cp's OEIS Frontend

A165715 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 {p(n,k)^b(A165713(n), k)}.

%I A165715 #7 Mar 11 2014 01:32:48
%S A165715 2,9,2,5,6,343,4,3,20,11,6,28561,14,75,2,17,12,19,10,21,88,529,6,125,
%T A165715 52,3,14,29,30,28629151,2,33,34,1225,6,37,38,351,20,41,84,43,44,15,
%U A165715 368,2209,18,7,10,153,4394,53,6,1375,14,57,58,59,30,51520374361,124,21,2
%N A165715 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 {p(n,k)^b(A165713(n), k)}.
%C A165715 A165713(n) = the smallest integer > n that is divisible by exactly the same number of distinct primes as n is.
%e A165715 12 = 2^2 * 3^1, which is divisible by 2 distinct primes. The next larger integer divisible by exactly 2 distinct primes is 14 = 2^1 * 7^1. Taking the primes from the factorization of 12 and the exponents from the factorization of 14, we have a(12) = 2^1 * 3^1 = 6.
%Y A165715 Cf. A165713, A165714.
%K A165715 nonn
%O A165715 2,1
%A A165715 _Leroy Quet_, Sep 24 2009
%E A165715 Extended by _Ray Chandler_, Mar 12 2010