Base series of Conjecture 3.4: $\frac{\binom{2k}{k}^2\binom{4k}{2k}}{(-12288)^k}(28k+3)=\frac{16}{\sqrt3\,\pi}$

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Base series of Conjecture 3.4: $\frac{\binom{2k}{k}^2\binom{4k}{2k}}{(-12288)^k}(28k+3)=\frac{16}{\sqrt3\,\pi}$ · Prove2Me