Here's a proof with a gap in it. Denote the set of finite sets of primes by P, and let f:N->P be the function that maps an integer to its set of prime factors. Note that this is the set, not the multiset, so that f(30) = f(60) = {2, 3, 5} Let A be the set of numbers that occur in A098550. Define g:P-> {power set of A} as the "set inverse" of f. That is, g(Q) = f^-1(Q) = the set of all numbers x that occur in A such that f(x) = P. Call a set of primes Q "good" if g(Q) is infinite, that is, if infinitely many numbers whose set of prime factors is exactly Q occur in A. Note that if Q is good, g(Q) includes all integers n with f(n) = Q, in ascending order, since when an element of Q is eligible to be the next element of the sequence, all small elements of Q are also eligible. Given two sets of primes X and Y, we say that X is a follower of Y if it is legal for a to directly follow b in the sequence, with f(a) = X and f(b) = Y, that is, if X and Y have non-empty intersection, and X is not a subset of Y. Lemma: If X is a follower of Y, and Y is good, then X is good. Proof: choose an element z of X. After each number a with f(a) = Y, the next element of the sequence will be the smallest remaining element b such that f(b) is a follower of Y. So after at most z numbers occur that are mapped by f to Y,, one of them will be followed by z. Theorem: Every element of P containing at least 2 elements is good. Proof: Suppose Z is not good. Then for any Y such that Z is a follower of Y, Y is not good either, by the lemma. So every good set is either a subset of Z, or disjoint from Z. But if X disjoint from Z is good, choose Y containing one element of X and one element of Z. Now Z follows Y which follows X, a contradiction. Similarly, if a subset X of Z is good, choose Y to contain an element of X and an element not in X, and again, Z follows Y follows X, a contradiction. The only remaining case is if *no* element of P is good. I'm convinced this is impossible, but my attempts to prove this keep getting annoyingly long, and I'm sure there's a simple proof that this is impossible. Can anyone help me with this last step in the proof? Andy
-- Andy.Latto@pobox.com