A168262 Intersection of A003418 and A116998.

1, 2, 6, 12, 60, 420, 840, 27720, 360360, 5354228880

Offset

1,2

Comments

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.

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.

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).)

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.

Links

Table of n, a(n) for n=1..10.

Eric Weisstein's World of Mathematics, Distinct Prime Factors

Crossrefs

Also intersection of A003418 and A060735, and of A003418 and A168264. (A168264 is a subsequence of A060735, which is a subsequence of A116998.)

See also A001221, A168263.

Sequence in context: A328459 A051451 A090951 * A085819 A069047 A095921

Adjacent sequences: A168259 A168260 A168261 * A168263 A168264 A168265

Keyword

fini,full,nonn

Author

Matthew Vandermast, Nov 23 2009

Status

approved