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Offset matches A019565.

Based on Selcoe's comment in A280864 regarding k in sequences S_r = { k = m*r : rad(m) | r }, squarefree r > 1, appearing in order. The appearance of r itself introduces the lineage S_r, followed by lpf(r)*r, etc., if A280864 is a permutation of natural numbers.

Conjecture: there are no zeros in this sequence, which is equivalent to the conjecture that A280864 is a permutation of natural numbers. Minor corollary: a(127) > 2^18.

Let Q be a fixed odd prime. It appears that with only finitely many exceptions, when there is a term A280864(k) = Q*p, p prime, then the next term in A280864, A280864(k+1), is (Q+1)*p.

The present sequence lists the exceptions in the case Q=5. It is quite likely that there are no further terms.

If Q=3, it appears that there are just five exceptions, 3, 11, 31, 59, 71.

If Q=7, the complete list of exceptions appears to be 3, 5, 7, 11, 23, 37, 43, 73, 79, 83, 181, 191, 193, 197, 199, 211, 227, 229, 233, 239, 251, 257, 263, 269, 277, 1021, 1069, 1103, 1153.

If Q=11, the complete list of exceptions appears to be 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 47, 53, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 127, 139, 149, 151, 167, 173, 181, 191, 193, 197, 199, 211, 227, 229, 233, 239, 251, 257, 263, 269, 277, 281, 293, 311, 353, 431, 557, 563, 571, 619, 1289, 1291, 1307, 1499, 1571, 1579, 1583, 1621, 1627, 2011, 2029, 2131, 2207, 2221, 2281, 2287, 2311, 2341, 2347, 2357, 2399, 2551.

All four of these searches were carried out using the first 100000 terms of A280864.

It is known that every even number appears in A280864, and that the powers of 2 appear in increasing order.

If it were known that A280864(k-1) were always even, this would prove that A280864 contains every odd number, and so complete the proof that A280864 contains every positive number.

By definition, all terms are squarefree (see A007947); repeated terms here are the squarefree kernels of A280864(n).

All even squarefree numbers appear infinitely often.

1 appears only at a(1).

Even terms appear consecutively in pairs, each pair followed by one or more odd terms.

Conjecture: all odd squarefree numbers > 1 appear infinitely often. If so, then A280864 is a permutation of the natural numbers.

Theorem: a(n) = b(n-1)*b(n) where b = A280738. - N. J. A. Sloane, Apr 11 2017

In a series of submissions (A280864, A280866, A375029, A375030) Rémy Sigrist studies sequences whose terms are locally connected via their prime factors, i.e., where neighbors influence each other in their divisibility.

Sigrist chooses IN = {1, 2,...} as domain. The triangle discussed here is based on Sigrist's idea but chooses IN = {0, 1, 2,...} as the domain. We recall that all numbers divide 0, but 0 only divides 0. Neither 0 nor 1 have prime factors.

For a row of the triangle T(n) = [T(n, k) | k=0..n] and for a term t in this row, let 't be its predecessor and t' its successor. The terms of a row are subject to the following two conditions:

(1) For all k in 1..n-1 and t = T(n, k), t has no prime factor, or it is not the case that there is a prime factor of t such that (p | 't) <=> (p | t'). Expressed more succinctly (in pseudo-Python):

all(not any((p | 't) <=> (p | t') for p in primefactors(t)) for k = 1..n-1).

(2) If t = T(n, n), then t has no prime factor or for all prime factors of t, (p | t) => (p | t').

The row itself must also satisfy two conditions:

(3) T(n) is a permutation of {0, 1, 2, ..., n}.

(4) T(n) is the lexicographically earliest list among all lists whose terms satisfy conditions (1) and (2).

From Peter Luschny, Aug 05 2024: (Start)

On StackExchange (see link), user Bubbler found by exhaustive analysis for n < 29 that only n <= 14 and n = 16 have a solution. Bubbler also states that "since the 4th Ramanujan prime is 29, there are at least four primes that are greater than n/2 (i.e., prime factors that appear only once) when n >= 29 but there are only 3 positions that such primes can go (both sides of 0 and the first element), which proves that there is no solution when n >= 29."

We set all terms of row 15 equal to 0 by convention to make the sequence finite and full. (End)

Suggested by the Yellowstone permutation A098550 except that now the key conditions in the definition have been reversed.

Let Ker(k), the kernel of k, denote the set of primes dividing k. Thus Ker(36) = {2,3}, Ker(1) = {}. Then Product_{p in Ker(k)} p = A000265(k), which is denoted by ker(k).

Theorem 1: For n>2, a(n) is the smallest number m not yet in the sequence such that

(i) Ker(m) intersect Ker(a(n-1)) is nonempty,

(ii) Ker(m) intersect Ker(a(n-2)) is empty, and

(iii) The set Ker(m) \ Ker(a(n-1)) is nonempty.

(Without condition (iii), every prime dividing m might also divide a(n-1), which would make it impossible to find a(n+1).)

Idea of proof: m always exists and is unique; no smaller choice for a(n) is possible; and taking a(n)=m does not lead to a contradiction. So a(n) must be m.

Theorem 2: For n>2, Ker(a(n)) contains at least two primes. (Immediate from Theorem 1, since a(n) must contain a prime in a(n-1) and a prime not in a(n-1).)

It follows that no odd prime p or even-or-odd prime power q^k, k>1, appears in the sequence. Obviously this sequence is not a permutation of the positive integers.

Theorem 3. For any M there is an n_0 such that n > n_0 implies a(n) > M. (This is a standard property of any sequence of distinct positive terms - see the Yellowstone paper).

Theorem 4. For any prime p, some term is divisible by p.

Proof. Take p=17 for concreteness. If 17 does not divide any term, then 19 cannot either (because the first time 19 appears, we could have used 17 instead).

So all terms are products only of 2,3,5,7,11,13. Go out a long way, use Theorem 2, and consider two huge successive terms, A*B, C*D, where Ker(B) = Ker(C) and Ker(A) intersect Ker(D) is empty. Either C or D must contain a huge prime power q^k, 2 <= q <= 13. If it is in C, replace it by q and multiply D by 17. If it is in D, replace it by 17. Either way we get a smaller legal candidate for C*D that is a multiple of 17. QED

Theorem 5. There are infinitely many even terms.

Proof. Suppose the prime p appears for the first times as a factor of a(n). Then we have a(n-1) = x*q^i, a(n) = q*p, where q

= 1. If q=2 then a(n) is even. So we may suppose q is odd. If x is odd then a(n+1) = 2*p. If x is even then obviously a(n-1) is even. So one of a(n-1), a(n), or a(n+1) is even for every prime p. So there are infinitely many even terms. QED - N. J. A. Sloane, Aug 28 2020

Theorem 6: For any prime p, infinitely many terms are divisible by p. - N. J. A. Sloane, Sep 09 2020. (I thought I had a proof that for any odd prime p, there is a term equal to 2p, but there was a gap in the argument. - N. J. A. Sloane, Sep 23 2020)

Theorem 7: There are infinitely many odd terms. - N. J. A. Sloane, Sep 12 2020

Conjecture 1: Every number with at least two distinct prime factors is in the sequence. In other words, apart from 1 and 2, this sequence is the complement of A000961.

[It seems very likely that the arguments used to prove Theorem 1 of the Yellowstone Permutation paper can be modified to prove the conjecture.]

The conditions permit us to start with a(1)=1, a(2)=2, and that does not lead to a contradiction, so those are the first two terms.

After 1, 2, the next term cannot be 4 or 5, but a(3) = 6 works.

For a(4), we can rule out 3, 4, 5, 7, 8, 9 11, 13 (powers of primes), and 10, 12, and 14 have a common factor with a(2). So a(4) = 15.

The graph of the first 100000 terms (see link) is similar to that of the Yellowstone permutation, but here the points lie on more lines.

The sequence has fixed points at n = 1, 2, 10, 90, 106, 150, 162, 246, 394, 398, 406, 410, ... (see A338050). - Scott R. Shannon, Aug 13 2020

The initial pattern of odd and even terms: (odd, even, even, odd), repeat, is misleading as it does not persist. (See A337644 for more about this point.)

Discussion of when primes first divide some term, from N. J. A. Sloane, Oct 21 2020: (Start)

When an odd prime p first divides a term of the Enots Wolley sequence (the present sequence), that term a(n) is equal to q*p where q

A337648), that there are precisely 34 instances when q = 3 (see A337649), and q>3 happens just once, at a(5) = 35 when q=5 and p=7.

We conjecture that even if p is introduced by some prime q>2, 2*p appears later.

Sequence A337275 lists the index k such that a(k) = 2*prime(n), or -1 if 2*prime(n) is missing, and A338074 lists the indices k such that a(k) is twice a prime.

Comparison of those two sequences shows that they appear to be essentially identical (see the table in A337275).

The differences between the two sequences are caused by the fact that although normally if p and q are odd primes with p < q, then 2p precedes 2q, this is not true for the following primes: (7,5), (31,29), and (109, 113, 107), which appear in the order shown. We conjecture that these are the only exceptions.

Combining the above observations, we conjecture that for n >= 755 (at which point we have seen all the primes <= 367), every prime p is introduced by 2*p, and the terms 2*p appear in their natural order.

(End)