

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(n2), but none with a(n1).


203



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
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OFFSET

1,2


COMMENTS

For n > 3, gcd(a(n), a(n1)) = 1 and gcd(a(n), a(n2)) > 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.
Some of the known properties (but see the abovementioned article for a fuller treatment):
1. The sequence is infinite. Proof: We can always take a(n) = a(n2)*p, where p is a prime that is larger than any prime dividing a(1), ..., a(n1). QED
2. At least onethird 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.
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 wellestablished, 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 everincreasing 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(10.5/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
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(n2)=v, a(n1)=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(m2)=v_1, a(m1)=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


LINKS

David L. Applegate, Hans Havermann, Bob Selcoe, Vladimir Shevelev, N. J. A. Sloane, and Reinhard Zumkeller, The Yellowstone Permutation, arXiv preprint arXiv:1501.01669, 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]
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.


MAPLE

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[n1]) = 1 and igcd(k, A[n2]) > 1 then
A[n]:= k;
B[k]:= 1;
break
fi
od:
if k > N then break fi
od:


MATHEMATICA

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 *)


PROG

(Haskell)
import Data.List (delete)
a098550 n = a098550_list !! (n1)
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
(PARI) 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)>1next; setsearch(u, k)&&next; gcd(k, a)==1next; 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
(Python)
from fractions 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:
l2, l1, b[i] = l1, i, 1
while s in b:
b.pop(s)
s += 1
break


CROSSREFS

The inverse permutation is in A098551.
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.


KEYWORD



AUTHOR



STATUS

approved



