[seqfan] Re: Nice nontrivial integer sequence

[Fred Lunnon]

  Christian Sievers (Mon, Mar 13, 2023):
<< Do we know that each position eventually gets stable?
In other words: are we sure we have a well defined infinite series? >>

  There are at least four distinct issues conflated here, depending on the
exact interpretation of the term `each position'; however the answer to all
of them appears currently to be negative.

  Immediately, if there exist natural  k  and  m > k, and some provisional
finite portion of sequence  a(n)  satisfying NNSI construction rules for
0 <= n <= m  such that for all  i,j <= k  there holds  a(i)+a(j) <= m ,
then the sequence _is_ well-defined for  0 <= n <= k .

  But that is all: AFAIK, it seems entirely possible that the NNSI sequence
is
actually finite, and there exists some  k  for which there is _no_ such  m
.
The behaviour of provisional member  a(n) = 2  during current computations
already suggests that most values actually `escape' to  n = oo ; without
some
constructive bounds on (final) values of  a(n) , there seems no reason to
rule
out that, from some index  m  onwards, _all_ values might do so.

  It is tempting to speculate concerning the eventual fate of such
escapers.
For instance 2 could well be a member of some distinct, bi-directional
infinite
sequence commencing
    ..., 13, 5, 2, 6, 15, ...
from which other values (say  3 ?) in turn escape; indeed, AFAIK there
might
even be finite values which escape infinitely often from an infinitely
valent
tree of similar sequences!

Fred Lunnon


On Mon, Mar 13, 2023 at 4:14 AM Christian Sievers <seqfan at duvers.de> wrote:

> On Sat, Mar 11, 2023 at 07:58:01PM -0600, Tim Peters wrote:
>
> > The page for A360477
> >
> >     https://oeis.org/A360447
> >
> > contains
> >
> >     FORMULA a(n) = a(n-1) + a(n+1) if a(n+1) < a(n)
> >
> > I expect "a formula" to show a way to compute a(n), but that's not what
> > this is. It appears to be a claim about 3 consecutive terms in some
> cases,
> > but it's not a true claim. I think the first place it fails is at n=18:
>
> Well, I have been excited and then disappointed by such OEIS formula
> entries before, so I feel you, but in general such entries are accepted.
>
> > Alas, I can't think of a true claim related to what it says, so I can't
> > suggest "a fix". I don't believe anyone has "a formula" for a(n) in the
> > usual meaning of the word. Perhaps this section should just be deleted?
>
> While I agree with its deletion, I want to note that there is a possible
> fix: just strengthen the condition by adding "and a(n-1) < a(n)".
>
> The proof needs to discuss the list in progress, so it's not reasonable
> to talk about a(n) as the position of a number changes, instead let's
> claim that a number is the sum of its neighbours if both are smaller.
> That's true when a number gets inserted between two numbers, and when it
> is appended to the end of the list, then any number that appears right
> to it will be bigger. Also, any number that appears as a new neighbour
> to a number will be bigger than it, so the precondition is not
> satisfied. (Much words about a triviality.)
>
> But I wonder about a different question:
> Do we know that each position eventually gets stable?
> In other words: are we sure we have a well defined infinite series?
>
>
> All the best
> Christian
>
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>