[seqfan] Re: A130011 and the definition of "slowest increasing".

[Charles Greathouse]

I'd like a more precise description of "slowest increasing".

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Mon, Jul 13, 2015 at 10:08 PM, M. F. Hasler <seqfan at hasler.fr> wrote:

> On Mon, Jul 13, 2015 at 5:39 PM, Frank Adams-Watters
> <franktaw at netscape.net> wrote:
> > Note, by the way, that neither of these always exists. One possible
> problem is that the
> > greedy algorithm takes you to a dead end, while an infinite extenuation
> is possible.
> >
> > Flor slowest increasing, there may be two possible sequences that keep
> switching
> > off which is the smallest at a given index.
>
> I agree. As usual in mathematics, the best (or rather: the only
> possibility to have something well defined) would be to have a precise
> definition of what is meant by potentially ambiguous terms, here: "the
> slowest increasing".
> So far I met the notion of "faster/slower increasing" only in the
> sense of asymptotic growth ( f = o(g) ), which is (I think) not enough
> here.
> So does it mean: for any other sequence b(), there is N such that for
> all n large enough, one has a(n) <= b(n) ?
> (I think it is possible that strict "<" cannot be required (see
> below), but I'm not sure.)
> Is this enough to have uniqueness? Must one add that, if there is
> equality a(n) = b(n) for all n large enough, one has to take the
> lexicographic earliest sequence with that property?
> Or the sequence with the smallest slope (first differences)? (The
> lexicographic earlier will need to have stronger growth to catch up
> with the other one, so it is not "slower increasing"...)
> If so, in the absolute (sup norm) or in the lexicographical sense?
>
> Is the keyword "easy" still justified for this version which
> " is more complicated algorithmically " ?
>
> Maximilian
>
>
> > -----Original Message-----
> > From: Lars Blomberg <lars.blomberg at visit.se>
> > To: 'Sequence Fanatics Discussion list' <seqfan at list.seqfan.eu>
> > Sent: Mon, Jul 13, 2015 4:28 pm
> > Subject: [seqfan] Re: A130011 and the definition of "slowest increasing".
> >
> >
> > Alois,
> >
> > Thank you for the clarification.
> >
> > /Lars
> >
> > -----Ursprungligt
> > meddelande-----
> > Från: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] För Heinz,
> > Alois
> > Skickat: den 13 juli 2015 15:23
> > Till: Sequence Fanatics Discussion
> > list
> > Ämne: [seqfan] Re: A130011 and the definition of "slowest increasing".
> >
> > Am
> > 13.07.2015 um 08:19 schrieb Lars Blomberg:
> >
> > >
> > > Could someone please define what
> > "slowest increasing" means?
> > >
> > > And what is the difference between "slowest
> > increasing" and
> > > "lexicographically first"?
> > >
> >
> > "lexicographically first" ist
> > the greedy approach.  Use the smallest a(n) that satisfies the condition
> given
> > a(1), ..., a(n-1).
> > And do not change it later.  This is easy
> > algorithmically.
> >
> > "slowest increasing" here means that you accept a larger than
> > greedy
> > a(n) if it is possible to get a smaller a(m) for a larger m>n.
> > This is
> > more complicated algorithmically.
> >
> > Please do not change the definition of
> > A130011.
> >
> > If you want to have a new sequence with the greedy approach, please
> > use a new A-number.
> >
> > Best,
> > Alois
>
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