= Solution
<Legendre formula> gives
$$
v_p{2n\choose n}=\sum_{j\geq1}(\lfloor2n/p^j\rfloor-2\lfloor n/p^j\rfloor),
$$
whose summands are zero or one. If the sum is at least $k$, some nonzero summand has index $j\geq k$, whence $p^k\leq p^j\leq2n$.
Grouping the von Mangoldt function by prime gives
$$
\psi(x)=\sum_{p\leq x}\lfloor\log x/\log p\rfloor\log p.
$$
Every prime-power contribution to $\log {2n\choose n}$ occurs in $\psi(2n)$ by the first part. Finally ${2n\choose n}=\prod_{j=1}^n(n+j)/j\geq2^n$, so $\psi(2n)\geq n\log2$.
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