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Let $\rho_{k}=\beta_{k}+i\gamma_{k}$ the $k-th$ non-trivial zero of the Riemann zeta funcion. We consider only the zeros with $\gamma_{k}>0$ . Then we have$$\gamma_{1}=14.13...$$ $$\gamma_{2}=21.02...$$ etc. so it seems that holds $$\gamma_{k}>k.$$ I checked the first 100 zeros and it seems to be true. Is it known? I haven't found anything like that on the internet.

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The statement is false. Recalling that $N\left(T\right) $ is the number of non-trivial zeros of the Riemann zeta function with imaginary part $0<\gamma\leq T $, we know that holds $$N\left(T\right)\sim\frac{T}{2\pi}\log\left(T\right) $$ then we have $$k=N\left(\gamma_{k}\right)\sim\frac{\gamma_{k}}{2\pi}\log\left(\gamma_{k}\right) $$ and so$$\log\left(k\right)\sim\log\left(\gamma_{k}\right) $$ hence $$\gamma_{k}\sim\frac{2\pi k}{\log\left(k\right)} $$ as $k\rightarrow\infty .$

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