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

[M. F. Hasler]

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