Abstract
according as Ic is not square-free or is the product of exactly v =v(k) distinct primes (with the understanding that k = 1 belongs to the second case, with v(1) =0). Whether i is or is not square-free, let v(n) denote the number of its distinct prime divisors (e. g., v(12) = 2). Then it is easily realized that 21'(n) is the number of the square-free divisors of n. Hence, if T(n) denotes the number of all divisors of n, then
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