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If a prime p divides a(n), then there is a run of exactly three terms (one of which is a(n)) that are divisible by p.

If "three" is changed to "two", we get A280864.

Conjecture: This is a permutation of the positive integers.

In other words, each multiple of a prime p has at least one neighbor that is also a multiple of p.

This sequence is similar to A280864; the first difference oocurs at n=42: a(42)=50 whereas A280864(42)=55.

Conjectured to be a permutation of the natural numbers.

Comment from N. J. A. Sloane, Jan 14 2017, with minor edits May 08 2024: (Start)

Theorem: This is a permutation of the natural numbers.

Proof. 1. By definition no term is repeated.

2. The sequence is clearly infinite, since we can always use a very large prime G to build the next term. (G*a(n) is always a candidate for a(n+1).)

3. So for any integer m, either it appears in the sequence, or there is an n_0 such that for n > n_0, a(n) > m. (Let w(m) = index where m appears, or -1 if m does not appear. Let W(m) = max{w(i): 1 <= i <= m}. Then for n > W(m), a(n) > m.)

4. For any prime p, there is a term a(n) that is divisible by p. For if not, no prime greater than p can appear either, or we could have used p instead. This would mean all terms are products of the primes less than p. Go out a long way in the sequence until the terms exceed p!. Then one of the finite set of numbers p times {a possibly empty product of distinct primes less than p} will be a smaller candidate for the next term. So p will appear. Contradiction.

5. Call a(n) free if all its prime factors already appeared in a(n-1). If a(n) is free then there are no constraints on a(n+1) and so a(n+1) will be the smallest number not yet in the sequence.

6. For a prime p, let a(n) be the first term divisible by p. Either a(n) = k*p, where k is a product of the distinct primes needed to link a(n) to a(n-1), in which case a(n+1) = p and is free; or else a(n) = p, in which case a(n-1) was free. So in either case there is a free term associated with the prime p. (Except if there are two consecutive primes, like 13 and 17, when we only get one free term, not two. We will ignore this complication! There cannot be three consecutive primes.)

7. Since there are infinitely many primes, the sequence contains infinitely many free terms, and so every number will eventually appear. For suppose k never appears. Consider the first free term a(n) > k. Then a(n+1) = k, a contradiction. QED

(End)

The second occurrence of two consecutive primes is at n = 1324 .. 1325: a(1324) = 691, a(1325) = 701. - Antti Karttunen, May 18 2018

Alternative definition: a(1,2) = 1,2 then for n > 2 a(n) is the least novel multiple of rad(a(n-2)*a(n-1))/rad(a(n-2)), where rad is A007947. - David James Sycamore, Jan 27 2024

Comments from N. J. A. Sloane, Nov 02 2020: (Start)

Alternatively, this is the lexicographically earliest infinite sequence of distinct positive numbers such that every prime is followed by its double.

Theorem: This is a permutation of the positive integers.

Proof. Sequence is clearly infinite, so for any k there is a number N_0(k) such that n >= N_0(k) implies a(n) > k.

Suppose m is missing. Consider a(n) for n = N_0(m). Then a(n) must be a prime p (otherwise it would have been m, which is missing), a(n+1) = 2*p, and a(n+2) = m, a contradiction. QED.

(End)

A toy model of A280864, A280985, and A127202.

Alternative definition: a(1,2) = 1,2. Let P(k) = rad(a(1)*a(2)*...*a(k)), then for n > 2, a(n) = P(n)/P(n-1), where rad is A007947. - David James Sycamore, Jan 27 2024

Like A336957 no prime or prime power can be a term as if it shared a prime factor with the previous term it would then not contain a prime factor not in the previous term. It is likely all other composite numbers appear.

Let s(m) = A280864(m). The sequence gives the lengths of successive segments when A280864 is split after any term s(m) where every prime divisor of s(m) has already divided s(m-1). (Then s(m+1) is the smallest number not yet in A280864 which is relatively prime to s(m).)

See the link "Properties of A280864" in A280864 for more information.

This is to A280864 as A115510 is to A064413 (EKG) and A252867 is to A098550 (Yellowstone).

These are the numbers that take the longest to appear in A280864.

It appears to consist of 1, 25, and the primes.

This is a finite version of the "cup of coffee" sequence A280864.

An obvious upper bound is a(n) <= n-C, where C is the number of primes p <= n such that 2*p > n. For example a(4) <= 3, since we cannot make use of the prime 3.

Note that for a prime p <= n/2, the number of terms in S that are divisible by p must be even. So if floor(n/p) is odd for any p <= n/2, we have a(n) <= n-C-1.

When going from n=p-1 to n=p (where p is a prime), the new p cannot be used, so a(p) = a(p-1) for p prime.

We can obtain a sequence of lower bounds by taking S to consist of the first A280774(m) terms of A280864, for m = 1,2,3,... This gives

a(m) >= max_{1 <= i <= A280774(m)} A280864(i).... (**)

For example, with m=4, we can take the first A280774(4) = 10 terms of A280864, that is, the sequence S = 1,2,4,3,6,8,5,10,12,9, to get a(12) >= 10. In fact equality holds: a(12) = 10 (see EXAMPLES below).

It was possible that equality would always hold in (**). (**) gives a(4) >= 3, a(8) >= 6, a(12) >= 10, a(16) >= 13 (equality holds up to this point), then a(28) >= 23, a(45) >= 27, ... However, Rémy Sigrist has shown that a(28) = 24.

a(24) = 19 so a(25) >= 19. Here is an argument that shows that a(25) = 20 is impossible. We can't use the big primes 13 17 19 23. There are 5 multiples of 5 we could use, 5 10 15 20 25, but we can only use 4 of them. There are 3 multiples of 7 we could use, namely 7 14 21, but we can only use 2 of them. These two lists are disjoint. So a(25) >= 25 - 4 - 2 = 19. - N. J. A. Sloane, Jul 27 2024

This sequence is a squarefree variant of A280864.

Conjecture: this sequence is a permutation of the squarefree numbers (A005117).