cp's OEIS Frontend

We conjecture that the sequence is infinite or, equivalently, A257112 contains all primes.

Records: 1, 2, 11, 55, 165, 231, 275, 299, 325, 1547, 1909, 7943, 8215, 14335, 18815, 35953, 70303, 81793, 114481, 190999, 954995, 2021021, 10105105, 22231231, ..., . - Robert G. Wilson v, May 05 2015

For n > 3, gcd(a(n), a(n-1)) = 1 and gcd(a(n), a(n-2)) > 1. (This is just a restatement of the definition.)

This is now known to be a permutation of the natural numbers: see the 2015 article by Applegate, Havermann, Selcoe, Shevelev, Sloane, and Zumkeller.

From N. J. A. Sloane, Nov 28 2014: (Start)

Some of the known properties (but see the above-mentioned article for a fuller treatment):

1. The sequence is infinite. Proof: We can always take a(n) = a(n-2)*p, where p is a prime that is larger than any prime dividing a(1), ..., a(n-1). QED

2. At least one-third of the terms are composite. Proof: The sequence cannot contain three consecutive primes. So at least one term in three is composite. QED

3. For any prime p, there is a term that is divisible by p. Proof: Suppose not. (i) No prime q > p can divide any term. For if a(n)=kq is the first multiple of q to appear, then we could have used kp < kq instead, a contradiction. So every term a(n) is a product of primes < p. (ii) Choose N such that a(n) > p^2 for all n > N. For n > N, let a(n)=bg, a(n+1)=c, a(n+2)=dg, where g=gcd(a(n),a(n+2)). Let q be the largest prime factor of g. We know q < p, so qp < p^2 < dg, so we could have used qp instead of dg, a contradiction. QED

3a. Let a(n_p) be the first term that is divisible by p (this is A251541). Then a(n_p) = q*p where q is a prime less than p. If p < r are primes then n_p < n_r. Proof: Immediate consequences of the definition.

4. (From David Applegate, Nov 27 2014) There are infinitely many even terms. Proof:

Suppose not. Then let 2x be the maximum even entry. Because the sequence is infinite, there exists an N such that for any n > N, a(n) is odd, and a(n) > x^2.

In addition, there must be some n > N such that a(n) < a(n+2). For that n, let g = gcd(a(n),a(n+2)), a(n) = bg, a(n+1)=c, a(n+2)=dg, with all of b,c,d,g relatively prime, and odd.

Since dg > bg, d > b >= 1, so d >= 3. Also, g >= 3.

Since a(n) = bg > x^2, one of b or g is > x.

Case 1: b > x. Then 2b > 2x, so 2b has not yet occurred in the sequence. And gcd(bg,2b)=b > x > 1, gcd(2b,c)=1, and since g >= 3, 2b < bg < dg. So a(n+2) should have been 2b instead of dg.

Case 2: g > x. Then 2g > 2x, so 2g has not yet occurred in the sequence. And gcd(bg,2g)=g > 1, gcd(2g,c)=1, and since d >= 3, 2g < dg. So a(n+2) should have been 2g instead of dg.

In either case, we derive a contradiction. QED

Conjectures:

5. For any prime p > 97, the first time we see p, it is in the subsequence a(n) = 2b, a(n+2) = 2p, a(n+4) = p for some n, b, where n is about 2.14*p and gcd(b,p)=1.

6. The value of |{k=1,..,n: a(k)<=k}|/n tends to 1/2. - Jon Perry, Nov 22 2014 [Comment edited by N. J. A. Sloane, Nov 23 2014 and Dec 26 2014]

7. Based on the first 250000 terms, I conjectured on Nov 30 2014 that a(n)/n <= (Pi/2)*log n.

8. The primes in the sequence appear in their natural order. This conjecture is very plausible but as yet there is no proof. - N. J. A. Sloane, Jan 29 2015

(End)

The only fixed points seem to be {1, 2, 3, 4, 12, 50, 86} - see A251411. Checked up to n=10^4. - L. Edson Jeffery, Nov 30 2014. No further terms up to 10^5 - M. F. Hasler, Dec 01 2014; up to 250000 - Reinhard Zumkeller; up to 300000 (see graph) - Hans Havermann, Dec 01 2014; up to 10^6 - Chai Wah Wu, Dec 06 2014; up to 10^8 - David Applegate, Dec 08 2014.

From N. J. A. Sloane, Dec 04 2014: (Start)

The first 250000 points lie on about 8 roughly straight lines, whose slopes are approximately 0.467, 0.957, 1.15, 1.43, 2.40, 3.38, 5.25 and 6.20.

The first six lines seem well-established, but the two lines with highest slope at present are rather sparse. Presumably as the number of points increases, there will be more and more lines of ever-increasing slopes.

These lines can be seen in the Havermann link. See the "slopes" link for a list of the first 250000 terms sorted according to slope (the four columns in the table give n, a(n), the slope a(n)/n, and the number of divisors of a(n), respectively).

The primes (with two divisors) all lie on the lowest line, and the lines of slopes 1.43 and higher essentially consist of the products of two primes (with four divisors).

(End)

The eight roughly straight lines mentioned above are actually curves. A good fit for the "line" with slope ~= 1.15 is a(n)~=n(1+1.0/log(n/24.2)), and a good fit for the other "lines" is a(n)~= (c/2)*n(1-0.5/log(n/3.67)), for c = 1,2,3,5,7,11,13. The first of these curves consists of most of the odd terms in the sequence. The second family consists of the primes (c=1), even terms (c=2), and c*prime (c=3,5,7,11,13,...). This functional form for the fit is motivated by the observed pattern (after the first 204 terms) of alternating even and odd terms, except for the sequence pattern 2*p, odd, p, even, q*p when reaching a prime (with q a prime < p). - Jon E. Schoenfield and David Applegate, Dec 15 2014

For a generalization, see the sequence of monomials of primes in the comment in A247225. - Vladimir Shevelev, Jan 19 2015

From Vladimir Shevelev, Feb 24 2015: (Start)

Let P be prime. Denote by S_P*P the first multiple of P appearing in the sequence. Then

1) For P >= 5, S_P is prime.

Indeed, let

a(n-2)=v, a(n-1)=w, a(n)=S_P*P. (*)

Note that gcd(v,P)=1. Therefore, by the definition of the sequence, S_P*P should be the smallest number such that gcd(v,S_P) > 1.

So S_P is the smallest prime factor of v.

2) The first multiples of all primes appear in the natural order.

Suppose not. Then there is a pair of primes P < Q such that S_Q*Q appears earlier than S_P*P. Let

a(m-2)=v_1, a(m-1)=w_1, a(m)=S_Q*Q. (**)

Then, as in (*), S_Q is the smallest prime factor of v_1. But this does not depend on Q. So S_Q*P is a smaller candidate in (**), a contradiction.

3) S_P < P.

Indeed, from (*) it follows that the first multiple of S_P appears earlier than the first multiple of P. So, by 2), S_P < P.

(End)

For any given set S of primes, the subsequence consisting of numbers whose prime factors are exactly the primes in S appears in increasing order. For example, if S = {2,3}, 6 appears first, in due course followed by 12, 18, 24, 36, 48, 54, 72, etc. The smallest numbers in each subsequence (i.e., those that appear first) are the squarefree numbers A005117(n), n > 1. - Bob Selcoe, Mar 06 2015

Start with smallest number which has not yet appeared and satisfies the conditions: a(3)=6; thereafter always choose smallest number which has not yet appeared and satisfies the conditions.

This is a two-dimensional spiral analog of EKG sequence A064413.

In A064413 we have initial terms in the positions 1,2.

In the two-dimensional case we have 4 sides.

So the initial TERMS are

5

4 1 2 (1)

3

But the POSITIONS in the spiral are indexed thus:

.

7--8--9--10

|

6 1--2

| |

5--4--3

.

So the initial terms, by (1), are a(1)=1, a(2)=2, a(4)=3, a(6)=4, a(8)=5.

Conjecture: The sequence is a permutation of the positive integers. - Vladimir Shevelev, May 06 2015

The sequence is conjectured to be a permutation of the positive integers, although it takes many terms for most primes to appear, e.g. a(1807) = 13, a(35156) = 179. The primes do not occur in their natural order. In the first 200000 terms the smallest unused prime is 181, while the smallest unused composite number is 11881, which is itself a prime power.

This is a variation of the EKG sequence A064413 where the numbers are written on the square spiral such that each new number must share a common factor with not only the previous number but also with the adjacent inner spiral number, in one of the four axial directions, if such a number exists. This additional restriction causes the numbers to violate some of the patterns the numbers form in the standard EKG sequence, e.g., an odd prime p does not need to be preceded by 2p or followed by 3p, and the primes do not appear in increasing order.

For the first 100000 terms the smallest unseen number is 433.

The sequence is an analog of EKG-sequence A064413 with more hard condition on neighbor terms. However, we conjecture that, as A064413, it is a permutation of the positive integers.

By the definition, every two neighbor terms are strictly connected numbers defined in comment in A257112.

The sequence is an analog of the Yellowstone sequence A098550 with a more stringent condition on neighboriong terms at distance 2. However, we conjecture that, as with A098550, this is a permutation of the positive integers.

By definition, every pair of neighboring terms at distance 2 are strictly connected numbers as defined in comment in A257112.

This sequence is a 2D square lattice version of the Enots Wolley sequence A336957. Like that sequence, for this sequence to be infinite no number can be a prime or prime power - the first term must therefore be 6. Likewise, for each of its orthogonal nearest neighbors, a new number must have at least one prime factor that is not a factor of that neighbor. When choosing each new number a further test must also be performed against the previous number that is two squares ahead of the current number and on the edge of the spiral - this number is outside the eight nearest neighbors of the number being determined. See the examples below.

Like A336957 is it conjectured that all nonprime powers eventually appear. In the first 10000 terms the largest value is a(6975) = 3005315.