cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A336957 The Enots Wolley sequence: the lexicographically earliest infinite sequence {a(n)} of distinct positive numbers such that, for n>2, a(n) has a common factor with a(n-1) but not with a(n-2).

Original entry on oeis.org

1, 2, 6, 15, 35, 14, 12, 33, 55, 10, 18, 21, 77, 22, 20, 45, 39, 26, 28, 63, 51, 34, 38, 57, 69, 46, 40, 65, 91, 42, 30, 85, 119, 56, 24, 75, 95, 76, 36, 87, 145, 50, 44, 99, 93, 62, 52, 117, 105, 70, 58, 261, 111, 74, 68, 153, 123, 82, 80, 115, 161, 84, 60, 155, 217, 98, 48, 129, 215, 100
Offset: 1

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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
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)

Crossrefs

A337007 and A337008 describe the overlap between successive terms.
See A337066 for when n appears, A337275 for when 2p appears, A337276 for when 2k appears, A337280 for when p first divides a term, A337644 for runs of three odd terms, A337645 & A338052 for smallest missing legal number, A337646 & A337647 for record high points, A338056 & A338057 for record high values for a(n)/n.
See A338053 & A338054 for the "early" terms.
Further properties of the present sequence are studied in A338062-A338071.
A338059 has the missing prime powers inserted (see also A338060, A338061).
See A338055, A338351 for variants.
A280864 is a different but very similar lexicographically earliest sequence.

Programs

  • Maple
    with(numtheory);
    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 2 do A[n]:= n: od:
    for n from 3 do
      for k from 3 to N do
        if B[k] = 0 and igcd(k, A[n-1]) > 1 and igcd(k, A[n-2]) = 1 then
              if nops(factorset(k) minus factorset(A[n-1])) > 0 then
           A[n]:= k;
           B[k]:= 1;
           break;
              fi;
        fi
      od:
      if k > N then break; fi;
    od:
    s1:=[seq(A[i], i=1..n-1)]; # N. J. A. Sloane, Sep 24 2020, based on Theorem 1 and Robert Israel's program for sequence A098550
  • Mathematica
    M = 1000;
    A[1] = 1; A[2] = 2;
    Clear[B]; B[_] = 0;
    For[n = 3, True, n++,
    For[k = 3, k <= M, k++,
    If[B[k] == 0 && GCD[k, A[n-1]] > 1 && GCD[k, A[n-2]] == 1, If[Length[ FactorInteger[k][[All, 1]] ~Complement~ FactorInteger[A[n-1]][[All, 1]]] > 0, A[n] = k; B[k] = 1; Break[]]]]; If[k > M, Break[]]];
    Array[A, n-1] (* Jean-François Alcover, Oct 20 2020, after Maple *)
  • Python
    from math import gcd
    from sympy import factorint
    from itertools import count, islice
    def agen(): # generator of terms
        a, seen, minan = [1, 2], {1, 2}, 3
        yield from a
        for n in count(3):
            an, fset = minan, set(factorint(a[-1]))
            while True:
                if an not in seen and gcd(an, a[-1])>1 and gcd(an, a[-2])==1:
                    if set(factorint(an)) - fset > set():
                        break
                an += 1
            a.append(an); seen.add(an); yield an
            while minan in seen: minan += 1
    print(list(islice(agen(), 70))) # Michael S. Branicky, Jan 22 2022

Extensions

Added "infinite" to definition. - N. J. A. Sloane, Sep 03 2020
Added Scott R. Shannon's name "Enots Wolley" (Yellowstone backwards) for this sequence to the definition, since that has been mentioned in several talks. - N. J. A. Sloane, Oct 11 2020

A338053 "Early" terms in A336957, in order of appearance.

Original entry on oeis.org

6, 15, 35, 14, 12, 33, 55, 18, 21, 77, 22, 20, 45, 39, 26, 28, 63, 51, 34, 38, 57, 69, 46, 40, 65, 91, 42, 85, 119, 56, 75, 95, 76, 87, 145, 50, 44, 99, 93, 62, 52, 117, 105, 70, 58, 261, 111, 74, 68, 153, 123, 82, 80, 115, 161, 84, 155, 217, 98, 129, 215, 100, 141, 235, 110, 147, 133
Offset: 1

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Author

Scott R. Shannon and N. J. A. Sloane, Oct 11 2020, following a suggestion from Christopher Landauer

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Comments

A term A336957(k) is early if A336957(k) > k; punctual if A336957(k) = k (see A338050); and late if A336957(k) < k.
It appears that the majority of terms are late.

Crossrefs

A338054 "Early" terms in A336957, arranged in increasing order.

Original entry on oeis.org

6, 12, 14, 15, 18, 20, 21, 22, 26, 28, 33, 34, 35, 38, 39, 40, 42, 44, 45, 46, 50, 51, 52, 55, 56, 57, 58, 62, 63, 65, 68, 69, 70, 74, 75, 76, 77, 80, 82, 84, 85, 87, 91, 93, 95, 96, 98, 99, 100, 102, 104, 105, 108, 110, 111, 112, 115, 116, 117, 118, 119, 122, 123, 124, 126, 129
Offset: 1

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Author

Scott R. Shannon and N. J. A. Sloane, Oct 11 2020, following a suggestion from Christopher Landauer

Keywords

Comments

A term A336957(k) is early if A336957(k) > k; punctual if A336957(k) = k (see A338050); and late if A336957(k) < k.
It appears that the majority of terms are late.

Crossrefs

A361105 Fixed points in A360519.

Original entry on oeis.org

1, 88, 92, 112, 116, 172, 268, 272, 324, 17242, 18650, 43208, 55828, 192434, 1497756
Offset: 1

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It is unknown if the sequence is infinite.

Crossrefs

Extensions

a(15) from Rémy Sigrist, Mar 04 2023
Showing 1-4 of 4 results.