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Omni-MATH / Proof that

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problem

Proof that m=1n5ω(m)k=1nnkτ(k)2m=1n5Ω(m). \sum_{m=1}^n5^{\omega (m)} \le \sum_{k=1}^n\lfloor \frac{n}{k} \rfloor \tau (k)^2 \le \sum_{m=1}^n5^{\Omega (m)} .
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Proof that
∑_(m=1)^(n)5^(ω(m))≤∑_(k=1)^(n)⌊(n)/(k)⌋τ(k)²≤∑_(m=1)^(n)5^(Ω(m)).
Original LaTeX notation
Proof that
$$ \sum_{m=1}^n5^{\omega (m)} \le \sum_{k=1}^n\lfloor \frac{n}{k} \rfloor \tau (k)^2  \le \sum_{m=1}^n5^{\Omega (m)} .$$

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