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It is well-known that for the $j$-invariant, we have $j\left(\frac{1+\sqrt{-n}}{2}\right)\in\mathbb R$ whenever $n>0$. Moreover, $j\left(\frac{1+\sqrt{-n}}{2}\right)=0$ for $n=3$ at the cusp. In fact, looking at this as a function from $\mathbb R^+\to\mathbb R$, it turns out that $n=3$ (and by symmetry also $n=\frac13$) is a triple zero. The natural question arises:

Is there a closed form for the 'residue' $$ \lim_{n\to3}\frac{J\left(\frac{1+\sqrt{-n}}{2}\right)}{(n-3)^3}\approx -137.55695117232957120439316005299679837316833\ \mathbf?$$

Note that the inverse symbolic calculator doesn't find anything for this or for its cube root.

Wolfgang
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I have just come about a closely related question with a beautiful proof, using Eisenstein series, of the identity (writing the residue explicitly)$$R:=\lim_{n\to3}\frac{\left(\frac{1+\sqrt{-3}}{2}-\frac{1+\sqrt{-n}}{2}\right)^3}{J\left(\frac{1+\sqrt{-n}}{2}\right)}=-\frac{3^9\,\Gamma^{18}\left(\frac{2}{3}\right)}{2^{12}\pi^9}i.$$ From there it follows directly that $$\lim_{n\to3}\frac{J\left(\frac{1+\sqrt{-n}}{2}\right)}{(n-3)^3}=-\frac{\sqrt{-3}^3}{R}=-\frac{2^{12}\pi^9}{\sqrt{3}^{15}\,\Gamma^{18}\left(\frac{2}{3}\right)}=-\frac{\sqrt{3}^{3}\Gamma^{18}\left(\frac{1}{3}\right)}{2^{6}\pi^9}=\color {blue}{-\frac{8}{27}\frac{ \Gamma^{9}\left(\frac{1}{3}\right)}{ \Gamma^{9}\left(\frac{2}{3}\right)}}.$$ (I prefer the very last form for esthetical reasons.)

Wolfgang
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