Register size: $\frac{1}{2^{2n+1}} \le \frac{1}{2N^2} \le \frac{1}{2r^2}$ · Prove2Me

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Register size: $\frac{1}{2^{2n+1}} \le \frac{1}{2N^2} \le \frac{1}{2r^2}$ · Prove2Me