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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