Vrijeme: 11:06

Sigma #4

Odredi (p je prost)

\sum_{n = 1}^{2025} (-1)^{\sum_{p \mid n} \sum_{i = 1}^{\infty} \lfloor \frac{n}{p^i} \rfloor - \lfloor \frac{n - 1}{p^i} \rfloor} \left \lfloor \frac{2025}{n} \right \rfloor.

Calculate (p is prime) \sum_{n = 1}^{2025} (-1)^{\sum_{ p \mid n} \sum_{i = 1}^{\infty} \lfloor \frac{n}{p^i} \rfloor - \lfloor \frac{n - 1}{p^i} \rfloor} \left \lfloor \frac{2025}{n} \right \rfloor.