hihi - yes, understood, that was just my equivalent computational *elimination *rule - in fact, if every prime dividing a(n) also divides a(n-1), then there is no valid a(n+1). because every non-trivial common factor between any prospective a(n+1) and a(n) will have at least one of those primes, so the prospective a(n+1) will also have a non-trivial common factor with a(n-1), and is therefore invalid - that means that any prospective value for a(n+1) needs to have at least one prime that does not divide a(n) (or else there is no valid a(n+2)) that made the computation simpler to program more later, chris On Wed, Aug 19, 2020 at 8:08 AM Neil Sloane <njasloane@gmail.com> wrote:
Chris, the definition of Enots Wolley (A336957) is Lexicographically earliest 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). It doesn't say that every prime dividing a(n) divides a(n-1). Look at Theorem 1 in A336957!
Best regards Neil
Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
On Tue, Aug 18, 2020 at 7:10 AM Christopher Landauer <topcycal@gmail.com> wrote:
Hihi - If every prime dividing a[n] also divides a[n-1], then there is no integer b that has gcd(b, a[n])>1 and gcd(b, a[n-1])=1, so a[n+1] does not exist
We have a[3]=6, so this criterion applies to b=9
more later, Chris
Sent from my iPhone
On Aug 18, 2020, at 02:00, Neil Sloane <njasloane@gmail.com> wrote:
if a(4) = 9 then a(5) does not exist. therefore a(4) > 9
Best regards Neil
Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
On Tue, Aug 18, 2020 at 2:47 AM Frank Stevenson < frankstevensonmobile@gmail.com> wrote:
Hi Niel,
I was trying to compute this sequence, but I am having problems understanding the concept of lexicographical earliest. Why is not 9 the 4th number in the sequence, but 15 is ? Why not 105, 1005, 100005 etc ? What am I missing ?
Regards, Frank
On Sun, Aug 16, 2020 at 5:56 AM Neil Sloane <njasloane@gmail.com> wrote:
Obviously this is an inverted version of the Yellowstone sequence A098550 ! The name Enots Wolley is for personal use only, it must not be mentioned in the OEIS! We frown on such made-up names.
Definition: Lexicographically earliest 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). 1, 2, 6, 15, 35, 14, 12, 33, 55, 10, 18, 21, 77, 22, 20, 45, 39, 26, 28, 63, ... The original idea was due to Scott, with a different sequence, but this is my (canonical!) version.
Could someone please prove the conjecture that this is a permutation of the set {1, all numbers with at least two distinct prime factors} ?
I can't even prove that every number 2*p (p prime) appears, or that there are infinitely many even terms (although I've found a dozen false proofs). It's a slippery problem.
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