Let the prime factorization of m be m = product p(m,k)^b(m,k), where p(m,j)<p(m,j+1) for all j, the p's are the distinct primes dividing m, and each b is a positive integer. Then a(n) = product_k {p(A165713(n), k)^b(n,k)}.

A165714

Let the prime factorization of m be m = product p(m,k)^b(m,k), where p(m,j)<p(m,j+1) for all j, the p's are the distinct primes dividing m, and each b is a positive integer. Then a(n) = product_k {p(A165713(n), k)^b(n,k)}.