PE512

题意

$$f(n) = \sum_{i=1}^n {\phi (n^i)} (mod \ n+1)$$

$$g(n) = \sum_{i=1}^n {f(i)}$$

Find $g(5 \times 10^8) = ?$

解法：

如果有 $n = p_1^{t_1} p_2^{t_2}... p_m^{t_m}$

从而 $\phi (n) = n (1 - \frac{1}{p_1}) (1 - \frac{1}{p_2})... (1 - \frac{1}{p_m})$

取$n‘ = n^i$

从而 $\phi(n^i) = n^{i-1} \phi(n)$

$$f(n) = (1 + n + n^2 + ... + n^{n-1}) (mod \ n+1) \phi (n)$$

$$f(n) = (1+n)(1+n^2+...+n^{i-2}) + [n\ is\ an\ odd] (mod \ n+1) \phi(n)$$

从而

$$g(n) = \sum_{1 \leq k \leq n, k\ is\ an\ odd}{\phi(k)}$$

用杜教筛。