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

[Neil Sloane]

Hans,  Here is the flaw in that argument.  When we
see a type I prime, say a(n) = p, after a(n-1) = x,
then gcd(x,p)=1.  If there were a smaller missing number than p,
we would use it instead of p provided it was relatively prime to x.
But that's the rub.

Say we are trying to prove that there are infinitely many powers of 3
in the sequence
(something I can't prove).  The sequence could block them
by preceding every large prime p by 3A, 3B.
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 at gmail.com



On Wed, Mar 22, 2017 at 9:15 AM,  <hv at crypt.org> wrote:
> 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
>
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