[seqfan] Re: Binary Complement Sequences

[Tom Duff]

I've been running 425720 for almost 3 days now -- about 31 billion
iterations --  and the size of the numbers it's working on seems to have
settled down. From eyeballing the iteration vs. size graph (
http://iq0.com/iterations-vs-size.png ) the numbers look like they're
oscillating between 300k and a million bits. There's only so many million
bit numbers, so eventually it will either find one that descends to zero,
or take off into the multi-million bit range. If it hits zero, it may be
after the heat death of the universe. I'll stop it after a few more days
and give a final report. (BTW, I've been updating the graph at
http://iq0.com/iterations-vs-size.png sporadically, when I think of it.)

On Tue, Dec 20, 2022 at 11:01 PM Joshua Searle (larry) <jprsearle at gmail.com>
wrote:

> > Allan Wechsler 16 December 2022 at 21:38:07 GMT
> >
> > When you start adding sequences to OEIS, please remember to add the
> > underlying function (triple, then complement). I think it starts 0, 1, 6,
> > 3, 0, 13, 10 ..., and is definitely not in OEIS already.
>
> > "M. F. Hasler” 17 December 2022 at 18:27:51 GMT
> >
> > The first sequence that should be submitted (if not already in OEIS) is
> the
> > function itself,
> > a(n) = binary complement of 3n,
>
> I’ve submitted the sequence, its A359194. I hadn’t submitted a sequence
> before now so once I’ve got the format down I’ll do the iteration based
> sequences.
>
> > "M. F. Hasler” 17 December 2022 at 18:27:51 GMT
> >
> > I think a key to understanding and proving that the sequence always
> > terminates are the "max values" (your 1b), so these should also be
> listed,
> > but maybe (also / rather) in increasing order without repetitions.
> Define
> > them as starting values that are the maximum of their orbit (or some
> other
> > way to say that all subsequent values are smaller).
> >
> > Then, to prove that all values terminate, it is sufficient (and
> equivalent)
> > that any starting values  eventually reaches such a max number".
> > I think it should be possible to prove by considering the binary
> expansion,
> > and in particular the initial digits of the numbers. My hunch is that it
> > should be sufficient to distinguish a finite number of patterns for the
> > initial digits, and what might possibly come after that.
> >
> > - Maximilian
>
> I could also add this sequence, which makes 10 total. This isn’t excessive
> is it?
>
> > Tom Duff 17 December 2022 at 20:30:54 GMT
> >
> > I have 425720 running on my desktop machine right now. It's been going
> for
> > just under an hour and after 1,572,000,000 iterations it's working on a
> > 125,000+ bit number. Graphing the length of the iterate vs the iteration
> > number (see___) gives a fairly ragged
> > curve, but it doesn't look like it's likely to come back down. Of course,
> > in the absense of a proof, anything might happen.
>
> Seems like 2-3x faster than my code. Should reach 10 billion in about a
> day if it scales the same way as mine. Since I’m replying 3 days after the
> fact, you may be at 20 odd billion or it may have terminated! Considering
> the jumps in step count between record holders, it is not improbable that
> it could reach 100 billion iterations before termination. Assuming it
> actually does of course.
>
> Given n balanced 1d random walks of constant step size, how far would you
> expect the last path to go before returning to the “0” point? I have no
> idea if this has any relevance to this problem but perhaps it could be used
> as a measuring stick?
>
>
> Joshua Searle
>
> (I am using the weekly digest to prevent my inbox being swamped)
>
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