A247225 -id:A247225 - cp's OEIS frontend

A098550 The Yellowstone permutation: a(n) = n if n <= 3, otherwise the smallest number not occurring earlier having at least one common factor with a(n-2), but none with a(n-1).

1, 2, 3, 4, 9, 8, 15, 14, 5, 6, 25, 12, 35, 16, 7, 10, 21, 20, 27, 22, 39, 11, 13, 33, 26, 45, 28, 51, 32, 17, 18, 85, 24, 55, 34, 65, 36, 91, 30, 49, 38, 63, 19, 42, 95, 44, 57, 40, 69, 50, 23, 48, 115, 52, 75, 46, 81, 56, 87, 62, 29, 31, 58, 93, 64, 99, 68, 77, 54, 119, 60

Offset: 1

Reinhard Zumkeller, Sep 14 2004

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

Paul Tek, Table of n, a(n) for n = 1..10000
David Applegate, C++ program for efficiently computing terms
David Applegate, Revised C++ program with option to print more variables
David L. Applegate, Hans Havermann, Bob Selcoe, Vladimir Shevelev, N. J. A. Sloane, and Reinhard Zumkeller, The Yellowstone Permutation, arXiv preprint arXiv:1501.01669 [math.NT], 2015. Also Journal of Integer Sequences, Vol. 18:6 (2015), Article 15.6.7
Brady Haran and N. J. A. Sloane, The Yellowstone Permutation, Numberphile video, Jan 29 2023. [At 4:07 when I say "we got twice 61" I should have said "we got about twice 61", and the image should show 120 rather than 122.- _N. J. A. Sloane_, Feb 02 2023]
Hans Havermann, Graph of first 300000 terms (the red line is y=x)
Hans Havermann, Loops and unresolved chains for map n -> A098550(n) trajectories
L. Edson Jeffery, Log plot of first 2484 terms.
Dariel I. Ojera, Unveiling the Properties and Relationship of Yellowstone Permutation Sequence, Psych. Educ.: Multidisc. J. (2024) Vol. 27, No. 2, 173-184.
Scott R. Shannon, Graph of 250000 terms, based on Reinhard Zumkeller data, plotted with colors indicating the least prime factor (lpf). Terms with an lpf of 2 are shown in white, terms with an lpf of 3,5,7,11,13,17,19 are shown as one of the seven rainbow colors from red to violet, and terms with an lpf >= 23 are shown in gray.
N. J. A. Sloane, First 250000 terms sorted according to slope a(n)/n
N. J. A. Sloane, Conant's Gasket, Recamán Variations, the Enots Wolley Sequence, and Stained Glass Windows, Experimental Math Seminar, Rutgers University, Sep 10 2020 (video of Zoom talk)
N. J. A. Sloane, The OEIS: A Fingerprint File for Mathematics, arXiv:2105.05111 [math.HO], 2021.
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, p. 16.
Reinhard Zumkeller, Table of n, a(n) for n = 1..250000
Index entries for sequences that are permutations of the natural numbers

Cf. A098548, A098551, A249943 (first time all 1..n appear), A251553.
The inverse permutation is in A098551.
A098552(n) = a(a(n)).
A251102(n) = GCD(a(n+2),a(n)).
Cf. A251101 (smallest prime factor), A251103 (largest prime factor), A251138 (number of distinct prime factors), A251140 (total number of prime factors), A251045 (squarefree part), A251089 (squarefree kernel), A250127 and A251415 (records for a(n)/n), A251411 (fixed points), A248647 (records).
Cf. also A251412 (trajectory of 11), A251556 (finite cycles), A251413 and A251414 (variant involving odd numbers), A249357 ("Fibonacci" variant).
Smallest missing numbers: A251416, A251417, A251546-A251552, A247253. See also A251557, A241558, A251559.
Indices of some pertinent subsequences: A251237 (even numbers), A251238 (odd numbers), A251391 (squarefree), A251541 and A251239 (primes), A251240 (squares of primes), A251241 (prime powers), A251393 (powers of 2), A251392 (nonprimes), A253297 (primes p for which some multiple k*p > 2*p precedes p).
Three arrays concerning the occurrences of multiples of primes: A251637, A251715, A251716.
Two sequences related to the numbers which immediately follow a prime: A253048, A253049. Seven sequences related to the numbers that appear two steps after primes: A251542, A251543, A251544, A251545, A253052, A253053, A253054. See also A253055 and A253056.
Other starting values: A251554, A251555.
See also A251756, A253297, A251662, A253572, A253573, A253591, A253593, A253588, A253590, A253609, A252865, A252867, A252868, A247225, A247942, A254003, A254077, A254669, A254670, A255509 (version with a priority for appearance of the primes), A255615, A255617, A256189, A256224, A256368, A256461.
See also A064413 (EKG sequence), A255582, A121216 (similar sequences), A257112 (two-dimensional analog).
See also A253765 and A253766 (bisections), A250299 (parity), A253768 (partial sums).
See A336957 for a variation.

import Data.List (delete)
a098550 n = a098550_list !! (n-1)
a098550_list = 1 : 2 : 3 : f 2 3 [4..] where
   f u v ws = g ws where
     g (x:xs) = if gcd x u > 1 && gcd x v == 1
                   then x : f v x (delete x ws) else g xs
-- Reinhard Zumkeller, Nov 21 2014

N:= 10^4: # to get a(1) to a(n) where a(n+1) is the first term > N
B:= Vector(N,datatype=integer[4]):
for n from 1 to 3 do A[n]:= n: od:
for n from 4 do
  for k from 4 to N do
    if B[k] = 0 and igcd(k,A[n-1]) = 1 and igcd(k,A[n-2]) > 1 then
       A[n]:= k;
       B[k]:= 1;
       break
    fi
  od:
  if k > N then break fi
od:
seq(A[i],i=1..n-1); # Robert Israel, Nov 21 2014

f[lst_List] := Block[{k = 4}, While[ GCD[ lst[[-2]], k] == 1 || GCD[ lst[[-1]], k] > 1 || MemberQ[lst, k], k++]; Append[lst, k]]; Nest[f, {1, 2, 3}, 68] (* Robert G. Wilson v, Nov 21 2014 *)
NN = Range[4, 1000]; a098550 = {1, 2, 3}; g = {-1}; While[g[[1]] != 0, g = Flatten[{FirstPosition[NN, v_ /; GCD[a098550[[-1]], v] == 1 && GCD[a098550[[-2]], v] > 1, 0]}]; If[g[[1]] != 0, d = NN[[g]]; a098550 = Flatten[Append[a098550, d[[1]]]]; NN = Delete[NN, g[[1]]]]]; Table[a098550[[n]], {n, 71}] (* L. Edson Jeffery, Jan 01 2015 *)

a(n, show=1, a=3, o=2, u=[])={n<3&&return(n); show&&print1("1, 2"); for(i=4,n, show&&print1(","a); u=setunion(u, Set(a)); while(#u>1 && u[2]==u[1]+1, u=vecextract(u,"^1")); for(k=u[1]+1, 9e9, gcd(k,o)>1||next; setsearch(u,k)&&next; gcd(k,a)==1||next; o=a; a=k; break)); a} \\ Replace "show" by "a+1==i" in the main loop to print only fixed points. - M. F. Hasler, Dec 01 2014

from math import gcd
A098550_list, l1, l2, s, b = [1,2,3], 3, 2, 4, {}
for _ in range(1,10**6):
    i = s
    while True:
        if not i in b and gcd(i,l1) == 1 and gcd(i,l2) > 1:
            A098550_list.append(i)
            l2, l1, b[i] = l1, i, 1
            while s in b:
                b.pop(s)
                s += 1
            break
        i += 1 # Chai Wah Wu, Dec 04 2014

A254077 a(n) = n if n <= 3, otherwise the smallest number not occurring earlier such that gcd(a(n),a(n-2)) > gcd(a(n),a(n-1)).

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 10, 12, 5, 14, 15, 7, 18, 21, 16, 27, 20, 33, 24, 11, 26, 22, 13, 28, 39, 32, 30, 44, 25, 34, 35, 17, 40, 51, 36, 68, 42, 52, 45, 38, 48, 19, 46, 57, 23, 54, 69, 50, 63, 55, 49, 60, 56, 65, 58, 70, 29, 62, 87, 31, 66, 93, 64, 75, 72, 85, 74, 80, 37, 76, 111

Offset: 1

Vladimir Shevelev, Jan 25 2015

nonn

Conjecture: The sequence is infinite (that is, a(n) always exists).
Ray Chandler reports that the sequence certainly exists for 10^7 terms, Apr 02 2015. John P. Linderman confirms this, and has extended the sequence to 12.9 million terms, Apr 09 2015. Extended to 50 million terms by John Mason Apr 21 2015. John P. Linderman reached 150 million terms on May 04 2015, 2.5 billion terms on Jun 29 2015, and 5 billion terms on Apr 07 2017 (see attached letter).
Note that if a(n) ever divides a(n+1), the sequence will terminate. This has not happened in the first 2.5 billion terms (see the Linderman links), but there have been some close calls. For example, at n=9671, a(9671) = 4973 = a prime p, and a(9672) = 9947 = 2p+1. Conversely, if a(n) never divides a(n+1), the sequence is infinite. - N. J. A. Sloane, Mar 22 2015 and Jun 06 2015
Conjecture: The sequence is a permutation of the natural numbers.
Conjectures: 1) For k>=3, except for k=5, if a(n) = prime(k), then a(n-2) = 2*prime(k) and a(n+2) = 3*prime(k). This conjecture was verified by Peter J. C. Moses for n <= 5000. - Vladimir Shevelev, Feb 09 2015. This conjecture verified for n <= 10^7. - Ray Chandler, Apr 02 2015. Extended to n <= 10^9. - John Mason, Jun 08 2016
2) For k>=3, except for k=4, if a(n)=prime(k)^2, then a(n-2) = prime(k)^2 + prime(k). This conjecture was verified by Peter J. C. Moses for n <= 35000. - Vladimir Shevelev, Feb 12 2015. This conjecture verified for n <= 394349. - Ray Chandler, Mar 07 2015. This conjecture is false - for n = 4488245, a(n) = 2137^2, but a(n-2) = 2137^2 + 2*2137. - Ray Chandler, Mar 30 2015. Next are n = 30655601, a(n) = 5581^2, but a(n-2) = 5581^2 + 2*5581, and n = 922447261, a(n) = 30577^2, but a(n-2) = 30577^2 + 2*30577. - John Mason, Sep 15 2016
Theorem: a(n) does not divide a(n-1). For suppose a(n-2)=x, a(n-1)=b*c, a(n)=c. Then gcd(x,c) <= c, and gcd(b*c,c) = c, which contradicts the definition of a(n). - N. J. A. Sloane, Mar 22 2015
Theorem: If a(n-2) is prime AND a(n) EXISTS then a(n) is a multiple of a(n-2). For: by sequence definition, assuming a(n-2) = p, prime, then gcd(a(n),p) > gcd(a(n),a(n-1)); hence gcd(a(n),p) > 1; but p is prime and has only 1 and itself as divisors; so gcd(a(n),p)=p, and so a(n) is a multiple of p. (Weaker than the similar conjecture above.) - John Mason, Apr 15 2015 [WORDS IN CAPS ADDED BY N. J. A. Sloane, Apr 16 2015]
Theorem: If a(n) EXISTS AND a(n) > a(n-2)/2 then a(n) is composite. For: suppose instead that a(n)=p, prime; then by sequence definition, gcd(p,a(n-2)) > gcd(p,a(n-1)); hence gcd(p,a(n-2))>1; hence a(n-2) is a multiple of p; but a(n-2) < 2p so we have a contradiction; hence a(n) is composite. This theorem improves the efficiency of some sequence generation algorithms. - John Mason, Apr 15 2015 [WORDS IN CAPS ADDED BY N. J. A. Sloane, Apr 16 2015] [Further corrected by John Mason, May 28 2017]
Theorem: If a(n-2)=mp for some prime p, and m divides a(n-1), then a(n), if it exists, is a multiple of p (generalization of previous theorem, which is the special case of m=1). See for example a(33)=17, a(35)=51, a(37)=68; a(37) is a multiple of 17 because a(36) is a multiple of 3, which is "m" in a(35). (It follows that if a(n-2) / gcd(a(n-2), a(n-1)) is p, prime, then a(n), if it exists, is a multiple of p. - John Mason, May 19 2015)
Proof: consider consecutive terms mp,y,z for prime p, and m dividing y. By sequence definition gcd(z,mp)>gcd(z,y). Suppose that z is not a multiple of p. Then gcd(z,mp)=gcd(z,m), and so gcd(z,m)>gcd(z,y). Since m divides y, then gcd(z,m)>gcd(z,mq) for q = y/m, but this is clearly impossible. Hence z is a multiple of p. - John Mason, Apr 17 2015
Theorem: The first occurrences of the primes as factors of terms of the sequence are in ascending order, and without gaps (that is, 2 precedes 3, 3 precedes 5 (factor of 10), 5 precedes 7 (factor of 14), ...).
Proof: Suppose a(n)=mp is the first term having p as a factor. Then the theory states that q, prime and gcd(mp,a(n-1)). As a(n) is the first to have p as a factor, p does not divide a(n-2) and a(n-1), and neither does q. Hence gcd(mp,a(n-2))=gcd(m,a(n-2)) and gcd(mp,a(n-1))= gcd(m,a(n-1)). Hence gcd(m,a(n-2)) > gcd(m,a(n-1)). Hence gcd(mq,a(n-2)) > gcd(mq,a(n-1)). Hence mq, < mp, would have satisfied the conditions of the sequence for a(n), which is a contradiction. Hence no such prime q exists. - John Mason, Apr 17 2015
Theorem: The first occurrence of a prime p as a factor of a term in the sequence is in a term that is not equal to p itself.
Proof: Suppose a(n)=p, prime, the first term having p as a factor. Then gcd(p,a(n-2))=1, and therefore cannot be greater than gcd(p,a(n-1)), a contradiction of the rules of sequence construction. - John Mason, Apr 17 2015
Conjecture. The primes exist as elementary terms of the sequence in ascending order. - John Mason, Apr 17 2015
John Mason reports that each prime p seems to appear at a term n which is approaching 2p, as p increases. See A256213. - N. J. A. Sloane, Apr 16 2015
Conjecture. For any n > 4, the lowest value x missing from a(1) thru a(n) is prime. - John Mason, Apr 29 2015. In fact, taking into account that, apparently, prime p appears in the sequence at position circa 2p, we may conjecture that the lowest k values missing from a(1) thru a(n) are prime, where k = pi(n)-pi(n/2) - see A000720. - John Mason, Jun 03 2015
Theorem: if a(n) = p for some prime p > 3, then a(n-2) is a multiple of p. As a direct consequence, if all prime factors of a(n-2) are already present in the sequence, then a(n), if it exists, is composite.
Proof: by sequence definition, unless p=2 or 3, gcd(p,a(n-2)) > gcd(p,a(n-1)), and hence gcd(p,a(n-2))>1, and hence a(n-2) is a multiple of p. - John Mason, May 19 2015
First differs from A255582 at a(29). - Omar E. Pol, May 21 2015
Conjecture. For n > 778, if a(n) < n, then a(n) is prime. This has been confirmed for n through 10^9. - John Mason, Jun 03 2015 [Corrected, following suggestion by John P. Linderman, by John Mason, May 28 2017]
Conjecture. As for "Conjecture 1" above, which is its mirror image, except for n=2,3,21, corresponding to primes 2,3,11, if a(n-2)=mp is the first occurrence of prime p as a factor in the sequence, then m=2 and a(n)=p. Also, if a(n-2)=mp is the first occurrence of prime p as a factor in the sequence, then p does not divide a(n-1). - John Mason, May 31 2016
Theorem 1: If a(n) is the first term having p (prime) as a factor, then a(n+1), if it exists, is not a multiple of p. For proof, see links. - John Mason, Jul 26 2016
Theorem 2: If a(n)=cp is the first occurrence of prime p as a factor (n >3), than c has exactly one distinct prime factor. In other words, c may be expressed as k^i for some prime k, and i > 0.
Corollary. If a(n)=cp is the first occurrence of prime p as a factor (n >3), and as a consequence c= k^i for some prime k, and i > 0, then k^i divides a(n-2) and k^(i-1) is the maximum power of k that divides a(n-1).
Theorem 3 . If a(n) = 2p is the first term having p (prime) as a factor, then a(n-1) is odd and a(n-2) is even.
Theorem 4. If a(n) = 2p is the first term having p (prime) as a factor, then a(n+2), if it exists, is either p or 2u for some integer u such that 2u < p. (Note that it is conjectured to be always p, and observation confirms the conjecture.)
Theorem 5, generalization of Theorem 4. If a(n) = cp is the first term having p (prime) as a factor (n >3), and as a consequence c=k^i for prime k and i>0, then a(n+2), if it exists, is either p or ku for some integer u such that ku < p. (Note that it is conjectured to be always p, and observation confirms the conjecture.)
Theorem 6. If a(n) = cp is first term having p, prime, as a factor (n >3), and a(n+2)=p, then a(n+3) exists, and is not a multiple of p, and so does not terminate the sequence.
Theorem 7. If a(n) = cp is first term having p, prime, as a factor (n >3), and a(n+2)=p, then a(n+4) exists and is 2p or 3p. Also, a(n+5) exists.
For proofs, see links. - John Mason, Aug 03 2016
Theorem 8: If the sequence is infinite, it is a permutation of the positive integers. For proof, see link. - John Mason, Sep 14 2016
Conjecture: After 2 and 3, no two primes are consecutive terms. This conjecture is derivable from the previous conjecture : "For k>=3, except for k=5, if a(n) = prime(k), then a(n-2) = 2*prime(k) ...". For, if sequence has terms z,2p,2q,p,q for primes p & q, then gcd(2q,z)>gcd(2q,2p)=2. Hence q divides z. So terms are mq,2p,2q,p,q. So we could have used q in place of 2q. - John Mason, May 28 2017

Peter J. C. Moses and Ray Chandler, Table of n, a(n) for n = 1..10000 (first 1000 terms from Peter J. C. Moses)
David L. Applegate, Hans Havermann, Bob Selcoe, Vladimir Shevelev, N. J. A. Sloane, and Reinhard Zumkeller, The Yellowstone Permutation, arXiv preprint arXiv:1501.01669 [math.NT], 2015.
John P. Linderman, Table of n, a(n) for n = 1..1000000 (about 13MB)
John P. Linderman, Table of n, a(n) for n = 1..12940331 (about 61MB)
John P. Linderman, Go program [Revised Jun 29 2015]
John P. Linderman, Notes on computation of first 5 billion terms
John Mason, Some observations
John Mason, Some proofs
John Mason, Proof of theorem: If the sequence is infinite, it is a permutation of the positive integers

Cf. A098550, A247225, A247942, A249167, A251604, A256974, A318826.
For indices of primes see A256213. Sequence mod 2 is A257585.
Changing > in definition to >= produces A255582 (which DOES exist).
Cf. A256528 (partial sums).

a254077 n = a254077_list !! (n-1)
a254077_list = 1 : 2 : 3 : f 2 3 [4..] where
   f u v ws = g ws where
     g (x:xs) = if gcd x u > gcd x v then x : f v x (delete x ws) else g xs
-- Reinhard Zumkeller, May 05 2015

f[n_] := Block[{s = Range@ n, j, k}, For[k = 4, k <= n, k++, j = 4; While[Nand[GCD[j, s[[k - 2]]] > GCD[j, s[[k - 1]]], !MemberQ[Take[s, k - 1], j]], j++]; s[[k]] = j]; s]; f@ 72 (* Michael De Vlieger, Apr 15 2015 *)

More terms from Peter J. C. Moses, Jan 25 2015

The sequence differs from A098550 from a(11) onward.

The sequence is a permutation of the natural numbers. The proof is similar to that for A098550 (with minor changes). - Vladimir Shevelev, Jan 14 2015

In particular, A098550(10)=6, A247225(14)=6.

On the other hand, it seems likely that 6 never appears in A247665 (which has a different although related definition). - N. J. A. Sloane, Apr 26 2015

cp's OEIS Frontend

A098550 The Yellowstone permutation: a(n) = n if n <= 3, otherwise the smallest number not occurring earlier having at least one common factor with a(n-2), but none with a(n-1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Python

A254077 a(n) = n if n <= 3, otherwise the smallest number not occurring earlier such that gcd(a(n),a(n-2)) > gcd(a(n),a(n-1)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Mathematica

Extensions

A247942 a(n) = n if n <= 3, otherwise the smallest number not occurring earlier having at least one common factor with a(n-2)*a(n-3), but none with a(n-1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

A254003 For n>=1, let B_n(k) be the sequence defined by b(1)=1, b(2)=p_1, ..., b(n+2)=p_(n+1), thereafter (for k>=n+3) the smallest number not occurring earlier having at least one common factor with b(k-(n+1)), but none with b(k-1)*b(k-2)*...*b(k-n); a(n) is m such that B_n(m)=6.

Views

Author

Keywords

Comments

Crossrefs

A255509 a(1)=1, a(2)=2, a(3)=3; for n>=4, a(n) is the maximal prime factor P_n of a(n-2) if P_n is not already a term, otherwise a(n) is the smallest not appeared earlier positive number x such that gcd(x,a(n-2))>1, gcd(x,a(n-1))=1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Extensions

A254003 For n>=1, let B_n(k) be the sequence defined by b(1)=1, b(2)=p_1, ..., b(n+2)=p_(n+1), thereafter (for k>=n+3) the smallest number not occurring earlier having at least one common factor with b(k-(n+1)), but none with b(k-1)b(k-2)...*b(k-n); a(n) is m such that B_n(m)=6.