[seqfan] Re: seqs whose |differences| are 1,2,3,4,...

[Robert Munafo]

That's close to what I'm getting (see the new thread based on my specific
algorithm, which may differ from the original problem)

I also get a 2*3^k "gap" size (I called them "clumps of holes", but it's the
same thing) but I differ on what the actual "gaps" are. You got 15-20, I got
14-19; instead of 45-62 I got 44-61, and so on.

Also of note are the sizes of the blocks of integers included by the
sequence (the "figure" to which the gaps are the "ground") which are: 8, 24,
72, 216, 648, ... or 8*3^k.

On Thu, Mar 11, 2010 at 07:22, Raff, Paul <paul at myraff.com> wrote:

> Hi all again,
>
> Like I said before, I should be doing other things and plan to focus
> more on this problem by early next week, but I did want to add my two
> cents. It's a good sequence!
>
> The sequence from my algorithm has the following gaps (i.e. numbers
> that aren't in the sequence): {5, 6}, {15, 16, 17, 18, 19, 20}, {45,
> 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62} .
> . . and the pattern continues. The gaps have size 2*3^k, which
> (superficially) reminds me of the Cantor Set.
>
> It's telling that 5 and 6 are at the very end of Ron's sequence - I
> imagine that an increase in Ron's parameter L will cause 5 and 6 (and
> 15 and 16, etc.) to be pushed farther to the back of the sequence,
> where in the limit they disappear to infinity and beyond.
>
> [...]
>
> On Thu, Mar 11, 2010 at 4:21 AM, Robert Munafo <mrob27 at gmail.com> wrote:
> > I believe this is the crucial insight:* "**hole-filling postponed without
> > limit" [1]*
> >
> > [...] if one establishes that all the holes can be filled in later, and
> > further that we only need to determine some conveniently-small finite
> > initial portion of the sequence, then we can see that given any finite N,
> > the first N terms of the following two sequences are the same (using
> wording
> > of [3] and changing "one-to-one" to "injective and unbounded"):
> >
> >  The lexicographically first one-to-one sequence a(n)
> >  satisfying |a(n+1)-a(n)| = n
> >
> >  The lexicographically first injective and unbounded sequence b(n)
> >  satisfying |b(n+1)-b(n)| = n
> >
> > [...]
> > Once we accept that, then the method of FTAW [6] seems to work: Use a
> > conditionally-greedy selection algorithm that does not accept any
> candidate
> > next term until a guaranteed "escape route" is found.
> >
> > [1] from Ron Hardin's message of Mar 11, 2010 at 01:53.
> >
> > [...]
> >
> > [3] from Franklin T. Adams-Watters's message of Mar 10, 2010 at 15:08.
>

-- 
 Robert Munafo  --  mrob.com