Additional values of Dedekind's $\eta$ function in radical form

Question

Can anyone confirm the following values of the $\eta $ function to increase the table of the post What is the exact value of $\eta(6i)$? ?

$\eta(6i)=\frac{1}{2^{11/8}3^{3/8}}\big(2-\sqrt{3}\big)^{1/24}\big(\sqrt{2}-3^{1/4}\big)^{1/4} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(7i)=\frac{1}{2^{13/8}7^{7/16}}\sqrt{\sqrt{5-\sqrt{7}}-\sqrt{-7+3\sqrt{7}}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(8i)=\frac{1}{2^{73/32}}\big(\sqrt{2}-1\big)^{1/8}\big(2^{1/4}-1\big)^{1/2} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(9i)$ = $\frac{1} {6} \big(\sqrt{6}\, (2+\sqrt{3})^{1/6} -3 \big)^{1/3} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(10i) = \frac {1} {2^{11/8} \sqrt{5} \varphi^{1/2}} \frac {5^{1/4}-1} {\sqrt{2}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

where $\varphi$ is golden ratio.

$\eta(11i)=\frac{1}{2^{13/12}*3^{1/4}*11^{11/24}}\Big(4*22^{1/3}-(306 \sqrt{33}-837\sqrt{3}-351\sqrt{11}+1490)^{1/3}-(-306 \sqrt{33}+837\sqrt{3}-351\sqrt{11}+1490)^{1/3}\Big)^{1/4}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

A modular equation of 11th degree of Dedekind's $\eta$ function.

$\eta(12i) = \frac {1} {2^{31/16} 3^{3/8}} (2+\sqrt{3})^{5/48} (\sqrt{2}-3^{1/4})^{3/8} (\sqrt{2}-1)^{1/4} (\sqrt{3}-\sqrt{2})^{1/4} (3^{1/4}-1)^{1/2} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$.

$\eta(13i)=\frac{1} {2 \sqrt{3} \sqrt{13} } \sqrt{ -5- (15 \sqrt{39}+39 \sqrt{3}-18 \sqrt{13}-91 )^{1/3}+ (15 \sqrt{39}+39 \sqrt{3}+18 \sqrt{13}+91)^{1/3}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

A modular equation of 13th degree of Dedekind’s $ \eta$ function.

$\eta(14i)=\frac{1} {2^{11/4} 7^{7/16}} \big(\sqrt{\sqrt{3\sqrt{7}-7}+\sqrt{5-\sqrt{7}}}-\sqrt{\sqrt{27\sqrt{7}-7}-\sqrt{7\sqrt{7}+21}}\big) \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(15i)=\frac{1} {4 \sqrt{10}* 3^{3/8}}(\sqrt{5}-2)^{1/2}(2-\sqrt{3})^{11/12} \big(\frac{\sqrt{4+\sqrt{15}}-15^{1/4}} {2} \big)^{2} (540^{1/4}+60^{1/4}+2) \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(16i)=\frac{1}{2^{177/64}}\big(\sqrt{2}-1\big)^{1/16}\big(2^{1/4}-1\big)^{1/4} \big(\sqrt{1+\sqrt{2}}-2^{5/8}\big)^{1/2}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(17i)=\frac{1}{4\sqrt{34}} \sqrt{272^{1/8}\big(\sqrt{61-7\sqrt{17}}-\sqrt{5\sqrt{17}+17}\big)-17^{3/4}+3\sqrt{17}-17^{1/4}-1} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

A modular equation of 17th degree of Dedekind’s $\eta$ function.

$\eta(18i)=\frac{1} {2^{91/72} 3} \frac{\big(1-(2. 108^{1/4}-2\sqrt{3}-2)^{1/3}\big)^{1/3}} {\big(( 3.12^{1/4}+108^{1/4}+2\sqrt{3}+4)^{1/3}+2^{1/3}\big)^{1/3}} \big((6\sqrt{3}+18)^{1/3}-3\big)^{1/3} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(18i) =\frac{1} {2^{11/8}*3^{2/3}}\Big(\big(-\frac{5*12^{1/4}}{6}+\frac{7*\sqrt{3}}{9}+\frac{108^{1/4}}{6}+\frac{2}{3}\big)^{1/3}-1\Big)^{1/3} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

Thanks to Vladimir Reshetnikov

A modular equation of 19th degree of Dedekind’s $\eta$ function.

$\eta(19i)=\frac{1}{20. 2^{7/4}.19^{3/8}.1203^{1/4}} \Big( 100680000 +7361892000 \gamma+76992000 \sqrt{19} \gamma -1888138300 \gamma^{2}+145028140\gamma^{3}-4533799 \gamma^{4}\Big)^{1/4} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

with

$$\small\begin{align} \alpha &=30 \sqrt{198616747730+65513019062 \sqrt{5}},\\ \beta &=30 \sqrt{198616747730-65513019062 \sqrt{5}},\\ \eta &=\sqrt[5]{11410567+2790935 \sqrt{5}+\alpha\;}+\sqrt[5]{11410567+2790935 \sqrt{5}-\alpha\;}\\ &+\sqrt[5]{11410567-2790935 \sqrt{5}+\beta\;}-\sqrt[5]{2790935 \sqrt{5}-11410567+\beta\;},\,\text{and}\\ \gamma& =8-\left(\frac{2}{19}\right)^{4/5} \eta. \end{align}$$

$\eta(20i)=\frac{1} {2^{29/16}.\sqrt{5}} (\sqrt{2}-1)^{1/2} (5^{1/4}- \sqrt{2})^{1/2} (\sqrt{10}-3)^{1/4} \big(\frac{5^{1/4}-1} {\sqrt{2}}\big)^{3/2} \varphi^{1/4} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(20i)=\frac{1} {\sqrt{5}.2^{31/16}}\Big(-102.5^{1/4}+6.5^{3/4}+69\sqrt{5}-14\sqrt{2}+6\sqrt{10}-21\Big)^{1/8}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(21i)=\frac{1} {2^{11/8} 7^{7/16}\sqrt{3}} \frac{z a} {b c^{2} d e^{1/3}} \big(1+2\sqrt{2} \frac{b^{3/2} d^{3/2} e^{1/2}} {a^{3/2} c^{6}} \big)^{1/4}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$a=\sqrt{3+\sqrt{7}}-252^{1/8}$$

$$b==\sqrt{3+\sqrt{7}}+252^{1/8}$$

$$c=\frac{\sqrt{4+\sqrt{7}}+7^{1/4}} {2}$$

$$d=\frac{\sqrt{7}+\sqrt{3}} {2}$$

$$e=2+\sqrt{3}$$

$$z=\sqrt{\sqrt{13+\sqrt{7}}+\sqrt{7+3\sqrt{7}}}$$.

$\eta(22i)=\frac{(a-b-c)^{1/4}} {2^{35/24}*3^{1/4}*11^{11/24}*\sqrt{G}}*\big(G^{12}-\sqrt{G^{24}-1}\big)^{1/8}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$G=\frac{r(s+t)+2}{3*\sqrt{2}};$$ $$r=(3*\sqrt{11}+11)^{1/3};$$ $$s=(3*\sqrt{11}+3*\sqrt{3}+4)^{1/3};$$ $$t=(3*\sqrt{11}-3*\sqrt{3}+4)^{1/3};$$ $$a=4*22^{1/3};$$ $$b=(306*\sqrt{33}-837*\sqrt{3}-351*\sqrt{11}+1490)^{1/3};$$ $$c=(-306*\sqrt{33}+837*\sqrt{3}-351*\sqrt{11}+1490)^{1/3};$$

Thanks to Vladimir Reshetnikov

A modular equation of 23rd degree of Dedekind’s $\eta$ function.

$\eta(23i)=\frac{1}{2^{13/12}\cdot 3^{1/4}\cdot \sqrt{23}} \Big(\alpha^{1/3}+\frac{B^{2}-A^{2}}{\alpha^{1/3}}+2B\Big)^{1/4}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

with $$\alpha=32\sqrt{23}(38-9\sqrt{3})+2^{2}\cdot3^{3}\cdot23(9\cdot\sqrt{3}-4)-9\sqrt{2}23^{1/4)}(-454\sqrt{3}+139\sqrt{23}+62\sqrt{69}+717)$$

$$A=\frac{\sqrt{3}23^{1/8}(8920771 \cdot 23^{1/4}-702648\cdot 23^{3/4}+12400065\cdot \sqrt{46}-50943951\cdot \sqrt{2})^{1/6}}{2^{2/3}}$$

$$B=\frac{23^{1/4}(9+3 \cdot \sqrt{23}-2\cdot 92^{1/4})}{2^{7/6}}$$

$\eta(24i) =\frac{\sqrt{d}*c^{1/4}*e^{1/8}*f^{1/4}} {2^{75/32}*3^{3/8}*a^{3/8}*b^{1/12}}*\sqrt{b^{1/16}*c^{3/8}-\sqrt{2a*f}*e^{1/4}}\phi^{1/4} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$a=\sqrt{2}-1; b=2-\sqrt{3}; c=\sqrt{2}-3^{1/4}; d=2^{1/4}-1; e=\sqrt{3}-\sqrt{2}; f=3^{1/4}-1$

$\eta(25i)=\frac{1} {10} \Big(A^{1/5}-B^{1/5}-1 \Big) \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$A=\sqrt {(\frac{9 \sqrt{5}}{2}+\frac{45}{2} )}+\frac{\sqrt{5}}{2}+\frac{9}{2}$

$B=\sqrt{\frac{9\sqrt{5}}{2}+\frac{45}{2} }-\frac{\sqrt{5}}{2}-\frac{9}{2}$

$\eta(25i)=\frac{1} {10} \Big(-1 + (4+\varphi+3*5^{1/4}\sqrt{\varphi})^{1/5}+(4+\varphi-3*5^{1/4}\sqrt{\varphi})^{1/5} \Big) \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(25i)=\frac{1} {40 \varphi^{10}} \big( 1+(4\varphi)^{1/5} \big( (3+\frac{5^{1/4}} { \varphi^{3/2}} )^{1/5} + (3-\frac{5^{1/4}} { \varphi^{3/2}} )^{1/5} )\big). \big(1+\varphi^{3} \big( 1-(4/\varphi)^{1/5} \big( (3+5^{1/4} \varphi^{3/2})^{1/5} + (3-5^{1/4}\varphi^{3/2})^{1/5} \big) \big)^{2}\big)^{2} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

where $\varphi$ is golden ratio.

$\eta(26i)=\frac{\sqrt{-A+B-5}} {\sqrt{39}*2^{11/8}*\sqrt{G}}*\Big(G^{12}-\sqrt{G^{24}-1}\Big)^{1/8}*\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$G=\frac{\sqrt{a+b+26}+\sqrt{a+b+14}} {2*\sqrt{3}}; $$ $$a=(10673-936*\sqrt{3})^{1/3};$$

$$b=(10673+936*\sqrt{3})^{1/3};$$ $$A=(15*\sqrt{39}+39*\sqrt{3}-18*\sqrt{13}-91)^{1/3};$$ $$B=(15*\sqrt{39}+39*\sqrt{3}+18*\sqrt{13}+91)^{1/3}.$$

$\eta(27i)=\frac{(\sqrt{3}-1)^{1/6}} {2^{13/12}*3^{95/72}}\frac{\Big(-3^{5/6}\big((48\sqrt{3}-72)^{1/3}+(16\sqrt{3}-16)^{1/3}+3\sqrt{3}-3\big)^{1/3}+(72-24\sqrt{3})^{1/3}+4^{1/3}\Big)^{1/3} } { \big((48\sqrt{3}-72)^{1/3}+(16\sqrt{3}-16)^{1/3}+3\sqrt{3}-3\big)^{1/9} }\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(27i)=\frac{\Big(-3^{5/6}*a^{5/9}+\big(\sqrt{2}-\sqrt{3}*a^{1/6}\big)^{1/3}*\big(2*\sqrt{3}*a^{1/6}+\sqrt{2}\big)\Big)^{1/3}} {3^{23/72}*a^{11/108}}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$a=2-\sqrt{3}$$

$\eta(27i)=\frac{a^{1/12}}{6*3^{1/24}}\Big(-1+\big(+1+ \frac{\big(\sqrt{3}a^{1/6}-\sqrt{2}\big)^{4}}{3^{5/2}a^{5/3}}\big)^{1/3}\Big)^{1/3}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$a=2-\sqrt{3}$$

$\eta(28i)=\frac{1}{2^{29/16}*7^{7/16}} \frac{ \sqrt{ 8*\sqrt{m-n}+ ( \sqrt{n+p}-\sqrt{q-r} )^{3}} -2*\sqrt{2}*(m-n)^{1/4} } {a^{1/4}*(m-n)^{1/12}}*\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$a=\sqrt{2}-1;m=\sqrt{83-31*\sqrt{7}};n=\sqrt{3*\sqrt{7}-7};p=\sqrt{5-\sqrt{7}};$$

$$q=\sqrt{27*\sqrt{7}-7};r=\sqrt{7*\sqrt{7}+21}.$$

$\eta(30i)=\frac{1}{\color{blue}{4}\sqrt{5}.2^{7/8}.3^{3/8}} \frac{c\Big(a^{4}b^{12}\varphi^{12}-\sqrt{ a^{8}b^{24}\varphi^{24}-1}\Big)^{1/8}} {\varphi^{2}a^{13/12}b^{5/2}}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

where $\varphi$ is golden ratio, $a=2+\sqrt{3}$ and $b=\frac{\sqrt{4+\sqrt{15}}+15^{1/4}}{2}$ and $c=540^{1/4}+60^{1/4}+2.$

$\eta(32i)=\frac{1} {8.2^{33/128}}\frac{ (2^{1/4}-1)^{1/8} (\sqrt{1+\sqrt{2}} – 2^{5/8} )^{5/4} } { ( \sqrt{2}+1)^{1/32} (2^{1/4}+1+2^{13/16} (\sqrt{2}+1)^{1/4} )^{1/2} } \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(35i)= \frac{\sqrt { \sqrt{7+3 \sqrt{7}} + \sqrt{13+ \sqrt{7}}}} {\sqrt{5} . 2^{11/8}.7^{7/16} \varphi^{2}. \sqrt{b}. c^{3}. d^{2}} \sqrt{1+2 \frac{ \varphi b ^{1/4} d} {c^{7/2} } } \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

where $\varphi$ is golden ratio

$$b=6+\sqrt{35}$$

$$c=\frac{\sqrt{4+\sqrt{7}}+7^{1/4}} {2}$$

$$d=\sqrt{\frac{43+15 \sqrt{7}+(8+3\sqrt{7})\sqrt{10 \sqrt{7}}} {8}}+\sqrt{\frac{35+15 \sqrt{7}+(8+3\sqrt{7}) \sqrt{10 \sqrt{7}}} {8}}$$.

$\eta(36i)=\frac{a^{1/12}}{2^{29/16}*3^{5/6}*b^{1/18}}\Big(\sqrt{a}*\big(\sqrt{2}-\sqrt{3}*b^{1/6}\big)-3^{1/4}*\sqrt{e}*b^{5/16}*c^{11/8}*d^{1/4}\Big)^{1/3}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$a=\sqrt{2}-1; b=2-\sqrt{3}; c=\sqrt{2}-3^{1/4}; d=\sqrt{3}-\sqrt{2}; e=3^{1/4}-1.$$

$\eta(36i)=\frac{a^{1/4}}{2^{95/48}*3^{5/6}}\Big(A^{1/3}-\sqrt{6}\Big)^{1/3} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$ $$A=d^{3}e^{6}b^{5/4}c^{9/2}+6\sqrt{6}$$ $$a=\sqrt{2}-1;b=2+\sqrt{3};c=\sqrt{2}-3^{1/4};d=\sqrt{3}-\sqrt{2};e=3^{1/4}-1$$

$\eta(40i)=\frac{\sqrt{d*e}} {2^{77/32}*\sqrt{5*\varphi}*a^{3/8}} (\sqrt{b}*a^{1/4}*c^{1/4}*\varphi^{3/4}-d^{3/2})*\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$a=\sqrt{2}-1; b=5^{1/4}-\sqrt{2}; c=\sqrt{10}-3; d=\frac{5^{1/4}-1}{\sqrt{2}};e=2^{1/4}-1; \varphi=\frac{\sqrt{5}+1}{2}.$$

$\eta(45i)=\frac{1} {12 \sqrt{5}} (\frac{\sqrt{5}-1} {2})^{5/2} (3+\sqrt{5}+(\sqrt{3}+\sqrt{5}+60^{1/4}) (2+\sqrt{3})^{1/3}\big(\frac{\sqrt{2}(\frac{\sqrt{5}+1} {2})^{2} (2-\sqrt{3})^{1/3} \frac{\sqrt{4+\sqrt{15}}-15^{1/4}} {2}-1} {\sqrt{2}(\frac{\sqrt{5}-1} {2})^{2} (2+\sqrt{3})^{1/3} \frac{\sqrt{4+\sqrt{15}}+15^{1/4}} {2}+1}\big)^{2/3}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(45i)=\frac{1} {2.3. \sqrt{5}.a^{2/9}.b^{2/3}.\varphi^{7/6}} \Big(\sqrt{2}.a.b^{3}-3.b^{2}.\varphi^{2}.a^{2/3}+2\varphi^{6}\Big)^{1/3}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$a=2+\sqrt{3}$$

$$b=\frac {\sqrt{4+\sqrt{15}}+15^{1/4}}{2}$$

$$\varphi=\frac{\sqrt{5}+1}{2}$$

$\eta(50i)=\frac{2^{1/8}}{40}\frac{Ab\Big(B^{2}\varphi^{3}+1\Big)^{3}}{d^{5}\varphi^{15}\Big(C^{2}-b\Big)} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$C=1+u^{1/5}+v^{1/5}$$

$$u=8d\big(+\sqrt{4-d^{2}}+\sqrt{2}d^{2}-d+\sqrt{2}\big)$$

$$v=8d\big(-\sqrt{4-d^{2}}+\sqrt{2}d^{2}-d+\sqrt{2}\big)$$

$$A=1+(4\varphi)^{1/5}\big(m^{1/5}+n^{1/5}\big)$$

$$m=3+\frac{5^{1/4}}{\varphi^{3/2}}$$

$$n=3-\frac{5^{1/4}}{\varphi^{3/2}}$$

$$B=1-(\frac{4}{\varphi})^{1/5}\big(r^{1/5}+s^{1/5}\big)$$

$$r=3+5^{1/4}\varphi^{3/2}$$

$$s=3-5^{1/4}\varphi^{3/2}$$

$$b=3+2*5^{1/4}$$

$$d=\frac{5^{1/4}+1}{\sqrt{2}}$$

$\varphi$ is golden ratio.

$\eta(50i)≅\frac{2^{1/8}\varphi}{25\cdot D^{3}}\Big(a^{1/5}+b^{1/5}-1\Big)^{3}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$a=4+\varphi+3\cdot 5^{1/4}\sqrt{\varphi}$$ $$b=4+\varphi-3\cdot 5^{1/4}\sqrt{\varphi}$$ $$D=\frac{5^{1/4}-1}{\sqrt{2}}$$

$\eta(54i)=\frac{2^{5/8}}{3^{79/72}a^{1/72}b^{1/12}}\Big(\big(b^{3}\sqrt{3a}+(f^{1/3}-1)^{4}\big)^{1/3}-b(3a)^{1/6}\Big)^{1/3}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$ $$a=2-\sqrt{3};b=\sqrt{2}-3^{1/4};f=-\frac{5*12^{1/4}}{6}+\frac{7*\sqrt{3}}{9}+\frac{108^{1/4}}{6}+\frac{2}{3}$$

$\eta(63i)=\frac{\sqrt{\sqrt{5-\sqrt{7}}-\sqrt{3\sqrt{7}-7}}} {2^{13/8}.3.7^{7/16}} \big( \frac{ 2 ( \sqrt{3+\sqrt{7}}-252^{1/4}) (\frac{\sqrt{4+\sqrt{7}}+7^{1/4}} {2})^{4}} {\sqrt{3+\sqrt{7}}+252^{1/4}) (\frac{\sqrt{7}+\sqrt{3}} {2}) (\frac{\sqrt{3}+1} {\sqrt{2}})^{2/3}} + \frac{\sqrt{2}( \sqrt{3+\sqrt{7}}+252^{1/4}) (\frac{\sqrt{7}+\sqrt{3}} {2})^{1/2} (\frac{\sqrt{3}+1} {\sqrt{2}})^{1/3}} {( \sqrt{3+\sqrt{7}}-252^{1/4}) (\frac{\sqrt{4+\sqrt{7}}+7^{1/4}} {2})^{2}}-3 \big)^{1/3} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$\eta(81i)=\frac{\Big(-3^{17/18}*a^{5/27}*b^{4/3}+\sqrt{ 2}\big(-3^{5/6}*a^{5/9}+\sqrt{2}*b^{1/3}*(\sqrt{6}*a^{1/6}+1)\big)^{1/3}*(\sqrt{2}*3^{5/6}*a^{5/9}+\sqrt{6}*a^{1/6}*b^{1/3}+b^{1/3})\Big)^{1/3} } {2*3^{49/27}*a^{19/162}*b^{1/9}}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$b=\sqrt{2}-\sqrt{3}*a^{1/6};$$ $$a=2-\sqrt{3}.$$

$\eta(90i)=\frac{D}{2^{11/8}3^{2/3}\sqrt{5\varphi}}\Big(-1+\Big(1+\frac{\sqrt{3}c^{12}\big(a^{4}b^{12}\varphi^{12}-\sqrt{a^{8}b^{24}\varphi^{24}-1}\big)^{3/2}}{2^{18}3^{2}D^{12}a^{13}b^{30}\varphi^{18}}\Big)^{1/3}\Big)^{1/3}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$a=2+\sqrt{3}$$

$$b=\frac{\sqrt{4+\sqrt{15}}+15^{1/4}}{2}$$

$$c=540^{1/4}+60^{1/4}+2$$

$$D=\frac{5^{1/4}-1}{\sqrt{2}}$$.

$\varphi$ is the golden ratio.

$\eta(100i)=\frac{2^{1/16}}{10}\frac{A b^{7/2}F^{11/2}}{G^{5/2}H^{1/4}d^{25/2}\varphi^{45/2}}\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}$

$$H=\sqrt{2}F^{4}d^{20}G^{4}+2048b^{4}\varphi^{20}$$

$$G=C^{2}-b$$

$$F=1+\varphi^{3}B^{2}$$

$$C=1+u^{1/5}+v^{1/5}$$

$$u=8d\big(+\sqrt{4-d^{2}}+\sqrt{2}d^{2}-d+\sqrt{2}\big)$$

$$v=8d\big(-\sqrt{4-d^{2}}+\sqrt{2}d^{2}-d+\sqrt{2}\big)$$

$$A=1+(4\varphi)^{1/5}\Big(\big(3+\frac{5^{1/4}}{\varphi^{3/2}}\big)^{1/5}+ (\big(3-\frac{5^{1/4}}{\varphi^{3/2}}\big)^{1/5}\Big)$$

$$B=1-(\frac{4}{\varphi})^{1/5}\Big(\big(3+5^{1/4}\varphi^{3/2}\big)^{1/5}+ (\big(3-5^{1/4}\varphi^{3/2}\big)^{1/5}\Big)$$

$$b=3+2*5^{1/4}$$

$$d=\frac{5^{1/4}+1}{\sqrt{2}}$$

$$=\varphi=\frac{\sqrt{5}+1}{2}$$

Encouraged by Gerald Edgar (https://math.stackexchange.com/users/442/gedgar)

I would like to implement the list of Dedekind’s $\eta$-function of the type $\eta(i\sqrt{n})$, with $n∈N$.

$\eta(i)=\frac{\Gamma\big(\frac{1}{4}\big)}{2\pi^{3/4}}$

$\eta(i\sqrt{2})=\frac{1}{2^{11/8}} \frac{\Big(\Gamma\big(\frac{1}{8}\big)\Gamma\big(\frac{3}{8}\big)\Big)^{1/2}}{\pi^{3/4}}$

$\eta(i\sqrt{3})=\frac{3^{1/8}}{2^{4/3}}\frac{\Big(\Gamma\big(\frac{1}{3}\big)\Big)^{3/2}}{\pi}$

$\eta(2i)=\frac{1}{2^{11/8}}\frac{\Gamma\big(\frac{1}{4}\big)}{\pi^{3/4}}$

$\eta(i\sqrt{5})=\frac{1}{2^{5/4}5^{1/4}\varphi^{1/8}}\frac{\Big(\Gamma\big(\frac{1}{20}\big)\Gamma\big(\frac{3}{20}\big) \Gamma\big(\frac{7}{20}\big) \Gamma\big(\frac{9}{20}\big)\Big)^{1/4}}{\pi^{3/4}}$

$\eta(i\sqrt{6})=\frac{1}{2^{3/2}3^{1/4}}\Big(\sqrt{2}-1\Big)^{1/12}\frac{\Big(\Gamma\big(\frac{1}{24}\big)\Gamma\big(\frac{5}{24}\big) \Gamma\big(\frac{7}{24}\big) \Gamma\big(\frac{11}{24}\big)\Big)^{1/4}}{\pi^{3/4}}$

$\eta(i\sqrt{7})=\frac{1}{2^{3/2}7^{1/8}}\frac{\Big(\Gamma\big(\frac{1}{7}\big)\Gamma\big(\frac{2}{7}\big) \Gamma\big(\frac{4}{7}\big) \Big)^{1/2}}{\pi}$

$\eta(2i\sqrt{2})=\frac{\Big(\sqrt{2}-1\Big)^{1/8}}{2^{7/4}}\frac{\Big(\Gamma\big(\frac{1}{8}\big)\Gamma\big(\frac{3}{8}\big)\Big)^{1/2}}{\pi^{3/4}}$

$\eta(3i)=\frac{\Big(2-\sqrt{3}\Big)^{1/12}}{2*3^{3/8}}\frac{\Gamma\big(\frac{1}{4}\big)}{\pi^{3/4}}$

$\eta(i\sqrt{10})=\frac{1}{4\big(5\varphi\big)^{1/4}}\frac{\Big(\Gamma\big(\frac{1}{40}\big)\Gamma\big(\frac{7}{40}\big) \Gamma\big(\frac{9}{40}\big) \Gamma\big(\frac{11}{40}\big) \Gamma\big(\frac{13}{40}\big) \Gamma\big(\frac{19}{40}\big) \Gamma\big(\frac{23}{40}\big) \Gamma\big(\frac{37}{40}\big)\Big)^{1/4}}{\pi^{5/4}}$

with $\varphi=\frac{\sqrt{5}+1}{2}$

$\eta(i\sqrt{11})=\frac{1}{3*2^{11/6}*11^{1/8}}\Big(\frac{f^{1/3}-g^{1/3}+2}{a^{1/3}+b^{1/3}}\Big)\frac{\Gamma\big(\frac{1}{11}\big)\Gamma\big(\frac{3}{11}\big) \Gamma\big(\frac{4}{11}\big) \Gamma\big(\frac{5}{11}\big) \Gamma\big(\frac{9}{11}\big) \Big)^{1/2}}{\pi^{3/2}}$

with $f=3\sqrt{33}+17$;$g=3\sqrt{33}-17$;$a=1+\sqrt{\frac{11}{27}}$;$b=1- \sqrt{\frac{11}{27}}$.

$\eta(2i\sqrt{3})=\frac{3^{1/8}}{2^{7/4}}\big(\frac{\sqrt{3}-1}{\sqrt{2}}\big)^{1/4}\frac{\Big(\Gamma\big(\frac{1}{3}\big)^{3/2}}{\pi}$

$\eta(i\sqrt{13})=\frac{1}{2^{9/4}*13^{1/4}}\Big(\frac{\sqrt{13}-3}{2}\Big)^{1/8}\frac{\Big(\Gamma\big(\frac{1}{52}\big)\Gamma\big(\frac{7}{52}\big) \Gamma\big(\frac{9}{52}\big) \Gamma\big(\frac{11}{52}\big) \Gamma\big(\frac{15}{52}\big) \Gamma\big(\frac{17}{52}\big) \Gamma\big(\frac{19}{52}\big) \Gamma\big(\frac{25}{52}\big) \Gamma\big(\frac{29}{52}\big) \Gamma\big(\frac{31}{52}\big) \Gamma\big(\frac{47}{52}\big) \Gamma\big(\frac{49}{52}\big)\Big)^{1/4}}{\pi^{7/4}}$

$\eta(i\sqrt{14})=\frac{1}{4*7^{1/4}}\Big(\sqrt{1+\sqrt{2}}-\sqrt{5-3\sqrt{2}}\Big)^{1/4} \frac{\Big(\Gamma\big(\frac{1}{56}\big)\Gamma\big(\frac{3}{56}\big) \Gamma\big(\frac{5}{56}\big) \Gamma\big(\frac{9}{56}\big) \Gamma\big(\frac{13}{56}\big) \Gamma\big(\frac{15}{56}\big) \Gamma\big(\frac{19}{56}\big) \Gamma\big(\frac{23}{56}\big) \Gamma\big(\frac{25}{56}\big) \Gamma\big(\frac{27}{56}\big) \Gamma\big(\frac{39}{56}\big) \Gamma\big(\frac{45}{56}\big)\Big)^{1/8}}{\pi}$

$\eta(i\sqrt{15})=\frac{1}{2^{5/4}3^{1/4}5^{1/4}}\Big(\frac{\sqrt{5}-1}{2}\Big)^{5/12}\frac{\Big(\Gamma\big(\frac{1}{15}\big)\Gamma\big(\frac{2}{15}\big) \Gamma\big(\frac{4}{15}\big) \Gamma\big(\frac{8}{15}\big)\Big)^{1/4}}{\pi^{3/4}}$

Answer 1

Using this answer one can easily verify the value of $\eta(9i)$.

We have by definition $$\eta(9i)=e^{-3\pi/4}\prod_{n=1}^{\infty} (1-e^{-18n\pi})$$ And using the answer linked above we can see that $$\eta(9i)=\frac{\sqrt[3]{\sqrt[3]{18+6\sqrt{3}}-3}}{6}\cdot\frac{\Gamma (1/4)}{\pi^{3/4}}$$ One can use a little bit of algebra to verify that $$\sqrt{6}(2+\sqrt{3})^{1/6}=\sqrt[3]{18+6\sqrt{3}}$$ and get the value of $\eta(9i)$ mentioned in your question.

The modular equation given in the linked answer can be used to evaluate $\eta(27i)$ given the values of $\eta(3i),\eta(9i)$ and in general one can get the values of $\eta(3^ni)$ in similar fashion. Using $\eta(2i),\eta(6i)$ one can also verify the value of $\eta(18i)$. You should use the value of $\eta(7i)$ (given in linked question in your post) and $\eta(63i)$ of your post together with Ramanujan's modular equation to get the value of $\eta(21i)$ and add it to your table.

Remaining set of values of eta function in your question require more effort to verify.

Answer 2

Values of $\eta(mi\sqrt{n})/\eta(i\sqrt{n})$ can be obtained up to a factor of elliptic functions. For example, let $$ k_{27}= \frac{\left ( 3-c_{27} \right ) }{c_{27}(1+c_{27})} \frac{\sqrt{6}-\sqrt{2} }{4},c_{27}=\sqrt{3+2^{4/3}(1+2^{1/3})}. $$ The singular value should be $$ K(k_{27}) =\frac{\sqrt{3+2^{4/3}(1+2^{1/3})}}{3^{3/4}\cdot2^{7/3}\pi} \Gamma\left ( \frac13 \right )^3. $$ And $\eta(i\sqrt{n})=\sqrt{\frac{2K(k_{n})}{\pi} } \cdot2^{-1/3}k_{n}^{1/6}(1-k_{n}^2)^{1/12}.$ For $n=17$, we calculate $$ k_{17} =\frac{\sqrt{1+c_{17}}-\sqrt{1-c_{17}} }{2}, c_{17}=20+5\sqrt{17}-8\sqrt{2+2\sqrt{17}}-6\sqrt{2\sqrt{17}-2} $$ and $$ K(k_{17})=\frac{\left ( \sqrt{17}-4 \right )^{1/8} \left ( 1+\sqrt{17} +\sqrt{2+2\sqrt{17} } \right )^{3/2}\sqrt{\pi} }{ 2^{9/2}\sqrt{17} } \left [ \prod_{m=1}^{68}\Gamma\left ( \frac{m}{68} \right )^{\left(\frac{-68}{m}\right)} \right ]^{1/8}. $$ $\left(\frac{m}{n}\right)$ is Legendre-Kronecker symbol.
For $n=18$ similarly, $$ k_{18}=\left ( \sqrt{2}- 1\right )^3 \left ( 2-\sqrt{3} \right )^2, $$ $$ K(k_{18}) =\frac{\left ( \sqrt{6}+\sqrt{2} -1 \right ) \left ( \sqrt{2}+1 \right )^{1/2} }{ 3\cdot2^{13/4}\sqrt{\pi} }\Gamma\left ( \frac18 \right )\Gamma\left ( \frac38 \right ). $$