摘要:
∑ i = 1 n ∑ j = 1 m gcd ( i , j ) k ∑ i = 1 n ∑ j = 1 m ∑ g g k [ gcd ( i , j ) = g ] ∑ g g k ∑ i = 1 ⌊ n g ⌋ ∑ j = 1 ⌊ m g ⌋ [ gcd ( i , j ) = 阅读全文
摘要:
引理 1 1 1: d ( i ⋅ j ) = ∑ x ∣ i ∑ y ∣ j [ gcd ( x , y ) = 1 ] d (i \cdot j) = \sum_{x \mid i} \sum_{y \mid j} [\gcd (x,y) = 1] d(i⋅j)=x∣i∑y∣j∑[gcd 阅读全文