[seqfan] Re: A new and stubborn "Yellowstone" sequence

[hv at crypt.org]

Neil Sloane <njasloane at gmail.com> wrote:
[...]
:I've been spending too much time lately on a new one, also from
:Remy Sigrist, A280864: the lexicographically earliest sequence with
:the property that if a prime p divides a(n) then p also divides
:exactly one of a(n-1) and a(n+1). It starts (beginning with a(1)):
:1, 2, 4, 3, 6, 8, 5, 10, 12, 9, 7, 14, 16, 11, 22, 18, 15, ...
:I can "almost" prove it is a permutation! Here is what I can prove
:(see A280864 for details):
:- every prime appears (naked)
:- every even number appears
:- there are infinitely many odd composite terms
:but I cannot prove that every odd number appears
[...]
:I have dozens of pages of notes on A280864, but I'll just mention a few things:
:Call a(n) "satisfied" if all its prime factors already divide a(n-1).
:If a(n) is satisfied, a(n+1) is the smallest missing number relatively
:prime to a(n).
:It is helpful to split the primes p into two classes:
:Type I. The first time that p divides any term is after a satisfied
:term a(n-1), with a(n) = p, a(n+1) = 2p, a(n+2) = smallest missing
:even number.
:Type II. The first time that p divides any term is when a(n) = b*p,
:b>1, a(n+1) = p (and is satisfied).
:The Type II primes (A280745) are
:13, 139, 379, 397, 647, 661, 967, 983, 997, 1021, 1063, 1109,...
:and there appear to be infinitely many of them. It appears that most
:primes are Type I, and I can prove there are infinitely many of Type
:I.

I've been thinking off and on about this, and had an insight this morning
that I think makes it trivial:

- a type I prime appears after a satisfied term, so when it appears it
is the least unused number
- a type II prime is satisfying, so the next term after it appears is
the least unused number
- every prime appears
- there are an infinite number of primes
- so the least unused number increases an infinite number of times.

Did I miss something?

Hugo