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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

First differs from A241559 at a(45).

If A237271(n) = 1 then a(n) = A241558(n) = A241559(n) = A000203(n).

If n is an odd prime then a(n) = (n + 1)/2 = A241558(n) = A241559(n).

For more information see A237593.

First differs from A241838 at a(45).

If A237271(n) = 1 then a(n) = A241558(n) = A241838(n) = A000203(n).

If n is an odd prime then a(n) = (n + 1)/2 = A241558(n) = A241838(n).

For more information see A237593.

Numbers n such that A243982(n) = 0.

First differs from A151991 at a(25).

Let n = 2^m * q with m >= 0 and q odd. Let c_n denote the count of regions in the symmetric representation of sigma(n), which is determined by the positions of 1's in the n-th row of A237048. The maximum of c_n occurs when odd and even positions of 1's alternate implying that all regions have width 1, denoted by w_n = 1. When m > 0 then sigma_0(n) > sigma_0(q) and c_n = sigma_0(n) is impossible. Therefore, exactly those odd n with w_n = 1 are in this sequence. Furthermore, since the 1's in A237048 represent the odd divisors of n, their odd-even alternation expresses the property 2*f < g for any two adjacent divisors f < g of odd number n; in other words, this sequence is also the complement of A090196 relative to the odd numbers. This last property permits computations of elements in this sequence faster than with function a244579, which is based on Dyck paths. - Hartmut F. W. Hoft, Oct 11 2015

From Hartmut F. W. Hoft, Dec 06 2016: (Start)

Also, integers n such that for any pair a < b of divisors of n the inequality 2*a < b holds (hence n is odd).

Let 1 = d_1 < ... < d_k = n be all (odd) divisors of n. The property 2*d_i < d_(i+1), for 1 <= i < k, is equivalent for the 1's in the n-th row of A249223 to be in positions 1 = d_1 < 2 < d_2 < 2*d_2 < ... < d_i <2*d_i < d_(i+1) < ... where 2*d_i represents the odd divisor e_i with d_i * e_i = n. In other words, the odd divisors are the number of parts in the symmetric representation of sigma(n). The rightmost 1 in the n-th row occurs in an odd (even) position when k is odd (even).

As a consequence this sequence is also the complement of A090196 in the set of odd numbers. (End)

a(3) > 36000000.

Also intersection of A241558 and A241559 (minimum = maximum) minus the union of A238443 and A239929 (number of parts <= 2).

If n is an odd prime (A065091) then a(n) = (n + 1)/2.

If n is a power of 2 (A000079) then a(n) = 2*n - 1.

If n is a perfect number (A000396) then a(n) = 1 assuming there are no odd perfect numbers.

a(n) is also the smallest number in the n-th row of the triangles A279391 and A280851.

a(n) is also the smallest nonzero term in the n-th row of triangle A296508.

The symmetric representation of sigma(n) has A001227(n) subparts.

For the definition of the "subpart" see A279387.

For a diagram with the subparts for the first 16 positive integers see A296508.

It appears that a(n) = 1 if and only if n is a hexagonal number (A000384). - Omar E. Pol, Sep 08 2021

The above conjecture is true. See A280851 for a proof. - Omar E. Pol, Mar 10 2022

If n is an odd prime (A065091) then a(n) = (n + 1)/2.

If n is a power of 2 (A000079) then a(n) = 2*n - 1.

If n is a perfect number (A000396) then a(n) = 2*n - 1, assuming there are no odd perfect numbers.

a(n) is also the largest element in the n-th row of the triangles A279391, A280851 and A296508.

The symmetric representation of sigma(n) has A001227(n) subparts.

For the definition of the "subpart" see A279387.

For a diagram with the subparts for the first 16 positive integers see A296508.

Conjecture 1: there are infinitely many pairs of the form a(x) = y; a(y) = x (see examples).

First differs from A351904 at a(11).

From Hartmut F. W. Hoft, Jun 10 2024: (Start)

For numbers less than or equal to a(2^20), (2^k, 2^(k+1) - 1), 0 <= k <= 19, are the only pairs satisfying a(a(x)) = x; the triple (36, 46, 91) is the only one satisfying a(a(a(x))) = x, and there are no proper order 4 quadruples and no order 5 quintuples, apart from fixed point 1.

Conjecture 2: Only the pairs x = 2^k and y = 2^(k+1) - 1, k >= 0, satisfy a(x) = y and a(y) = x.

A repeated number d in this sequence determines a pair of distinct indices u and v such that d = a(u) = a(v). This means that d is the smallest number for which parts of sizes u and v occur in the symmetric representation of sigma(d), SRS(d). There are 5507 such pairs less than a(2^20). (End)

When the symmetric representation of sigma(n) has an even number of parts, e.g., for every prime, the two middle parts have equal size so either one may be chosen.

Since a(45) = 32 while A241558(45) = 23 the two sequences are different, indeed both respective complements of the sequences, A241558 in a and a in A241558, are infinite as the symmetric representations of the following two subsequences of this sequence show:

(1) n = 5*3^k, k>1, has the 3 parts ( (5*3^k + 1)/2, 4*(3^k - 1), (5*3^k + 1)/2 ) with the middle the largest part.

(2) n = p^2, p > 2 prime, has the 3 parts ( (p^2 + 1)/2, p, (p^2 + 1)/2 ) with the middle the smallest part.

The parts of the symmetric representation of sigma are in A237270.

a(n) = A000203(n) if and only if n is a member of A174973.