cp's OEIS Frontend

A165714 Let the prime factorization of m be m = product p(m,k)^b(m,k), where p(m,j) 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)}.

%I A165714 #7 Mar 11 2014 01:32:48
%S A165714 3,2,25,7,10,2,27,121,6,13,28,2,15,6,83521,19,50,23,63,22,6,5,104,9,
%T A165714 14,24389,99,31,42,2,69343957,34,35,6,1444,41,39,10,88,43,30,47,45,92,
%U A165714 6,7,80,2809,867,26,12,59,6655,14,513,58,62,61,132,2,21,325,90458382169,34
%N 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)}.
%C A165714 A165713(n) = the smallest integer > n that is divisible by exactly the same number of distinct primes as n is.
%e A165714 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 14 and the exponents from the factorization of 12, we have a(12) = 2^2 * 7^1 = 28.
%Y A165714 Cf. A165713, A165715.
%K A165714 nonn
%O A165714 2,1
%A A165714 _Leroy Quet_, Sep 24 2009
%E A165714 Extended by _Ray Chandler_, Mar 12 2010