%I #2 Mar 30 2012 17:27:19
%S 1,2,6,12,60,420,840,27720,360360,5354228880
%N Intersection of A003418 and A116998.
%C If, for some prime p, A045948(p) > p^2, then all members of the sequence are less than A003418(p). (Let p_(n) be a prime for which the inequality is satisfied, and let p_(n+1) be the smallest prime > (p_(n))^2. No number smaller than A003418(p_(n+1)) can belong to this sequence. However, for any p_(n) that satisfies the inequality, so does p_(n+1), leading to an endless cycle.) This inequality is first satisfied at p=53, as A045948(53)=5040 > 53^2=2809.
%C Proof: It follows from the definitions of p_(n) and p_(n+1), and from Bertrand's Postulate, that 2(A045948(p_(n))) > 2((p_(n))^2) > p_(n+1). Therefore 2((A045948(p_(n)))^2 > (p_(n+1))^2.
%C Since any prime that divides A003418(p_(n)) divides A003418(p_(n+1)) at least twice as often, A045948(p_(n+1)) cannot be less than the product of (A045948(p_n))^2 and A034386(p_(n)). (The latter term greatly exceeds 2 for any actual p_(n).)
%C Therefore A045948(p_(n+1)) > 2((A045948(p_n))^2 > (p_(n+1))^2, and p_(n+1) satisfies the inequality, implying that no number smaller than A003418(p_(n+2)) can belong to this sequence.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DistinctPrimeFactors.html">Distinct Prime Factors</a>
%Y Also intersection of A003418 and A060735, and of A003418 and A168264. (A168264 is a subsequence of A060735, which is a subsequence of A116998.)
%Y See also A001221, A168263.
%K fini,full,nonn
%O 1,2
%A _Matthew Vandermast_, Nov 23 2009