-2x+\sqrt{2})$$ or $$x^6-4x^4+4x^2-2$$ so that $b=a^2=G_{11}^4$ is a root of $$g(x) =x^3-4x^2+4x-2\tag{14}$$ Next ingredient we need is the expression for $nP(q^{2n})-P(q^2)$ for $n=11$ provided by Ramanujan as $$nP(q^{2n})-P(q^2)=\frac{8KL}{\pi^2}\left\{2(1+kl+k'l')+\sqrt{kl} +\sqrt{k'l'} - \sqrt{klk'l'} \right\}\tag{15}$$ and here we can again assume $k$ being of degree $11$ over $l$ so that $l$ corresponds to $q$ and $k$ to $q^{n} $ and $K'/K=nL'/L$. Putting $q=e^{-\pi/\sqrt{11}}$ so that $l=k'$ and $L/K=K'/K=\sqrt {11}$ we get $$nP(q^{2n})-P(q^2)=\sqrt{11}\left(\frac{2K}{\pi}\right)^2(4(1+2kk')+4\sqrt{kk'}-2kk')$$ or $$nP(q^{2n})-P(q^2)=\sqrt{11}\left(\frac{2K}{\pi}\right)^2(4(1+G^{-12})+2\sqrt{2}G^{-6}-G^{-12})$$ or $$nP(q^{2n})-P(q^2)=\sqrt{11}\left(\frac{2K}{\pi}\right)^2(4+3b^{-3}+2\sqrt{2}a^{-3})$$ Using $\sqrt{2}=a^3-2a$ we get $$nP(q^{2n})-P(q^2)=\sqrt{11}\left(\frac{2K}{\pi}\right)^2\frac{6b^3-4b^2+3}{b^3}$$ We need another identity $$nP(q^{2n})+P(q^2)=\frac{6\sqrt{n}}{\pi}$$ which holds for $q=e^{-\pi/\sqrt{n}} $. Using this identity together with previous equation we get $$2P(e^{-2\pi\sqrt {11}})=\frac{6}{\pi\sqrt{11}}+\frac{1}{\sqrt{11}} \left(\frac{2K}{\pi}\right)^2\frac{6b^3-4b^2+3}{b^3}\tag{16}$$ Next we use the identity $$2P(q^2)-P(-q)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)$$ with $q=e^{-\pi\sqrt{11}}$ to get $$P(-q)=\frac{6}{\pi\sqrt{11}}+\left(\frac {2K}{\pi}\right)^2\left(\frac{6b^3-4b^2+3}{b^3\sqrt{11}}-(1-2k^2)\right)$$ or $$\sqrt{n} P_n-\frac{6}{\pi}=\left(\frac {2K}{\pi}\right)^2\left(\frac{6b^3-4b^2+3}{b^3}-\frac{\sqrt{11(b^6-1)}}{b^3}\right)\tag{17}$$ as $$(1-2k^2)^2=1-G^{-24}=1-b^{-6}$$ To evaluate $Q_n$ we note that $$Q(-q) =\left(\frac{2K}{\pi}\right) ^4(1-4G^{-24})=\left(\frac{2K}{\pi}\right) ^4\frac{b^6-4}{b^6}$$ and hence $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n}P_n-\frac{6}{\pi}\right)=\frac{6b^3-4b^2+3-\sqrt{11(b^6-1)}}{\sqrt {b^6-4}}.$$ The above expression can be simplified using equation $(14)$ which says that $$b^3-4b^2+4b-2=0$$ or $$b(b^2+4)=4b^2+2.$$ On squaring the above equation we get $$b^2(b^2+4)^2=(4b^2+2)^2$$ so that $b^2=G_{11}^8$ is a root of $$x(x+4)^2=(4x+2)^2$$ or $$x^3-8x^2-4=0.$$ This means that $$\sqrt {b^6-4}=\sqrt{8b^4}=2\sqrt {2}b^2.$$ We can thus prove that $$\frac{6b^3-4b^2+3-\sqrt{11(b^6-1)}}{\sqrt{b^6-4}}=\sqrt {2}$$ if $$\sqrt{11(b^6-1)}=6b^3-8b^2+3$$ or $$11(b^6-1)=(6(b^3-4b^2+4b-2)+16b^2-24b+15)^2$$ or $$11(8b^4+3)=(16b^2-24b+15)^2$$ or $$88b^4+33=256b^4+576b^2+480b^2+225-768b^3-720b$$ or $$168b^4-768b^3+1056b^2-720b+192=0$$ or $$7b^4-32b^3+44b^2-30b+8=0$$ or $$(7b-4)(b^3-4b^2+4b-2)=0$$ and this is true via equation $(14)$.
A similar calculation can be done for $n=19$ by noting that $2^{1/4}G_{19}$ is the real root of $x^3-2x-2$ and $$nP(q^{2n})-P(q^2)=\frac{24KL}{\pi^2}\left\{1+kl+k'l'+\sqrt{kl}+\sqrt{k'l'}-\sqrt {klk'l'} \right\}$$ and we can verify that $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)=\sqrt{6}$$ for $n=19$.
More generally if there is a formula of type $$nP(q^{2n})-P(q^2)=\frac {4KL}{\pi^2} \cdot A_n(k, l) $$ then $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)=\frac{G_n^{12}\cdot A_n(k, k') - \sqrt {n(G_n^{24}-1)}} {\sqrt{G_n^{24}-4} } $$ Ramanujan also gave formulas of type $$nP(-q^n) - P(-q) =\frac{4KL}{\pi^2}\cdot B_n(k,l)$$ for some odd positive integer values of $n$ and if such a formula is available then $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)=\frac{G_n^{12}\cdot B_n(k, k') } {2\sqrt {G_n^{24}-4} }$$
The case for $n=27$ can be handled with more ease using modular equations of degree $3$. Let the moduli $k, l$ and elliptic integrals $K, L$ correspond to nomes $q^3,q$ respectively so that $k$ is of degree $3$ over $l$. Ramanujan established that $$3P(q^6)-P(q^2)=\frac{4KL}{\pi^2}\left\{1+kl+k'l'\right\}\tag{18}$$ Using Landen transformation we can replace $q^2$ by $q$, $K$ by $(1+k)K$, $L$ by $(1+l)L$, $k$ by $2\sqrt{k}/(1+k)$, $l$ by $2\sqrt{l}/(1+l)$ in above formula to get $$3P(q^3)-P(q)=\frac{4(1+k)(1+l)KL}{\pi^2}\left\{1+\frac{4\sqrt {kl}} {(1+k)(1+l)} +\frac{(1-k)(1-l)}{(1+k)(1+l)}\right\}$$ and after some simplification we get $$3P(q^3)-P(q)=\frac{8KL}{\pi^2}\left\{1+\sqrt{kl}\right\}^2$$ Next we change $q$ to $-q$. This changes $K$ to $k'K$ and $L$ to $l'L$. To see how this transforms the expression $\sqrt{kl} $ let us observe that $$\sqrt{kl} =4q\frac{\psi(q^2)\psi(q^6)}{\phi(q)\phi(q^3)}$$ where $$\phi(q) =\sum_{n\in\mathbb{Z}} q^{n^2}$$ and $$\psi(q) =\sum_{n\geq 0}q^{n(n+1)/2}$$ are Ramanujan theta functions. The function $\phi(q) $ is same as $\vartheta_3(q)$ of Jacobi and one can now observe that changing $q$ to $-q$ changes $\sqrt {kl} $ to $-\sqrt {kl} /\sqrt{k'l'} $ and hence we get $$3P(-q^3)-P(-q)=\frac{8KL}{\pi^2}\left\{\sqrt{k'l'}-\sqrt{kl}\right\}^2\tag{19}$$ We are going to use this equation repeatedly in what follows. We will also need the following formulas related to multiplier $m=L/K$ established by Ramanujan: \begin{align} m&=\left\{1+4\left(\frac{(kk')^3}{ll'}\right)^{1/4}\right\} ^{1/2}\tag{20a}\\ k^{2} &= \frac{(m - 1)^{3}(m + 3)}{16m}\tag{20b} \\ l^{2} &= \frac{(m - 1)(m + 3)^{3}}{16m^{3}}\tag{20c} \\ k'^{2} &= \frac{(m + 1)^{3}(3 - m)}{16m}\tag{20d} \\ l'^{2} &= \frac{(m + 1)(3 - m)^{3}}{16m^{3}}\tag{20e} \end{align} Next let us now use moduli related to $n=27$ and let $k, l, r, s$ be elliptic moduli and $K, L, R, S$ be elliptic integrals corresponding to nomes $q^{27},q^9,q^3,q=\exp(-\pi/3\sqrt{3})$ respectively so that $s=k', r=l'$ and further $k, l, r$ are of degree $3$ over $l, r, s$ respectively. Also we have $$l=\frac{\sqrt{3}-1}{2\sqrt{2}},l'=\frac{\sqrt{3}+1}{2\sqrt{2}}$$ so that $ll'=1/4$. Using $(19)$ repeatedly we have $$3P(-q^3)-P(-q)=\frac{8RS}{\pi^2}(\sqrt{r's'}-\sqrt{rs})^2=\frac{72KL}{\pi^2}(\sqrt{k'l'}-\sqrt{kl} )^2\tag{21}$$ and $$3P(-q^9)-P(-q^3)=\frac{8LR}{\pi^2}(\sqrt{l'r'}-\sqrt{lr})^2=0\tag{22}$$ as $lr=l'r'$ and $$3P(-q^{27})-P(-q^9)=\frac{8KL}{\pi^2}(\sqrt{k'l'}-\sqrt{kl})^2\tag{23}$$ Multiplying equation $(23)$ by $9$, $(22)$ by $3$ and then adding these to $(21)$ we get $$27P(-q^{27})-P(-q)=\frac{144KL}{\pi^2}(\sqrt{k'l'}-\sqrt{kl})^2\tag{24}$$ Noting that $k$ is of degree $3$ over $l$ we have using equations $(20b),\dots,(20e)$ $$\sqrt {kl} - \sqrt{k'l'} =\frac{(m-1)(m+3)-(m+1)(3-m)}{4m}=\frac{m^2-3}{2m}$$ and hence $$27P(-q^{27})-P(-q)=\frac{4K^2}{\pi^2}\cdot\frac{9(m^2-3)^2}{m}$$ Next we note that the value $2kk'$ equals $G_{27}^{-12}$ and it is known that $$G_{27}=\frac{\sqrt[12]{2}} {\sqrt[3]{\sqrt[3]{2}-1}} $$ and hence $$2kk'=\frac{(a-1)^4}{2}$$ where $a=2^{1/3}$. We also note that $ll'=1/4$ and use the equation $(20a) $ to get $$m^2=1+2^{7/4}(2kk')^{3/4}=1+2(a-1)^3=1+2(a^3-3a^2+3a-1)=3+6a-6a^2$$ or $$m=\sqrt{3}\sqrt{1+2a-2a^2}$$ and thus $$\frac{9(m^2-3)^2}{m}=\frac{324(a^2+2a-4)}{\sqrt{3}\sqrt{1+2a-2a^2}}$$ and finally we have $$27P(-q^{27})-P(-q)=\frac{4K^2}{\pi^2}\cdot\frac{324(a^2+2a-4)}{\sqrt{3}\sqrt{1+2a-2a^2}}\tag{25}$$ We also have the identity $$27P(-q^{27})+P(-q)=\frac{12\sqrt{n}}{\pi}=\frac{36\sqrt{3}}{\pi}$$ From the above two equations we get $$\sqrt {n} P_n-\frac{6}{\pi}=\sqrt{27}P(-q^{27})-\frac{6}{\pi}=\frac{4K^2}{\pi^2}\cdot\frac{18(a^2+2a-4)}{\sqrt{1+2a-2a^2}}$$ Next we have $$Q_n=\left(\frac{2K}{\pi}\right)^4(1-4G_{27}^{-24})$$ or $$\sqrt{Q_n} =\left(\frac{2K}{\pi}\right)^2\sqrt{1-(a-1)^8}$$ and thus $$\frac{1}{\sqrt{Q_n}}\left(\sqrt {n} P_n-\frac{6}{\pi}\right)=\frac{18(a^2+2a-4)}{\sqrt{(1+2a-2a^2) (1-(a-1)^8) }}\tag{26} $$ The square of right hand side is a rational function of $a$ and one can use the relation $a^3=2$ to reduce numerator and denominator each to quadratic polynomials in $a$. First we deal with square of denominator which equals $$(1+2a-2a^2)(1-(a-1)^8)$$ which is same as $$(1+2a-2a^2)(-a^8 + 8a^7 - 28a^6 + 56a^5 - 70a^4 + 56a^3 - 28a^2 + 8a)$$ Reducing the above using $a^3=2$ we get $$(1+2a-2a^2)(-4a^2+32a-112+112a^2-140a+112-28a^2+8a)$$ or $$20a(1+2a-2a^2)(4a-5)=20a(-8a^3 + 18a^2 - 6a - 5) $$ and the above finally equals $$60a(6a^2-2a-7)=60(6a^3-2a^2-7a)=60(12-7a-2a^2)$$ The square of numerator of right side of $(26)$ is $$324(a^2+2a-4)^2=324(a^4+4a^2+16+4a^3-16a-8a^2)$$ which reduces to $$648(12-7a-2a^2)$$ Thus the square of right hand side of $(26)$ equals $648/60=54/5$ and hence the expression on right hand side of $(26)$ equals $\sqrt{54/5}=3\sqrt{6/5}$.
There is also an empirical approach which works nicely for $n=11,19,43,67,163$. If we numerically evaluate the expression $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)$$ for $n=11,19$ then it is easy to spot that these values are $\sqrt{2},\sqrt{6}$ respectively. Next for each of the values of $n$ mentioned in last paragraph one can prove that $G_n^8$ is a root of a polynomial $$x^3-a_nx^2-4$$ where $a_n$ is a positive integer. Further we can write $a_n=b_n^2\cdot c_n$ where $c_n$ is square free so that $$\sqrt{G_n^{24}-4}=\sqrt{x^3-4}=b_n x\sqrt{c_n}, x=G_n^8$$ and we can evaluate the expression $$\sqrt{\frac{c_n}{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)$$ numerically and confirm that it is an almost integer (depending on precision of numerical calculation) and let us say that the nearest integer is $d_n$. Then the expression in question is $d_n/\sqrt{c_n} $. Moreover for $n=43,67,163$ the expressions $P_n, Q_n$ are practically equal to $1$ and hence one just needs to evaluate the much simpler expression $$\sqrt{c_n} \left(\sqrt{n} - \frac{6}{\pi}\right)$$ and it turns out to be an almost integer. For completeness sake let us note that $c_n$ equals $2,6,15, 330,10005$ for $n=11,19,43,67,163$ respectively. Having guessed the value of the expression under question empirically we can then try to prove it analytically.
Now I show how the information in the page from Ramanujan's lost notebook can be used to obtain series for $1/\pi$. It is also important to understand that Ramanujan presented the values in table in exactly the form needed to get these series. In particular the coefficient of $Q_n^3$ has a factor $n$ and Ramanujan specifically mentions that for $n=163$ we have $$53360^3+1=3^3\cdot 7^2\cdot 11^2\cdot 19^2\cdot 127^2\cdot 163$$ The factorization for coefficient of $Q_n^3$ is needed to get the desired series as will be shown below.
Let us start with hypergeometric identity $$\left(\frac{2K}{\pi}\right)^2=\frac{1}{\sqrt{1-4G^{-24}}}{}_3F_2\left(\frac{1}{6},\frac{5}{6},\frac{1}{2};1,1;-\frac{27G^{-24}}{(1-4G^{-24})^3}\right)\tag{27}$$ Also let us use the simplified notation $$f(x) ={}_3F_2\left(\frac{1}{6},\frac{5}{6},\frac{1}{2};1,1;x\right),g(x) =xf'(x) \tag{28}$$ in what follows.
Using derivatives of elliptic integrals and the equation $(5a)$ above we can prove that $$P(-q) = \left(\frac{2K}{\pi}\right)^2(1-2k^2)+3kk'^2\frac{d}{dk} \left(\frac{2K}{\pi}\right)^2\tag{29}$$ Applying the above formula on equation $(27)$ and in the process doing reasonable amount of symbol shunting we arrive at the following beautiful result $$P(-q) =\frac{(1-2k^2)(1+8G^{-24})}{(1-4G^{-24})^{3/2}}\left(f(v(k))+6g(v(k))\right),v(k)=-\frac{27G^{-24}}{(1-4G^{-24})^3}\tag{30}$$ Next let $n>3$ be an odd positive integer and we use $q=e^{-\pi\sqrt{n}} $ so that $$P_n=P(-q), Q_n=Q(-q), R_n=R(-q), G_n=G$$ If we carefully observe the formula $(30)$ we notice at once that the first factor on right is $$\frac{(1-2k^2)(1+8G_n^{-24})}{(1-4G_n^{-24})^{3/2}}=\sqrt {\frac{R_n^2} {Q_n^3}}$$ and the values of the right hand side are given using the relationship between $Q_n, R_n$ provided by Ramanujan. It should be noted that the factorization of coefficient of $Q_n^3$ (given in relation between $Q_n, R_n$) helps in finding the square root on right side of the above equation.
Let us now set $$p_n=\frac{1}{\sqrt{Q_n}}\left(\sqrt{n}P_n-\frac{6}{\pi}\right)$$ so that $$P_n=\frac{6}{\pi\sqrt{n}}+\frac{p_n\sqrt{Q_n}}{\sqrt{n}}=\frac{6}{\pi\sqrt{n}}+\frac{p_n}{\sqrt{n}}\left(\frac{2K}{\pi}\right)^2\sqrt{1-4G_n^{-24}}$$ and using equation $(29), (30)$ we get $$P_n= \frac{6}{\pi\sqrt{n}}+\frac{p_n}{\sqrt{n}}\cdot f(v(k)) \tag{31}$$ Comparing equations $(30),(31)$ and using the values $p_n$ and the relations between $Q_n, R_n$ we get desired series for $1/\pi$ as $$\frac{6}{\pi}=\left(\sqrt{\frac{nR_n^2}{Q_n^3}}-p_n\right)f(v(k))+6\sqrt{\frac{nR_n^2}{Q_n^3}}g(v(k))\tag{32}$$ For $n=11,19,43,67,163$ the relation between $Q_n, R_n$ and value $v(k) $ is easily found. One can prove with some effort that for each of these values of $n$ the value $x_n=G_n^8$ is a root of $$x^3-u_nx^2-4=0$$ where $u_n$ is a positive integer and $y_n=x_n^3=G_n^{24}$ is a root of $$y^3 - (u_n^3 + 12)y^2 + 48y - 64=0$$ The values of $u_n$ are $$u_n=8,24,240,1320,160080$$ for the above values of $n$ respectively. Let us now note that $$v(k) =-\frac{27G_n^{-24}}{(1-4G_n^{-24})^3}=-\frac{27y_n^2}{(y_n-4)^3}=-\frac{27}{u_n^3}$$ And further we have \begin{align} \frac{R_n^2} {Q_n^3}& =\frac{(1-2k^2)^2(1+8G_n^{-24})^2}{(1-4G_n^{-24})^{3}}\notag\\ &=\frac{(1-G_n^{-24})(1+8G_n^{-24})^2} {(1-4G_n^{-24}) ^3} \notag\\ &=\frac{y_n^3+15y_n^2+48y_n-64} {(y_n-4)^3 } \notag\\ &=1+\frac{27}{u_n^3}\notag \end{align} so that using the values of $u_n$ the relations between $Q_n, R_n$ are available.
Using these values the series $(32)$ becomes $$\frac{6\sqrt{u_n}}{\pi}=\left(\frac{\sqrt{n(u_n^3+27)}}{u_n}-p_n\sqrt{u_n}\right)f\left(-\frac{27}{u_n^3}\right)+6\cdot\frac{\sqrt{n(u_n^3+27)}}{u_n}\cdot g \left(-\frac{27}{u_n^3}\right) \tag{33}$$ It should be noted that the expressions $$\sqrt{n(u_n^3+27)},p_n\sqrt{u_n}$$ are positive integers for these particular values of $n$ (we have discussed the empirical approach to evaluate the value of $p_n$ using the fact that $p_n\sqrt{u_n} $ is a positive integer earlier). For $n=35$ these expressions are algebraic integers from field $\mathbb{Q} (\sqrt{5})$.
We exhibit two series for $1/\pi$ based on above approach for $n=11,163$. For $n=11$ we have $u_n=8,p_n=\sqrt{2}$ and $$ \frac{\sqrt{n(u_n^3+27)}}{u_n}-p_n\sqrt{u_n}=\frac{\sqrt{11\cdot 539}}{8}-4=\frac{45}{8}$$ and $$ \frac{\sqrt{n(u_n^3+27)}}{u_n}=\frac{77}{8}$$ and using equation $(33)$ we get the desired series as $$\frac{12\sqrt{2}}{\pi}=\frac{45}{8}f(-27/512)+\frac{6\cdot 77}{8}g(-27/512) $$ or $$\frac{32\sqrt{2}}{\pi}=15f(-27/512)+154g(-27/512)$$ or $$\frac{32\sqrt{2}}{\pi}=\sum_{j\geq 0}(-1)^j\frac{(1/6)_j(5/6)_j(1/2)_j} {(j!) ^3}(15+154j)\left(\frac{3} {8}\right)^{3j}$$ For $n=163$ we have $$u_n=160080,p_n=362\sqrt{\frac{3}{3335}}$$ so that $$ \frac{\sqrt{n(u_n^3+27)}}{u_n}-p_n\sqrt{u_n}=\frac{\sqrt{163\cdot 27(53360^3+1)}}{160080}-362\sqrt{\frac{3}{3335}}\sqrt{160080}\tag{34}$$ Now we use the factorization $$53360^3+1=3^3\cdot 7^2\cdot 11^2\cdot 19^2\cdot 127^2\cdot 163$$ given by Ramanujan to simplify the expression in $(34)$ as $$\frac{27\cdot 163\cdot 7\cdot 11\cdot 19\cdot 127}{160080}-362\cdot 12=\frac{40774227} {53360}$$ and we have $$\frac{\sqrt{n(u_n^3+27)}}{u_n} =\frac{272570067}{53360}$$ and using equation $(33)$ we get $$\frac{6\sqrt {160080}}{\pi} =\frac{40774227}{53360}f(-1/53360^3)+\frac{272570067\cdot 6}{53360}g(-1/53360^3)$$ or $$\frac{106720\sqrt{160080}}{\pi}=\sum_{j\geq 0}(-1)^j\frac{(1/6)_j(5/6)_j(1/2)_j}{(j!)^3}\frac{13591409+545140134j} {53360^{3j}}$$ which is the famous series by Chudnovsky brothers.
Note: The above presentation is based on a thread on MathOverflow I posted sometime ago.
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A page from Ramanujan's Lost Notebook
On page 211 of his Lost Notebook (original manuscript) Ramanujan provides a table of relations between his functions $Q,R$ and values of a cryptic expression alongside certain positive integers |
| Page 211 from the Lost Notebook |
Evaluation of $P_n=P(-q) $
Let us observe that $$P(-q) =24q\frac {d} {dq} \log\eta(-q) =q\frac{d} {dq} \log\eta^{24}(-q) =q\frac{d} {dq} \log(-\eta^{24}(-q))$$ and we have \begin{align} -\eta^{24}(-q)&=q\prod_{j\geq 1}(1-(-q)^j)^{24}\notag\\ &=q\prod_{j\geq 1}(1+q^{2j-1})^{24}(1-q^{2j})^{24}\notag\\ &=\left(q^{-1}\prod_{j\geq 1}(1+q^{2j-1})^{24}\right) \left(q^2\prod_{j\geq 1}(1-q^{2j}) ^{24}\right) \notag\\ &=2^6(2kk')^{-2}\eta^{24}(q^2)\notag\\ &=2^4(kk')^{-2}\cdot 2^{-8}(2K/\pi)^{12}(kk')^{4}\notag\\ &=2^{-4}\left(\frac{2K}{\pi}\right)^{12} (kk') ^2\notag \end{align} Using logarithmic differentiation and noting that $$\frac{dq} {dk} =\frac{\pi^2q}{2kk'^2K^2}$$ we arrive at $$P(-q) =\left(\frac {2K}{\pi}\right)^2\left(\frac{6E}{K}+4k^2-5\right)$$ mentioned earlier. A similar procedure leads us to $$P(q^2)= \left(\frac {2K}{\pi}\right)^2\left(\frac{3E}{K}+k^2-2\right)$$ and then we get $$2P(q^2)-P(-q)= \left(\frac {2K}{\pi}\right)^2\left(1-2k^2\right)$$ which is also mentioned earlier.Let us now take another elliptic modulus $l$ which is of degree $n$ over $k$ so that its nome is $q^n$ and the corresponding elliptic integral is denoted by $L$ and $L'/L=nK'/K$. Then we have $$\eta^{24}(-q^n)=-2^{-4}(2L/\pi)^{12}(ll')^{2}$$ Next we have $$nP(-q^n) =q\frac{d} {dq} \log\eta^{24}(-q^n)$$ so that \begin{align} nP(-q^n) - P(-q) &=q\frac{d} {dq} \log\frac{\eta^{24}(-q^n)}{\eta^{24}(-q)} \notag\\ &=q\frac{dk} {dq} \frac{d} {dk} \log\left\{\left(\frac{L} {K} \right) ^{12}\left(\frac{ll'} {kk'} \right) ^2\right\} \notag\\ &=\left(\frac{2K}{\pi}\right) ^2kk'^2\frac{d} {dk} \log\left\{\left(\frac{L} {K} \right) ^{6}\frac{ll'} {kk'} \right\} \notag\\ &=\frac{4KL}{\pi^2}mkk'^2\frac{d} {dk} \log\frac{ll'} {m^6kk'}\notag \end{align} where $m=K/L$ is the multiplier. Since $m$ as well as $l$ are algebraic functions of $k$ it follows that $$nP(-q^n) - P(-q) =\frac{4KL} {\pi^2}B_n(l, k)\tag{9} $$ where $B_n(l, k) $ is an algebraic function of $l, k$. If we put $l=k'$ so that $L=K'$ and $K'/K=1/\sqrt{n}$ the $q=e^{-\pi/\sqrt {n}} $ and we get $$nP(-e^{-\pi\sqrt{n}} ) - P(-e^{-\pi/\sqrt{n}} ) =\left(\frac{2L}{\pi}\right)^2B_n\tag{10}$$ where $L$ corresponds to nome $e^{-\pi\sqrt{n}} $ and $B_n$ is some algebraic number dependent on $n$.
Next note that $$\frac{\eta^{24}(-e^{-\pi\sqrt{n}})} {\eta^{24}(-e^{-\pi/\sqrt{n}}) } =\left(\frac{K'} {K} \right) ^{12}=n^{-6}$$ and logarithmic differentiation with respect to $n$ gives us $$nP(-e^{-\pi\sqrt{n}}) +P(-e^{-\pi/\sqrt{n}}) =\frac{12\sqrt{n}}{\pi}\tag{11}$$ Using equations $(10),(11)$ we see that $$P(-e^{-\pi\sqrt{n}}) =\frac{6}{\pi\sqrt{n}}+\left(\frac{2K}{\pi} \right) ^2\frac{B_n}{2n}$$ where elliptic integral $K$ corresponds to nome $e^{-\pi\sqrt{n}} $. Ramanujan perhaps calculated certain values of this expression for different values of $n$ and figured out that multiplying the above expression by $\sqrt{n} $ and further dividing the result by $\sqrt{Q(-q)} $ to get rid of factor $(2K/\pi) ^2 $ leads to very simple algebraic numbers. And this probably led to the table of values of the expression $(\sqrt{n} P_n-(6/\pi)) /\sqrt{Q_n} $.
Ramanujan gave the value $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n}P_n-\frac{6}{\pi}\right)=\sqrt{2}$$ for $n=11$ and we show the calculations needed to obtain this value. We need two ingredients here with the first one being the modular equation of degree $11$ given by Ramanujan $$\sqrt{kl} +\sqrt{k'l'} +2(4klk'l')^{1/6}=1\tag{12}$$ The equation is symmetric in $k, l$ and hence works when modulus $l$ is of degree $11$ over modulus $k$ as well as when $k$ is of degree $11$ over $l$. Let us assume the latter here and let $k, l$ correspond to nomes $e^{-\pi\sqrt{11}},e^{-\pi/\sqrt{11}} $ respectively so that $l=k'$ and then the equation $(12)$ reduces to $$2\sqrt{kk'}+2(2kk')^{1/3}=1$$ Let us further observe that $G=G_{11}=(2kk')^{-1/12}$ and hence the above equation can be written in terms of $G$ as $$2(G^{-12}/2)^{1/2}+2G^{-4}=1$$ or $$\sqrt{2}+2G^{2}=G^6$$ so that $a=G^2=G_{11}^2$ is the root of $$f(x) = x^3-2x-\sqrt{2}\tag{13}$$ It can be checked that this equation has no roots in $\mathbb{Q} (\sqrt{2})$ and hence $a$ is of degree $3$ over $\mathbb{Q} (\sqrt{2})$ and hence of degree $6$ over $\mathbb{Q} $.
The minimal polynomial for $a$ over rationals is then $$(x^3-2x-\sqrt{2})(x^3-2x+\sqrt{2})$$ or $$x^6-4x^4+4x^2-2$$ so that $b=a^2=G_{11}^4$ is a root of $$g(x) =x^3-4x^2+4x-2\tag{14}$$ Next ingredient we need is the expression for $nP(q^{2n})-P(q^2)$ for $n=11$ provided by Ramanujan as $$nP(q^{2n})-P(q^2)=\frac{8KL}{\pi^2}\left\{2(1+kl+k'l')+\sqrt{kl} +\sqrt{k'l'} - \sqrt{klk'l'} \right\}\tag{15}$$ and here we can again assume $k$ being of degree $11$ over $l$ so that $l$ corresponds to $q$ and $k$ to $q^{n} $ and $K'/K=nL'/L$. Putting $q=e^{-\pi/\sqrt{11}}$ so that $l=k'$ and $L/K=K'/K=\sqrt {11}$ we get $$nP(q^{2n})-P(q^2)=\sqrt{11}\left(\frac{2K}{\pi}\right)^2(4(1+2kk')+4\sqrt{kk'}-2kk')$$ or $$nP(q^{2n})-P(q^2)=\sqrt{11}\left(\frac{2K}{\pi}\right)^2(4(1+G^{-12})+2\sqrt{2}G^{-6}-G^{-12})$$ or $$nP(q^{2n})-P(q^2)=\sqrt{11}\left(\frac{2K}{\pi}\right)^2(4+3b^{-3}+2\sqrt{2}a^{-3})$$ Using $\sqrt{2}=a^3-2a$ we get $$nP(q^{2n})-P(q^2)=\sqrt{11}\left(\frac{2K}{\pi}\right)^2\frac{6b^3-4b^2+3}{b^3}$$ We need another identity $$nP(q^{2n})+P(q^2)=\frac{6\sqrt{n}}{\pi}$$ which holds for $q=e^{-\pi/\sqrt{n}} $. Using this identity together with previous equation we get $$2P(e^{-2\pi\sqrt {11}})=\frac{6}{\pi\sqrt{11}}+\frac{1}{\sqrt{11}} \left(\frac{2K}{\pi}\right)^2\frac{6b^3-4b^2+3}{b^3}\tag{16}$$ Next we use the identity $$2P(q^2)-P(-q)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)$$ with $q=e^{-\pi\sqrt{11}}$ to get $$P(-q)=\frac{6}{\pi\sqrt{11}}+\left(\frac {2K}{\pi}\right)^2\left(\frac{6b^3-4b^2+3}{b^3\sqrt{11}}-(1-2k^2)\right)$$ or $$\sqrt{n} P_n-\frac{6}{\pi}=\left(\frac {2K}{\pi}\right)^2\left(\frac{6b^3-4b^2+3}{b^3}-\frac{\sqrt{11(b^6-1)}}{b^3}\right)\tag{17}$$ as $$(1-2k^2)^2=1-G^{-24}=1-b^{-6}$$ To evaluate $Q_n$ we note that $$Q(-q) =\left(\frac{2K}{\pi}\right) ^4(1-4G^{-24})=\left(\frac{2K}{\pi}\right) ^4\frac{b^6-4}{b^6}$$ and hence $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n}P_n-\frac{6}{\pi}\right)=\frac{6b^3-4b^2+3-\sqrt{11(b^6-1)}}{\sqrt {b^6-4}}.$$ The above expression can be simplified using equation $(14)$ which says that $$b^3-4b^2+4b-2=0$$ or $$b(b^2+4)=4b^2+2.$$ On squaring the above equation we get $$b^2(b^2+4)^2=(4b^2+2)^2$$ so that $b^2=G_{11}^8$ is a root of $$x(x+4)^2=(4x+2)^2$$ or $$x^3-8x^2-4=0.$$ This means that $$\sqrt {b^6-4}=\sqrt{8b^4}=2\sqrt {2}b^2.$$ We can thus prove that $$\frac{6b^3-4b^2+3-\sqrt{11(b^6-1)}}{\sqrt{b^6-4}}=\sqrt {2}$$ if $$\sqrt{11(b^6-1)}=6b^3-8b^2+3$$ or $$11(b^6-1)=(6(b^3-4b^2+4b-2)+16b^2-24b+15)^2$$ or $$11(8b^4+3)=(16b^2-24b+15)^2$$ or $$88b^4+33=256b^4+576b^2+480b^2+225-768b^3-720b$$ or $$168b^4-768b^3+1056b^2-720b+192=0$$ or $$7b^4-32b^3+44b^2-30b+8=0$$ or $$(7b-4)(b^3-4b^2+4b-2)=0$$ and this is true via equation $(14)$.
A similar calculation can be done for $n=19$ by noting that $2^{1/4}G_{19}$ is the real root of $x^3-2x-2$ and $$nP(q^{2n})-P(q^2)=\frac{24KL}{\pi^2}\left\{1+kl+k'l'+\sqrt{kl}+\sqrt{k'l'}-\sqrt {klk'l'} \right\}$$ and we can verify that $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)=\sqrt{6}$$ for $n=19$.
More generally if there is a formula of type $$nP(q^{2n})-P(q^2)=\frac {4KL}{\pi^2} \cdot A_n(k, l) $$ then $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)=\frac{G_n^{12}\cdot A_n(k, k') - \sqrt {n(G_n^{24}-1)}} {\sqrt{G_n^{24}-4} } $$ Ramanujan also gave formulas of type $$nP(-q^n) - P(-q) =\frac{4KL}{\pi^2}\cdot B_n(k,l)$$ for some odd positive integer values of $n$ and if such a formula is available then $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)=\frac{G_n^{12}\cdot B_n(k, k') } {2\sqrt {G_n^{24}-4} }$$
The case for $n=27$ can be handled with more ease using modular equations of degree $3$. Let the moduli $k, l$ and elliptic integrals $K, L$ correspond to nomes $q^3,q$ respectively so that $k$ is of degree $3$ over $l$. Ramanujan established that $$3P(q^6)-P(q^2)=\frac{4KL}{\pi^2}\left\{1+kl+k'l'\right\}\tag{18}$$ Using Landen transformation we can replace $q^2$ by $q$, $K$ by $(1+k)K$, $L$ by $(1+l)L$, $k$ by $2\sqrt{k}/(1+k)$, $l$ by $2\sqrt{l}/(1+l)$ in above formula to get $$3P(q^3)-P(q)=\frac{4(1+k)(1+l)KL}{\pi^2}\left\{1+\frac{4\sqrt {kl}} {(1+k)(1+l)} +\frac{(1-k)(1-l)}{(1+k)(1+l)}\right\}$$ and after some simplification we get $$3P(q^3)-P(q)=\frac{8KL}{\pi^2}\left\{1+\sqrt{kl}\right\}^2$$ Next we change $q$ to $-q$. This changes $K$ to $k'K$ and $L$ to $l'L$. To see how this transforms the expression $\sqrt{kl} $ let us observe that $$\sqrt{kl} =4q\frac{\psi(q^2)\psi(q^6)}{\phi(q)\phi(q^3)}$$ where $$\phi(q) =\sum_{n\in\mathbb{Z}} q^{n^2}$$ and $$\psi(q) =\sum_{n\geq 0}q^{n(n+1)/2}$$ are Ramanujan theta functions. The function $\phi(q) $ is same as $\vartheta_3(q)$ of Jacobi and one can now observe that changing $q$ to $-q$ changes $\sqrt {kl} $ to $-\sqrt {kl} /\sqrt{k'l'} $ and hence we get $$3P(-q^3)-P(-q)=\frac{8KL}{\pi^2}\left\{\sqrt{k'l'}-\sqrt{kl}\right\}^2\tag{19}$$ We are going to use this equation repeatedly in what follows. We will also need the following formulas related to multiplier $m=L/K$ established by Ramanujan: \begin{align} m&=\left\{1+4\left(\frac{(kk')^3}{ll'}\right)^{1/4}\right\} ^{1/2}\tag{20a}\\ k^{2} &= \frac{(m - 1)^{3}(m + 3)}{16m}\tag{20b} \\ l^{2} &= \frac{(m - 1)(m + 3)^{3}}{16m^{3}}\tag{20c} \\ k'^{2} &= \frac{(m + 1)^{3}(3 - m)}{16m}\tag{20d} \\ l'^{2} &= \frac{(m + 1)(3 - m)^{3}}{16m^{3}}\tag{20e} \end{align} Next let us now use moduli related to $n=27$ and let $k, l, r, s$ be elliptic moduli and $K, L, R, S$ be elliptic integrals corresponding to nomes $q^{27},q^9,q^3,q=\exp(-\pi/3\sqrt{3})$ respectively so that $s=k', r=l'$ and further $k, l, r$ are of degree $3$ over $l, r, s$ respectively. Also we have $$l=\frac{\sqrt{3}-1}{2\sqrt{2}},l'=\frac{\sqrt{3}+1}{2\sqrt{2}}$$ so that $ll'=1/4$. Using $(19)$ repeatedly we have $$3P(-q^3)-P(-q)=\frac{8RS}{\pi^2}(\sqrt{r's'}-\sqrt{rs})^2=\frac{72KL}{\pi^2}(\sqrt{k'l'}-\sqrt{kl} )^2\tag{21}$$ and $$3P(-q^9)-P(-q^3)=\frac{8LR}{\pi^2}(\sqrt{l'r'}-\sqrt{lr})^2=0\tag{22}$$ as $lr=l'r'$ and $$3P(-q^{27})-P(-q^9)=\frac{8KL}{\pi^2}(\sqrt{k'l'}-\sqrt{kl})^2\tag{23}$$ Multiplying equation $(23)$ by $9$, $(22)$ by $3$ and then adding these to $(21)$ we get $$27P(-q^{27})-P(-q)=\frac{144KL}{\pi^2}(\sqrt{k'l'}-\sqrt{kl})^2\tag{24}$$ Noting that $k$ is of degree $3$ over $l$ we have using equations $(20b),\dots,(20e)$ $$\sqrt {kl} - \sqrt{k'l'} =\frac{(m-1)(m+3)-(m+1)(3-m)}{4m}=\frac{m^2-3}{2m}$$ and hence $$27P(-q^{27})-P(-q)=\frac{4K^2}{\pi^2}\cdot\frac{9(m^2-3)^2}{m}$$ Next we note that the value $2kk'$ equals $G_{27}^{-12}$ and it is known that $$G_{27}=\frac{\sqrt[12]{2}} {\sqrt[3]{\sqrt[3]{2}-1}} $$ and hence $$2kk'=\frac{(a-1)^4}{2}$$ where $a=2^{1/3}$. We also note that $ll'=1/4$ and use the equation $(20a) $ to get $$m^2=1+2^{7/4}(2kk')^{3/4}=1+2(a-1)^3=1+2(a^3-3a^2+3a-1)=3+6a-6a^2$$ or $$m=\sqrt{3}\sqrt{1+2a-2a^2}$$ and thus $$\frac{9(m^2-3)^2}{m}=\frac{324(a^2+2a-4)}{\sqrt{3}\sqrt{1+2a-2a^2}}$$ and finally we have $$27P(-q^{27})-P(-q)=\frac{4K^2}{\pi^2}\cdot\frac{324(a^2+2a-4)}{\sqrt{3}\sqrt{1+2a-2a^2}}\tag{25}$$ We also have the identity $$27P(-q^{27})+P(-q)=\frac{12\sqrt{n}}{\pi}=\frac{36\sqrt{3}}{\pi}$$ From the above two equations we get $$\sqrt {n} P_n-\frac{6}{\pi}=\sqrt{27}P(-q^{27})-\frac{6}{\pi}=\frac{4K^2}{\pi^2}\cdot\frac{18(a^2+2a-4)}{\sqrt{1+2a-2a^2}}$$ Next we have $$Q_n=\left(\frac{2K}{\pi}\right)^4(1-4G_{27}^{-24})$$ or $$\sqrt{Q_n} =\left(\frac{2K}{\pi}\right)^2\sqrt{1-(a-1)^8}$$ and thus $$\frac{1}{\sqrt{Q_n}}\left(\sqrt {n} P_n-\frac{6}{\pi}\right)=\frac{18(a^2+2a-4)}{\sqrt{(1+2a-2a^2) (1-(a-1)^8) }}\tag{26} $$ The square of right hand side is a rational function of $a$ and one can use the relation $a^3=2$ to reduce numerator and denominator each to quadratic polynomials in $a$. First we deal with square of denominator which equals $$(1+2a-2a^2)(1-(a-1)^8)$$ which is same as $$(1+2a-2a^2)(-a^8 + 8a^7 - 28a^6 + 56a^5 - 70a^4 + 56a^3 - 28a^2 + 8a)$$ Reducing the above using $a^3=2$ we get $$(1+2a-2a^2)(-4a^2+32a-112+112a^2-140a+112-28a^2+8a)$$ or $$20a(1+2a-2a^2)(4a-5)=20a(-8a^3 + 18a^2 - 6a - 5) $$ and the above finally equals $$60a(6a^2-2a-7)=60(6a^3-2a^2-7a)=60(12-7a-2a^2)$$ The square of numerator of right side of $(26)$ is $$324(a^2+2a-4)^2=324(a^4+4a^2+16+4a^3-16a-8a^2)$$ which reduces to $$648(12-7a-2a^2)$$ Thus the square of right hand side of $(26)$ equals $648/60=54/5$ and hence the expression on right hand side of $(26)$ equals $\sqrt{54/5}=3\sqrt{6/5}$.
There is also an empirical approach which works nicely for $n=11,19,43,67,163$. If we numerically evaluate the expression $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)$$ for $n=11,19$ then it is easy to spot that these values are $\sqrt{2},\sqrt{6}$ respectively. Next for each of the values of $n$ mentioned in last paragraph one can prove that $G_n^8$ is a root of a polynomial $$x^3-a_nx^2-4$$ where $a_n$ is a positive integer. Further we can write $a_n=b_n^2\cdot c_n$ where $c_n$ is square free so that $$\sqrt{G_n^{24}-4}=\sqrt{x^3-4}=b_n x\sqrt{c_n}, x=G_n^8$$ and we can evaluate the expression $$\sqrt{\frac{c_n}{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)$$ numerically and confirm that it is an almost integer (depending on precision of numerical calculation) and let us say that the nearest integer is $d_n$. Then the expression in question is $d_n/\sqrt{c_n} $. Moreover for $n=43,67,163$ the expressions $P_n, Q_n$ are practically equal to $1$ and hence one just needs to evaluate the much simpler expression $$\sqrt{c_n} \left(\sqrt{n} - \frac{6}{\pi}\right)$$ and it turns out to be an almost integer. For completeness sake let us note that $c_n$ equals $2,6,15, 330,10005$ for $n=11,19,43,67,163$ respectively. Having guessed the value of the expression under question empirically we can then try to prove it analytically.
$P(-q)$ and Chudnovsky series for $1/\pi$
The values of $P_n=P(-q)$ given in table above indicate that Ramanujan had managed to discover some more series for $1/\pi$ (apart from those listed in his famous paper Modular equations and approximations to $\pi$), although he never wrote them explicitly. And these series famously include the one given by Chudnovsky brothers.Now I show how the information in the page from Ramanujan's lost notebook can be used to obtain series for $1/\pi$. It is also important to understand that Ramanujan presented the values in table in exactly the form needed to get these series. In particular the coefficient of $Q_n^3$ has a factor $n$ and Ramanujan specifically mentions that for $n=163$ we have $$53360^3+1=3^3\cdot 7^2\cdot 11^2\cdot 19^2\cdot 127^2\cdot 163$$ The factorization for coefficient of $Q_n^3$ is needed to get the desired series as will be shown below.
Let us start with hypergeometric identity $$\left(\frac{2K}{\pi}\right)^2=\frac{1}{\sqrt{1-4G^{-24}}}{}_3F_2\left(\frac{1}{6},\frac{5}{6},\frac{1}{2};1,1;-\frac{27G^{-24}}{(1-4G^{-24})^3}\right)\tag{27}$$ Also let us use the simplified notation $$f(x) ={}_3F_2\left(\frac{1}{6},\frac{5}{6},\frac{1}{2};1,1;x\right),g(x) =xf'(x) \tag{28}$$ in what follows.
Using derivatives of elliptic integrals and the equation $(5a)$ above we can prove that $$P(-q) = \left(\frac{2K}{\pi}\right)^2(1-2k^2)+3kk'^2\frac{d}{dk} \left(\frac{2K}{\pi}\right)^2\tag{29}$$ Applying the above formula on equation $(27)$ and in the process doing reasonable amount of symbol shunting we arrive at the following beautiful result $$P(-q) =\frac{(1-2k^2)(1+8G^{-24})}{(1-4G^{-24})^{3/2}}\left(f(v(k))+6g(v(k))\right),v(k)=-\frac{27G^{-24}}{(1-4G^{-24})^3}\tag{30}$$ Next let $n>3$ be an odd positive integer and we use $q=e^{-\pi\sqrt{n}} $ so that $$P_n=P(-q), Q_n=Q(-q), R_n=R(-q), G_n=G$$ If we carefully observe the formula $(30)$ we notice at once that the first factor on right is $$\frac{(1-2k^2)(1+8G_n^{-24})}{(1-4G_n^{-24})^{3/2}}=\sqrt {\frac{R_n^2} {Q_n^3}}$$ and the values of the right hand side are given using the relationship between $Q_n, R_n$ provided by Ramanujan. It should be noted that the factorization of coefficient of $Q_n^3$ (given in relation between $Q_n, R_n$) helps in finding the square root on right side of the above equation.
Let us now set $$p_n=\frac{1}{\sqrt{Q_n}}\left(\sqrt{n}P_n-\frac{6}{\pi}\right)$$ so that $$P_n=\frac{6}{\pi\sqrt{n}}+\frac{p_n\sqrt{Q_n}}{\sqrt{n}}=\frac{6}{\pi\sqrt{n}}+\frac{p_n}{\sqrt{n}}\left(\frac{2K}{\pi}\right)^2\sqrt{1-4G_n^{-24}}$$ and using equation $(29), (30)$ we get $$P_n= \frac{6}{\pi\sqrt{n}}+\frac{p_n}{\sqrt{n}}\cdot f(v(k)) \tag{31}$$ Comparing equations $(30),(31)$ and using the values $p_n$ and the relations between $Q_n, R_n$ we get desired series for $1/\pi$ as $$\frac{6}{\pi}=\left(\sqrt{\frac{nR_n^2}{Q_n^3}}-p_n\right)f(v(k))+6\sqrt{\frac{nR_n^2}{Q_n^3}}g(v(k))\tag{32}$$ For $n=11,19,43,67,163$ the relation between $Q_n, R_n$ and value $v(k) $ is easily found. One can prove with some effort that for each of these values of $n$ the value $x_n=G_n^8$ is a root of $$x^3-u_nx^2-4=0$$ where $u_n$ is a positive integer and $y_n=x_n^3=G_n^{24}$ is a root of $$y^3 - (u_n^3 + 12)y^2 + 48y - 64=0$$ The values of $u_n$ are $$u_n=8,24,240,1320,160080$$ for the above values of $n$ respectively. Let us now note that $$v(k) =-\frac{27G_n^{-24}}{(1-4G_n^{-24})^3}=-\frac{27y_n^2}{(y_n-4)^3}=-\frac{27}{u_n^3}$$ And further we have \begin{align} \frac{R_n^2} {Q_n^3}& =\frac{(1-2k^2)^2(1+8G_n^{-24})^2}{(1-4G_n^{-24})^{3}}\notag\\ &=\frac{(1-G_n^{-24})(1+8G_n^{-24})^2} {(1-4G_n^{-24}) ^3} \notag\\ &=\frac{y_n^3+15y_n^2+48y_n-64} {(y_n-4)^3 } \notag\\ &=1+\frac{27}{u_n^3}\notag \end{align} so that using the values of $u_n$ the relations between $Q_n, R_n$ are available.
Using these values the series $(32)$ becomes $$\frac{6\sqrt{u_n}}{\pi}=\left(\frac{\sqrt{n(u_n^3+27)}}{u_n}-p_n\sqrt{u_n}\right)f\left(-\frac{27}{u_n^3}\right)+6\cdot\frac{\sqrt{n(u_n^3+27)}}{u_n}\cdot g \left(-\frac{27}{u_n^3}\right) \tag{33}$$ It should be noted that the expressions $$\sqrt{n(u_n^3+27)},p_n\sqrt{u_n}$$ are positive integers for these particular values of $n$ (we have discussed the empirical approach to evaluate the value of $p_n$ using the fact that $p_n\sqrt{u_n} $ is a positive integer earlier). For $n=35$ these expressions are algebraic integers from field $\mathbb{Q} (\sqrt{5})$.
We exhibit two series for $1/\pi$ based on above approach for $n=11,163$. For $n=11$ we have $u_n=8,p_n=\sqrt{2}$ and $$ \frac{\sqrt{n(u_n^3+27)}}{u_n}-p_n\sqrt{u_n}=\frac{\sqrt{11\cdot 539}}{8}-4=\frac{45}{8}$$ and $$ \frac{\sqrt{n(u_n^3+27)}}{u_n}=\frac{77}{8}$$ and using equation $(33)$ we get the desired series as $$\frac{12\sqrt{2}}{\pi}=\frac{45}{8}f(-27/512)+\frac{6\cdot 77}{8}g(-27/512) $$ or $$\frac{32\sqrt{2}}{\pi}=15f(-27/512)+154g(-27/512)$$ or $$\frac{32\sqrt{2}}{\pi}=\sum_{j\geq 0}(-1)^j\frac{(1/6)_j(5/6)_j(1/2)_j} {(j!) ^3}(15+154j)\left(\frac{3} {8}\right)^{3j}$$ For $n=163$ we have $$u_n=160080,p_n=362\sqrt{\frac{3}{3335}}$$ so that $$ \frac{\sqrt{n(u_n^3+27)}}{u_n}-p_n\sqrt{u_n}=\frac{\sqrt{163\cdot 27(53360^3+1)}}{160080}-362\sqrt{\frac{3}{3335}}\sqrt{160080}\tag{34}$$ Now we use the factorization $$53360^3+1=3^3\cdot 7^2\cdot 11^2\cdot 19^2\cdot 127^2\cdot 163$$ given by Ramanujan to simplify the expression in $(34)$ as $$\frac{27\cdot 163\cdot 7\cdot 11\cdot 19\cdot 127}{160080}-362\cdot 12=\frac{40774227} {53360}$$ and we have $$\frac{\sqrt{n(u_n^3+27)}}{u_n} =\frac{272570067}{53360}$$ and using equation $(33)$ we get $$\frac{6\sqrt {160080}}{\pi} =\frac{40774227}{53360}f(-1/53360^3)+\frac{272570067\cdot 6}{53360}g(-1/53360^3)$$ or $$\frac{106720\sqrt{160080}}{\pi}=\sum_{j\geq 0}(-1)^j\frac{(1/6)_j(5/6)_j(1/2)_j}{(j!)^3}\frac{13591409+545140134j} {53360^{3j}}$$ which is the famous series by Chudnovsky brothers.
Note: The above presentation is based on a thread on MathOverflow I posted sometime ago.
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