In the previous post we have handled the evaluation of $P_n=P(-e^{-\pi\sqrt{n}}) $ for $n=11,27$. We will evaluate $P_n$ for $n=19,35$ in the current post and also discuss an empirical approach for $n=43,67,163$. Finally we will use the information in table given by Ramanujan to obtain certain series for $1/\pi$ (including the famous one by Chudnovsky brothers).
Next we need the formula given by Ramanujan as $$nP(q^{2n})-P(q^2)=\frac{24KL}{\pi^2}\left\{1+kl+k'l'+\sqrt{kl}+\sqrt{k'l'}-\sqrt{klk'l'}\right\}\tag{2} $$ where $n=19$ and $k$ is of degree $19$ over $l$ so that nome $q$ corresponds to $l$ and $q^n$ to $k$. Putting $q=e^{-\pi/\sqrt{19}}$ so that $q^n=e^{-\pi\sqrt{19}}$ and $k=l', k'=l$ we get $$nP(q^{2n})-P(q^2)=6\left(\frac{2K}{\pi}\right)^2\frac{K'}{K}(1+kk'+2\sqrt{kk'})$$ or $$nP(q^{2n})-P(q^2)=6\sqrt{19}\left(\frac{2K}{\pi}\right)^2(1+(a^2/4)+a)=\frac{3\sqrt{19}}{2}\left(\frac{2K}{\pi}\right)^2(a+2)^2$$ Combining the above equation with the identity $$nP(q^{2n})+P(q^2)=\frac{6\sqrt{n}}{\pi}, q=e^{-\pi/\sqrt {n}}\tag{3} $$ we get $$2P(q^{2n})=\frac{6}{\pi\sqrt{n}}+\frac{3}{2\sqrt{19}}\left(\frac{2K}{\pi}\right)^2(a+2)^2$$ Switching notation such that $k, K$ correspond to nome $q=e^{-\pi\sqrt {19}}$ we get $$2P(q^2)-\frac{6}{\pi\sqrt {n}} =\frac{3}{2\sqrt{19}}\left(\frac{2K}{\pi}\right) ^2(a+2)^2$$ Using the identity $$2P(q^2)-P(-q)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)=\left(\frac{2K}{\pi}\right)^2\sqrt{1-G^{-24}}\tag{4}$$ we get $$\sqrt{n} P(-q) - \frac{6}{\pi}=\left(\frac{2K}{\pi}\right)^2\frac{3(a+2)^2-\sqrt {19(4-a^4)}}{2}$$ And then using identity $$Q(-q) = \left(\frac{2K}{\pi}\right)^4(1-4G^{-24})$$ we get $$\frac{1}{\sqrt{Q_n}}\left (\sqrt{n}P_n-\frac{6}{\pi}\right) =\frac{3(a+2)^2-\sqrt{19(4-a^4)}} {2\sqrt{1-a^4}}\tag{5}$$ To simplify the expression on right we evaluate some integral powers of $a$ \begin{align} a^3&=-5a^2-7a+1\notag\\ a^4&=-5a^3-7a^2+a=18a^2+36a-5\notag\\ \frac{1}{a}&=a^2+5a+7\notag\\ \frac{1}{a^2}&=a+5+\frac{7}{a}=7a^2+36a+54\notag \end{align} We can observe that $$\frac{1}{a^2}=a^2+6(a^2+6a+9)=a^2+6(a+3)^2$$ and thus on multiplying by $a^2$ we get $$1-a^4=6a^2(a+3)^2$$ ie $$\sqrt{1-a^4}=a(a+3)\sqrt{6}$$ Further $$4-a^4=9(1-4a-2a^2)$$ and hence we can write the expression on right of equation $(5)$ as $$\frac{3}{2}\cdot\frac{(a+2)^2-\sqrt{19(1-4a-2a^2)}}{a(a+3)\sqrt{6}}$$ and this equals the proposed value $\sqrt{6}$ if $$\sqrt{19(1-4a-2a^2)}=(a+2)^2-4a(a+3)=-3a^2-8a+4$$ Using numerical value of $a=0.16071$ we can confirm that the right hand side of above equation is positive and the above equation will hold true if the squares of both sides are equal ie if $$19(1-4a-2a^2)=(3a^2+8a-4)^2$$ or $$19(1-4a-2a^2)=9a^4+64a^2+16+48a^3-24a^2-64a$$ or $$3-12a-78a^2=9a^4+48a^3$$ or $$1-4a-26a^2=3a^4+16a^3$$ or $$1-4a-26a^2=3(18a^2+36a-5)+16(-5a^2-7a+1)$$ or $$1-4a-26a^2=1-4a-26a^2$$ which is true.
The case $n=35$ is a bit complicated as I haven't been able to find a suitable modular equation of degree $35$ which can be used to evaluate the class invariant $G_{35}$. Instead I use the table by Ramanujan to conclude that $G_{35}^8$ is a root of $$x^3-(60+28\sqrt {5})x^2-4\tag{6}$$ and its numerical value is around $122.61017$.
To simplify typing we use the symbol $a=\sqrt {5}$ in handling the case $n=35$. Next we can note that $b=G_{35}^4$ is a root of $$x^6-(60+28a)x^4-4$$ and the above polynomial can be factored as $$(x^3-(6+2a)x^2-2(1+a)x-2)(x^3+(6+2a)x^2-2(1+a)x+2)$$ in polynomial ring $\mathbb{Q} (a) [x] $. The above factorization was done using Magma calculator online and one can verify it manually. The first factor has a positive root and thus $b=G_{35}^4$ is a root of $$x^3-(6+2a)x^2-2(1+a)x-2\tag{7}$$ Replacing $x$ with $x^2$ doesn't lead to any further factorization and thus $G_{35}^2$ is of degree $6$ over $\mathbb{Q} (a) $.
Luckily we can note that $2b$ is a root of $$x^3-(12+4a)x^2-(8+8a)x-16$$ and replacing $x$ with $x^2$ gives us the factorization (again via Magma) $$(x^3-4x^2+2(1-a)x-4)(x^3+4x^2+2(1-a)x+4)$$ with first factor having a positive root and thus $\sqrt {2}G_{35}^2$ is a root of $$x^3-4x^2+2(1-a)x-4$$ It follows that $c=G_{35}^2$ is a root of $$x^3-2\sqrt{2}x^2+(1-a)x-\sqrt{2}$$ and hence $$\sqrt {2}=\frac{c(c^2+1-a)}{2c^2+1}=\frac{c(b+1-a)}{2b+1}$$ as $b=c^2$. Therefore $$\frac{\sqrt{2}} {c} =\sqrt{\frac{2} {b}}=\frac{b+1-a}{2b+1}\tag{8}$$ is a rational function of $b$ with coefficients in $\mathbb{Q} (a) $ and this fact will be used later.
We also need to evaluate the expression $1/(2b+1)$ as a polynomial in $b$. While one can use Magma for this, the verification by hand is bit lengthy and hence we use algebra directly. We note that $(2b+1)$ is a root of $$(x-1)^3-(12+4a)(x-1)^2-(8+8a)(x-1)-16 $$ or $$x^3-(15+4a)x^2+19x-(21-4a)$$ and hence $$\frac{21-4a}{2b+1}=(2b+1)^2-(15+4a)(2b+1)+19=4b^2-(26+8a)b+5-4a$$ and noting that $$(21-4a)(21+4a)=19^2 $$ we get $$\frac{1}{2b+1}=\frac{1}{19^2}((84+16a)b^2-(706+272a)b+25-64a)$$ Next we evaluate the expression $\sqrt{2/b}=(b+1-a)/(2b+1)$ and this turns out to be simpler in form than $1/(2b+1)$ and we have $$\sqrt{\frac{2}{b}}=\frac{b+1-a}{2b+1}=\frac{1}{19}(-2(1+2a)b^2+(53+30a)b+3(9-a))\tag{9}$$
We now use the formula given by Ramanujan $$nP(q^{2n})-P(q^2)=\frac{4KL}{\pi^2}\left\{2(\sqrt{kl}+\sqrt{k'l'}-\sqrt {klk'l'}) +(4klk'l')^{-1/6}(1-\sqrt{kl}-\sqrt{k'l'})^3\right \}$$ for $n=35$. Here we assume $k$ to be of degree $35$ over $l$ with $q$ being nome corresponding to $l$ and $q^n$ the nome corresponding to $k$. Putting $q=e^{-\pi/\sqrt {35}}$ so that $q^n=e^{-\pi\sqrt{35}}$ and $k=l', l=k'$ and $L/K=K'/K=\sqrt {35}$ we get $$nP(q^{2n})-P(q^2)=\sqrt {35}\left (\frac{2K}{\pi}\right) ^2\left(2(2\sqrt {kk'} - kk') +(4k^2k'^2)^{-1/6}(1-2\sqrt{kk'})^{3}\right)$$ Using $2kk'=G_{35}^{-12}$ we get $$nP(q^{2n})-P(q^2)=\sqrt {35}\left(\frac{2K}{\pi}\right)^2\left(2(\sqrt{2}G_{35}^{-6}-G_{35}^{-12}/2)+G_{35}^4(1-\sqrt{2}G_{35}^{-6})^3\right)$$ We now have $$\sqrt{2}G_{35}^{-6}=\frac{\sqrt {2}}{G_{35}^2\cdot G_{35}^4}=\frac{1}{b}\sqrt{\frac{2}{b}}$$ and then $$nP(q^{2n})-P(q^2)=\sqrt {35}\left(\frac{2K}{\pi}\right)^2\left\{2\left(\frac{1}{b}\sqrt{\frac{2}{b}}-\frac{1}{2b^3}\right)+b\left(1- \frac{1}{b}\sqrt{\frac{2}{b}} \right)^3\right\}$$ Combining this with identity $$nP(q^{2n})+P(q^2)=\frac{6\sqrt {n}} {\pi}, q=e^{-\pi/\sqrt{n}} $$ we get $$2P(q^{2n})=\frac{6}{\pi\sqrt{n}}+\frac{1}{\sqrt{35}}\left(\frac{2K}{\pi}\right)^2\left\{2\left(\frac{1}{b}\sqrt{\frac{2}{b}}-\frac{1}{2b^3}\right)+b\left(1-\frac{1}{b}\sqrt{\frac{2}{b}}\right)^3\right\}$$ Changing the notation a bit so that $q=e^{-\pi\sqrt{35}}$ corresponds to $k$ we have $$2P(q^2)=\frac{6}{\pi\sqrt{n}}+\frac{1}{\sqrt{35}}\left(\frac{2K}{\pi}\right)^2\left\{2\left(\frac{1}{b}\sqrt{\frac{2}{b}}-\frac{1}{2b^3}\right)+b\left(1-\frac{1}{b}\sqrt{\frac{2}{b}}\right)^3\right\}$$ Using the identity $$2P(q^2)-P(-q)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)=\left(\frac{2K}{\pi}\right)^2\sqrt{1-G^{-24}}$$ we get $$\sqrt{n} P(-q)-\frac{6}{\pi}=\frac{1}{b^3}\left(\frac{2K}{\pi}\right)^2\left\{2b^2\sqrt{\frac{2}{b}}-1+b\left(b-\sqrt{\frac{2}{b}}\right)^3-\sqrt{35(b^6-1)}\right\}$$ and finally using $$Q(-q) =\left(\frac{2K}{\pi}\right)^4(1-4G^{-24})$$ we get $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n}P_n-\frac{6}{\pi}\right)=\frac{1}{\sqrt{b^6-4}}\left\{2b^2\sqrt{\frac{2}{b}}-1+b\left(b-\sqrt{\frac{2}{b}}\right)^3-\sqrt{35(b^6-1)}\right\}$$ We have to show that this expression equals $$(2+\sqrt{5})\sqrt{\frac{2}{\sqrt{5}}}$$ Using equation $(6)$ we have $$b^6-4=(60+28a)b^4$$ and $$\sqrt{(60+28a)(2/a)}=\sqrt{56+24a}=6+2a$$ and thus we have to establish that $$2b^2\sqrt{\frac{2}{b}}-1+b\left(b-\sqrt{\frac{2}{b}}\right)^3-\sqrt{35(b^6-1)}=(2+a) (6+2a)b^2=(22+10a)b^2$$ or $$2b^2\sqrt{\frac{2}{b}}+b\left(b-\sqrt{\frac{2}{b}}\right)^3-(22+10a)b^2-1=\sqrt{35(b^6-1)}$$ We have $$b-\sqrt{\frac{2}{b}}=\frac{1}{19}(2(1+2a)b^2-2(17+15a)b-3(9-a))$$ and $$2b^2\sqrt{\frac{2} {b}} =\frac{1}{19}(2(9-a)b^2+4(9-a)b+4(1+2a))$$ Next we have $$\left(b-\sqrt{\frac{2}{b}} \right) ^2=\frac{1}{19}(4(9-a)b^2-8(10+a)b-2(15+11a))$$ and $$\left(b-\sqrt{\frac{2}{b}}\right) ^3=\frac{4}{19}((15+11a)b^2+2(15+11a)b+2(8-3a))$$ and $$b\left(b-\sqrt{\frac{2}{b}}\right) ^3=\frac{8}{19}((115+59a)b^2+(78+23a)b+(15+11a))$$ Using these expressions above we get $$2b^2\sqrt{\frac{2}{b}}+b\left(b-\sqrt{\frac{2}{b}}\right)^3-(22+10a)b^2-1\\=\frac{1}{19}((520+280a)b^2+(660+180a)b+(105+96a))$$ The expression on right is a positive real number (because $a, b$ are positive as well) and hence our job is done if we show that square of this number equals $$35(b^6-1)=35((22+10a)b^4+3)\\=35((7120+3184a)b^2 + (4280+1912a)b + 1283+576a) $$ I haven't verified this using hand calculation and relied on Magma for the same. All other calculations dealing with $a, b$ have been verified both via pen / paper and Magma. One should also note that Magma's function Sqrt when applied to $35(b^6-1)$ gives the negative square root. This is because number field calculations are not treated as calculations in an ordered field and hence if one wishes to use such functions one should try both positive and negative values and if needed check using numerical evaluation of the expressions as well.
The Magma code below creates the field extensions $K=\mathbb{Q} (a), L=K(b) $ and then one can evaluate expressions in field $L$ as polynomials in $b$ with coefficients in $K$. The code can be run using online Magma calculator.
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 $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)=\frac{G_n^{12}\cdot B_n(k, k') } {2\sqrt {G_n^{24}-4} }$$
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.
For $n=35$ we evaluate the expression $$\sqrt{60+28\sqrt {5}}\left(\sqrt {35}-\frac{6}{\pi}\right)$$ as $44.3606214$ and its decimal digits match that of $10\sqrt {5}$. Subtracting $10\sqrt{5}$ from the result we get $21.9999416$ and thus the expected value of above expression is $22+10\sqrt{5}$ and thus $P_n$ equals $$\frac{22+10\sqrt{5}}{\sqrt{60+28\sqrt{5}}}=(2+\sqrt{5})\sqrt {\frac{2}{\sqrt{5}}}$$ Having guessed the value of $P_n$ 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 $$\left(\frac{2K}{\pi}\right)^2=\frac{1}{\sqrt{1-4G^{-24}}}{}_3F_2\left(\frac{1}{6},\frac{5}{6},\frac{1}{2};1,1;-\frac{27G^{-24}}{(1-4G^{-24})^3}\right)\tag{10}$$ Also let us use the simplified notation $$f(x) ={}_3F_2\left(\frac{1}{6},\frac{5}{6},\frac{1}{2};1,1;x\right),g(x) =xf'(x) \tag{11}$$ in what follows.
Using derivatives of elliptic integrals we can prove that $$P(-q) = \left(\frac{2K}{\pi}\right)^2(1-2k^2)+3kk'^2\frac{d}{dk} \left(\frac{2K}{\pi}\right)^2\tag{12}$$ Applying the above formula on equation $(10)$ and in the process doing reasonable amount of symbol shunting we arrive at the following beautiful result $$P(-q) =\frac{(1-2k^2)(1+8G^{-24})}{(1-4G^{-24})^{3/2}}\left(f(v(k))+6g(v(k))\right),v(k)=-\frac{27G^{-24}}{(1-4G^{-24})^3}\tag{13}$$ 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 $(13)$ 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 $(12), (13)$ we get $$P_n= \frac{6}{\pi\sqrt{n}}+\frac{p_n}{\sqrt{n}}\cdot f(v(k)) \tag{14}$$ Comparing equations $(13),(14)$ and using the values $p_n$ and the relations between $Q_n, R_n$ we get desired series for $1/\pi$ as $$\frac{6}{\pi}=\left(\sqrt{\frac{nR_n^2}{Q_n^3}}-p_n\right)f(v(k))+6\sqrt{\frac{nR_n^2}{Q_n^3}}g(v(k))\tag{15}$$ 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 $(15)$ becomes $$\frac{6u_n\sqrt{u_n}}{\pi}\\=\left(\sqrt{n(u_n^3+27)}-p_nu_n\sqrt{u_n}\right)f\left(-\frac{27}{u_n^3}\right)+6\sqrt{n(u_n^3+27)}\cdot g \left(-\frac{27}{u_n^3}\right) \tag{16}$$ 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 $n=35$ we have $u_n=60+28\sqrt{5}$ and these expressions are algebraic integers from field $\mathbb{Q} (\sqrt{5})$. For $n=27$ we have $u_n^3=192000$ and $$\sqrt {n(u_n^3+27)},p_n\sqrt{u_n^3}$$ are positive integers.
The above formula holds for all positive integers $n>3$ with $u_n=(G_n^{24}-4)/G_n^{16}$ and in practice one uses odd values of $n$ because the evaluation and form of $G_n$ is simpler for odd $n$. Thus for example we have $u_n=15/4,p_n=\sqrt{3/5}$ for $n=7$. It turns out that for $n=7,11,19,27,43,67,163$ the series for $1/\pi$ involve only one quadratic irrationality and other numbers appearing in the series are rational.
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 $$ \sqrt{n(u_n^3+27)}-p_nu_n\sqrt{u_n}=\sqrt{11\cdot 539}-32=45$$ and $$ \sqrt{n(u_n^3+27)}=77$$ and using equation $(16)$ we get the desired series as $$\frac{96\sqrt{2}}{\pi}=45f(-27/512)+6\cdot 77g(-27/512) $$ 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 $$\sqrt{n(u_n^3+27)}-p_nu_n\sqrt{u_n}\\=\sqrt{163\cdot 27(53360^3+1)}-160080\cdot 362\sqrt{\frac{3}{3335}}\sqrt{160080}\tag{17}$$ 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 $(17)$ as $$27\cdot 163\cdot 7\cdot 11\cdot 19\cdot 127-160080\cdot 362\cdot 12=122322681$$ and we have $$\sqrt{n(u_n^3+27)} =817710201$$ and using equation $(16)$ we get $$\frac{6\cdot 160080\sqrt {160080}}{\pi} =122322681f(-1/53360^3)+817710201\cdot 6g(-1/53360^3)$$ 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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Evaluation of $P_n=P(-e^{-\pi\sqrt{n}}) =P(-q) $
To handle the case $n=19$ we start with a complicated modular equation of degree $19$ $$(\sqrt{kl} +\sqrt {k'l'} - 1)^5+ 112(\sqrt{kl}+\sqrt{k'l'}-1)^2\sqrt{klk'l'}\\-256(\sqrt{klk'l'}-\sqrt{kl} - \sqrt{k'l'})\sqrt{klk'l'} =0$$ Putting $l=k', k=l'$ we get $$(2\sqrt{kk'}-1)^5+112(2\sqrt{kk'}-1)^2kk'-256kk'(kk'-2\sqrt{kk'})=0$$ Putting $a=\sqrt{2}G_{19}^{-6}=2\sqrt{kk'}$ we get $$(a-1)^5+28(a-1)^2a^2-16a^4+64a^3=0$$ or $$a^5+7a^4+18a^3+18a^2+5a-1=0$$ Luckily $a=-1$ turns out to be a root and hence there is a factor $(a+1)$ involved here. We then have $$(a+1)(a^4+6a^3+12a^2+6a-1)=0$$ Since $a>0$ we have $$a^4+6a^3+12a^2+6a-1=0$$ Again we get a factor of $(a+1)$ and then $$(a+1)(a^3+5a^2+7a-1)=0$$ Since $a>0$ it turns out that $a=\sqrt{2}G_{19}^{-6}$ is a root of $$x^3+5x^2+7x-1\tag{1}$$ and the above polynomial is irreducible as neither $1$ nor $-1$ is a root.Next we need the formula given by Ramanujan as $$nP(q^{2n})-P(q^2)=\frac{24KL}{\pi^2}\left\{1+kl+k'l'+\sqrt{kl}+\sqrt{k'l'}-\sqrt{klk'l'}\right\}\tag{2} $$ where $n=19$ and $k$ is of degree $19$ over $l$ so that nome $q$ corresponds to $l$ and $q^n$ to $k$. Putting $q=e^{-\pi/\sqrt{19}}$ so that $q^n=e^{-\pi\sqrt{19}}$ and $k=l', k'=l$ we get $$nP(q^{2n})-P(q^2)=6\left(\frac{2K}{\pi}\right)^2\frac{K'}{K}(1+kk'+2\sqrt{kk'})$$ or $$nP(q^{2n})-P(q^2)=6\sqrt{19}\left(\frac{2K}{\pi}\right)^2(1+(a^2/4)+a)=\frac{3\sqrt{19}}{2}\left(\frac{2K}{\pi}\right)^2(a+2)^2$$ Combining the above equation with the identity $$nP(q^{2n})+P(q^2)=\frac{6\sqrt{n}}{\pi}, q=e^{-\pi/\sqrt {n}}\tag{3} $$ we get $$2P(q^{2n})=\frac{6}{\pi\sqrt{n}}+\frac{3}{2\sqrt{19}}\left(\frac{2K}{\pi}\right)^2(a+2)^2$$ Switching notation such that $k, K$ correspond to nome $q=e^{-\pi\sqrt {19}}$ we get $$2P(q^2)-\frac{6}{\pi\sqrt {n}} =\frac{3}{2\sqrt{19}}\left(\frac{2K}{\pi}\right) ^2(a+2)^2$$ Using the identity $$2P(q^2)-P(-q)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)=\left(\frac{2K}{\pi}\right)^2\sqrt{1-G^{-24}}\tag{4}$$ we get $$\sqrt{n} P(-q) - \frac{6}{\pi}=\left(\frac{2K}{\pi}\right)^2\frac{3(a+2)^2-\sqrt {19(4-a^4)}}{2}$$ And then using identity $$Q(-q) = \left(\frac{2K}{\pi}\right)^4(1-4G^{-24})$$ we get $$\frac{1}{\sqrt{Q_n}}\left (\sqrt{n}P_n-\frac{6}{\pi}\right) =\frac{3(a+2)^2-\sqrt{19(4-a^4)}} {2\sqrt{1-a^4}}\tag{5}$$ To simplify the expression on right we evaluate some integral powers of $a$ \begin{align} a^3&=-5a^2-7a+1\notag\\ a^4&=-5a^3-7a^2+a=18a^2+36a-5\notag\\ \frac{1}{a}&=a^2+5a+7\notag\\ \frac{1}{a^2}&=a+5+\frac{7}{a}=7a^2+36a+54\notag \end{align} We can observe that $$\frac{1}{a^2}=a^2+6(a^2+6a+9)=a^2+6(a+3)^2$$ and thus on multiplying by $a^2$ we get $$1-a^4=6a^2(a+3)^2$$ ie $$\sqrt{1-a^4}=a(a+3)\sqrt{6}$$ Further $$4-a^4=9(1-4a-2a^2)$$ and hence we can write the expression on right of equation $(5)$ as $$\frac{3}{2}\cdot\frac{(a+2)^2-\sqrt{19(1-4a-2a^2)}}{a(a+3)\sqrt{6}}$$ and this equals the proposed value $\sqrt{6}$ if $$\sqrt{19(1-4a-2a^2)}=(a+2)^2-4a(a+3)=-3a^2-8a+4$$ Using numerical value of $a=0.16071$ we can confirm that the right hand side of above equation is positive and the above equation will hold true if the squares of both sides are equal ie if $$19(1-4a-2a^2)=(3a^2+8a-4)^2$$ or $$19(1-4a-2a^2)=9a^4+64a^2+16+48a^3-24a^2-64a$$ or $$3-12a-78a^2=9a^4+48a^3$$ or $$1-4a-26a^2=3a^4+16a^3$$ or $$1-4a-26a^2=3(18a^2+36a-5)+16(-5a^2-7a+1)$$ or $$1-4a-26a^2=1-4a-26a^2$$ which is true.
The case $n=35$ is a bit complicated as I haven't been able to find a suitable modular equation of degree $35$ which can be used to evaluate the class invariant $G_{35}$. Instead I use the table by Ramanujan to conclude that $G_{35}^8$ is a root of $$x^3-(60+28\sqrt {5})x^2-4\tag{6}$$ and its numerical value is around $122.61017$.
To simplify typing we use the symbol $a=\sqrt {5}$ in handling the case $n=35$. Next we can note that $b=G_{35}^4$ is a root of $$x^6-(60+28a)x^4-4$$ and the above polynomial can be factored as $$(x^3-(6+2a)x^2-2(1+a)x-2)(x^3+(6+2a)x^2-2(1+a)x+2)$$ in polynomial ring $\mathbb{Q} (a) [x] $. The above factorization was done using Magma calculator online and one can verify it manually. The first factor has a positive root and thus $b=G_{35}^4$ is a root of $$x^3-(6+2a)x^2-2(1+a)x-2\tag{7}$$ Replacing $x$ with $x^2$ doesn't lead to any further factorization and thus $G_{35}^2$ is of degree $6$ over $\mathbb{Q} (a) $.
Luckily we can note that $2b$ is a root of $$x^3-(12+4a)x^2-(8+8a)x-16$$ and replacing $x$ with $x^2$ gives us the factorization (again via Magma) $$(x^3-4x^2+2(1-a)x-4)(x^3+4x^2+2(1-a)x+4)$$ with first factor having a positive root and thus $\sqrt {2}G_{35}^2$ is a root of $$x^3-4x^2+2(1-a)x-4$$ It follows that $c=G_{35}^2$ is a root of $$x^3-2\sqrt{2}x^2+(1-a)x-\sqrt{2}$$ and hence $$\sqrt {2}=\frac{c(c^2+1-a)}{2c^2+1}=\frac{c(b+1-a)}{2b+1}$$ as $b=c^2$. Therefore $$\frac{\sqrt{2}} {c} =\sqrt{\frac{2} {b}}=\frac{b+1-a}{2b+1}\tag{8}$$ is a rational function of $b$ with coefficients in $\mathbb{Q} (a) $ and this fact will be used later.
We also need to evaluate the expression $1/(2b+1)$ as a polynomial in $b$. While one can use Magma for this, the verification by hand is bit lengthy and hence we use algebra directly. We note that $(2b+1)$ is a root of $$(x-1)^3-(12+4a)(x-1)^2-(8+8a)(x-1)-16 $$ or $$x^3-(15+4a)x^2+19x-(21-4a)$$ and hence $$\frac{21-4a}{2b+1}=(2b+1)^2-(15+4a)(2b+1)+19=4b^2-(26+8a)b+5-4a$$ and noting that $$(21-4a)(21+4a)=19^2 $$ we get $$\frac{1}{2b+1}=\frac{1}{19^2}((84+16a)b^2-(706+272a)b+25-64a)$$ Next we evaluate the expression $\sqrt{2/b}=(b+1-a)/(2b+1)$ and this turns out to be simpler in form than $1/(2b+1)$ and we have $$\sqrt{\frac{2}{b}}=\frac{b+1-a}{2b+1}=\frac{1}{19}(-2(1+2a)b^2+(53+30a)b+3(9-a))\tag{9}$$
We now use the formula given by Ramanujan $$nP(q^{2n})-P(q^2)=\frac{4KL}{\pi^2}\left\{2(\sqrt{kl}+\sqrt{k'l'}-\sqrt {klk'l'}) +(4klk'l')^{-1/6}(1-\sqrt{kl}-\sqrt{k'l'})^3\right \}$$ for $n=35$. Here we assume $k$ to be of degree $35$ over $l$ with $q$ being nome corresponding to $l$ and $q^n$ the nome corresponding to $k$. Putting $q=e^{-\pi/\sqrt {35}}$ so that $q^n=e^{-\pi\sqrt{35}}$ and $k=l', l=k'$ and $L/K=K'/K=\sqrt {35}$ we get $$nP(q^{2n})-P(q^2)=\sqrt {35}\left (\frac{2K}{\pi}\right) ^2\left(2(2\sqrt {kk'} - kk') +(4k^2k'^2)^{-1/6}(1-2\sqrt{kk'})^{3}\right)$$ Using $2kk'=G_{35}^{-12}$ we get $$nP(q^{2n})-P(q^2)=\sqrt {35}\left(\frac{2K}{\pi}\right)^2\left(2(\sqrt{2}G_{35}^{-6}-G_{35}^{-12}/2)+G_{35}^4(1-\sqrt{2}G_{35}^{-6})^3\right)$$ We now have $$\sqrt{2}G_{35}^{-6}=\frac{\sqrt {2}}{G_{35}^2\cdot G_{35}^4}=\frac{1}{b}\sqrt{\frac{2}{b}}$$ and then $$nP(q^{2n})-P(q^2)=\sqrt {35}\left(\frac{2K}{\pi}\right)^2\left\{2\left(\frac{1}{b}\sqrt{\frac{2}{b}}-\frac{1}{2b^3}\right)+b\left(1- \frac{1}{b}\sqrt{\frac{2}{b}} \right)^3\right\}$$ Combining this with identity $$nP(q^{2n})+P(q^2)=\frac{6\sqrt {n}} {\pi}, q=e^{-\pi/\sqrt{n}} $$ we get $$2P(q^{2n})=\frac{6}{\pi\sqrt{n}}+\frac{1}{\sqrt{35}}\left(\frac{2K}{\pi}\right)^2\left\{2\left(\frac{1}{b}\sqrt{\frac{2}{b}}-\frac{1}{2b^3}\right)+b\left(1-\frac{1}{b}\sqrt{\frac{2}{b}}\right)^3\right\}$$ Changing the notation a bit so that $q=e^{-\pi\sqrt{35}}$ corresponds to $k$ we have $$2P(q^2)=\frac{6}{\pi\sqrt{n}}+\frac{1}{\sqrt{35}}\left(\frac{2K}{\pi}\right)^2\left\{2\left(\frac{1}{b}\sqrt{\frac{2}{b}}-\frac{1}{2b^3}\right)+b\left(1-\frac{1}{b}\sqrt{\frac{2}{b}}\right)^3\right\}$$ Using the identity $$2P(q^2)-P(-q)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)=\left(\frac{2K}{\pi}\right)^2\sqrt{1-G^{-24}}$$ we get $$\sqrt{n} P(-q)-\frac{6}{\pi}=\frac{1}{b^3}\left(\frac{2K}{\pi}\right)^2\left\{2b^2\sqrt{\frac{2}{b}}-1+b\left(b-\sqrt{\frac{2}{b}}\right)^3-\sqrt{35(b^6-1)}\right\}$$ and finally using $$Q(-q) =\left(\frac{2K}{\pi}\right)^4(1-4G^{-24})$$ we get $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n}P_n-\frac{6}{\pi}\right)=\frac{1}{\sqrt{b^6-4}}\left\{2b^2\sqrt{\frac{2}{b}}-1+b\left(b-\sqrt{\frac{2}{b}}\right)^3-\sqrt{35(b^6-1)}\right\}$$ We have to show that this expression equals $$(2+\sqrt{5})\sqrt{\frac{2}{\sqrt{5}}}$$ Using equation $(6)$ we have $$b^6-4=(60+28a)b^4$$ and $$\sqrt{(60+28a)(2/a)}=\sqrt{56+24a}=6+2a$$ and thus we have to establish that $$2b^2\sqrt{\frac{2}{b}}-1+b\left(b-\sqrt{\frac{2}{b}}\right)^3-\sqrt{35(b^6-1)}=(2+a) (6+2a)b^2=(22+10a)b^2$$ or $$2b^2\sqrt{\frac{2}{b}}+b\left(b-\sqrt{\frac{2}{b}}\right)^3-(22+10a)b^2-1=\sqrt{35(b^6-1)}$$ We have $$b-\sqrt{\frac{2}{b}}=\frac{1}{19}(2(1+2a)b^2-2(17+15a)b-3(9-a))$$ and $$2b^2\sqrt{\frac{2} {b}} =\frac{1}{19}(2(9-a)b^2+4(9-a)b+4(1+2a))$$ Next we have $$\left(b-\sqrt{\frac{2}{b}} \right) ^2=\frac{1}{19}(4(9-a)b^2-8(10+a)b-2(15+11a))$$ and $$\left(b-\sqrt{\frac{2}{b}}\right) ^3=\frac{4}{19}((15+11a)b^2+2(15+11a)b+2(8-3a))$$ and $$b\left(b-\sqrt{\frac{2}{b}}\right) ^3=\frac{8}{19}((115+59a)b^2+(78+23a)b+(15+11a))$$ Using these expressions above we get $$2b^2\sqrt{\frac{2}{b}}+b\left(b-\sqrt{\frac{2}{b}}\right)^3-(22+10a)b^2-1\\=\frac{1}{19}((520+280a)b^2+(660+180a)b+(105+96a))$$ The expression on right is a positive real number (because $a, b$ are positive as well) and hence our job is done if we show that square of this number equals $$35(b^6-1)=35((22+10a)b^4+3)\\=35((7120+3184a)b^2 + (4280+1912a)b + 1283+576a) $$ I haven't verified this using hand calculation and relied on Magma for the same. All other calculations dealing with $a, b$ have been verified both via pen / paper and Magma. One should also note that Magma's function Sqrt when applied to $35(b^6-1)$ gives the negative square root. This is because number field calculations are not treated as calculations in an ordered field and hence if one wishes to use such functions one should try both positive and negative values and if needed check using numerical evaluation of the expressions as well.
The Magma code below creates the field extensions $K=\mathbb{Q} (a), L=K(b) $ and then one can evaluate expressions in field $L$ as polynomials in $b$ with coefficients in $K$. The code can be run using online Magma calculator.
R<x>:= PolynomialRing(Integers()) ; f := x^2-5; K<a> := NumberField(f) ; T<y> := PolynomialRing(K) ; Factorization(y^6-(60+28*a)*y^4-4); g := y^3-(6+2*a)*y^2-2*(1+a)*y-2; Factorization(y^6-(12+4*a)*y^4-(8+8*a)*y^2-16); L<b> := ext<K|g>; Sqrt(2/b); (b+1-a)/(2*b+1); 2*b^2*Sqrt(2/b)+b*(b-Sqrt(2/b))^3-(22+10*a)*b^2-1; Sqrt(35*(b^6-1));
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 $$\frac{1}{\sqrt{Q_n}}\left(\sqrt{n} P_n-\frac{6}{\pi}\right)=\frac{G_n^{12}\cdot B_n(k, k') } {2\sqrt {G_n^{24}-4} }$$
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.
For $n=35$ we evaluate the expression $$\sqrt{60+28\sqrt {5}}\left(\sqrt {35}-\frac{6}{\pi}\right)$$ as $44.3606214$ and its decimal digits match that of $10\sqrt {5}$. Subtracting $10\sqrt{5}$ from the result we get $21.9999416$ and thus the expected value of above expression is $22+10\sqrt{5}$ and thus $P_n$ equals $$\frac{22+10\sqrt{5}}{\sqrt{60+28\sqrt{5}}}=(2+\sqrt{5})\sqrt {\frac{2}{\sqrt{5}}}$$ Having guessed the value of $P_n$ empirically we can then try to prove it analytically.
$P(-q)$ and Chudnovsky series for $1/\pi$
The values of $P_n=P(-q),Q_n=Q(-q),R_n=R(-q),q=e^{-\pi\sqrt{n}}$ in the table given by Ramanujan 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 $$\left(\frac{2K}{\pi}\right)^2=\frac{1}{\sqrt{1-4G^{-24}}}{}_3F_2\left(\frac{1}{6},\frac{5}{6},\frac{1}{2};1,1;-\frac{27G^{-24}}{(1-4G^{-24})^3}\right)\tag{10}$$ Also let us use the simplified notation $$f(x) ={}_3F_2\left(\frac{1}{6},\frac{5}{6},\frac{1}{2};1,1;x\right),g(x) =xf'(x) \tag{11}$$ in what follows.
Using derivatives of elliptic integrals we can prove that $$P(-q) = \left(\frac{2K}{\pi}\right)^2(1-2k^2)+3kk'^2\frac{d}{dk} \left(\frac{2K}{\pi}\right)^2\tag{12}$$ Applying the above formula on equation $(10)$ and in the process doing reasonable amount of symbol shunting we arrive at the following beautiful result $$P(-q) =\frac{(1-2k^2)(1+8G^{-24})}{(1-4G^{-24})^{3/2}}\left(f(v(k))+6g(v(k))\right),v(k)=-\frac{27G^{-24}}{(1-4G^{-24})^3}\tag{13}$$ 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 $(13)$ 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 $(12), (13)$ we get $$P_n= \frac{6}{\pi\sqrt{n}}+\frac{p_n}{\sqrt{n}}\cdot f(v(k)) \tag{14}$$ Comparing equations $(13),(14)$ and using the values $p_n$ and the relations between $Q_n, R_n$ we get desired series for $1/\pi$ as $$\frac{6}{\pi}=\left(\sqrt{\frac{nR_n^2}{Q_n^3}}-p_n\right)f(v(k))+6\sqrt{\frac{nR_n^2}{Q_n^3}}g(v(k))\tag{15}$$ 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 $(15)$ becomes $$\frac{6u_n\sqrt{u_n}}{\pi}\\=\left(\sqrt{n(u_n^3+27)}-p_nu_n\sqrt{u_n}\right)f\left(-\frac{27}{u_n^3}\right)+6\sqrt{n(u_n^3+27)}\cdot g \left(-\frac{27}{u_n^3}\right) \tag{16}$$ 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 $n=35$ we have $u_n=60+28\sqrt{5}$ and these expressions are algebraic integers from field $\mathbb{Q} (\sqrt{5})$. For $n=27$ we have $u_n^3=192000$ and $$\sqrt {n(u_n^3+27)},p_n\sqrt{u_n^3}$$ are positive integers.
The above formula holds for all positive integers $n>3$ with $u_n=(G_n^{24}-4)/G_n^{16}$ and in practice one uses odd values of $n$ because the evaluation and form of $G_n$ is simpler for odd $n$. Thus for example we have $u_n=15/4,p_n=\sqrt{3/5}$ for $n=7$. It turns out that for $n=7,11,19,27,43,67,163$ the series for $1/\pi$ involve only one quadratic irrationality and other numbers appearing in the series are rational.
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 $$ \sqrt{n(u_n^3+27)}-p_nu_n\sqrt{u_n}=\sqrt{11\cdot 539}-32=45$$ and $$ \sqrt{n(u_n^3+27)}=77$$ and using equation $(16)$ we get the desired series as $$\frac{96\sqrt{2}}{\pi}=45f(-27/512)+6\cdot 77g(-27/512) $$ 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 $$\sqrt{n(u_n^3+27)}-p_nu_n\sqrt{u_n}\\=\sqrt{163\cdot 27(53360^3+1)}-160080\cdot 362\sqrt{\frac{3}{3335}}\sqrt{160080}\tag{17}$$ 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 $(17)$ as $$27\cdot 163\cdot 7\cdot 11\cdot 19\cdot 127-160080\cdot 362\cdot 12=122322681$$ and we have $$\sqrt{n(u_n^3+27)} =817710201$$ and using equation $(16)$ we get $$\frac{6\cdot 160080\sqrt {160080}}{\pi} =122322681f(-1/53360^3)+817710201\cdot 6g(-1/53360^3)$$ 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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