[seqfan] Re: Iterates of n - A002828(n) ?
Charles Greathouse
charles.greathouse at case.edu
Tue Aug 20 18:51:00 CEST 2013
The conjecture is true. It suffices to prove it for m^2, m^2+1, and m^2+2
since all larger numbers will reduce to one of these. m^2 is the sum of one
square, so that takes it to m^2-1 as desired. m^2+1 is the sum of two
squares m^2 and 1^2, so that goes to m^2-1 for m > 0. m^2+2 is the sum of
three squares, m^2 + 1^2 + 1^2, so it will go to at least m^2-1.
Charles Greathouse
Analyst/Programmer
Case Western Reserve University
On Tue, Aug 20, 2013 at 10:26 AM, Antti Karttunen <antti.karttunen at gmail.com
> wrote:
> On Mon, Aug 19, 2013, *L. Edson Jeffery* <lejeffery2 at gmail.com>
> <seqfan%
> 40list.seqfan.eu?Subject=Re%3A%20%5Bseqfan%5D%20Re%3A%20Fwd%3A%20Iterates%20of%20n%20-%20A002828%28n%29%20%3F&In-Reply-To=%3CCAGRLqMgJ1BunNuV3BuK1Os91Dm_L-%3Djy_7PnNuNQOAm_Hx1yiw%40mail.gmail.com%3E
> >
> wrote:
>
> >
> >
> > *in *http://list.seqfan.eu/pipermail/seqfan/2013-August/011586.html
> >
>
>
> >
> > - Previous message: [seqfan] Fwd: Iterates of n - A002828(n) ?<
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> > - *Messages sorted by:* [ date ]<
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> > [ thread ]<
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> > [ subject ]<
> http://list.seqfan.eu/pipermail/seqfan/2013-August/subject.html#11586>
> > [ author ]<
> http://list.seqfan.eu/pipermail/seqfan/2013-August/author.html#11586>
> >
> > ------------------------------
> >
> > AK>...will the trajectory visit every square-1 (A005563) < k before it
> > reaches zero?
> >
> > Is it sufficient to prove, for any n, that max(square-1 < n) is in the
> > trajectory?
> >
> > Yes, just that.
>
> >
> > AK>Which would imply that there would be only one infinite sequence a_n
> > such that a(n-1) = a(n) - the least number of squares that add up to
> a(n).
> >
> > This should be essentially the first column in the first triangle below.
> >
> > AK>Then, would there be any regularity in the derived sequence counting
> the
> > number of iterations needed to hop from A005563(n+1) to A005563(n)?
> >
> > According to the second triangle below, it appears that your hopping
> > sequence should be {1,1,2,2,4,4,5,...}, unless I misunderstood or
> > miscalculated.
> >
> > Yes, exactly that, the first differences of the row lengths 1, 1, 3, 5,
> 9,
> 13,18, ...
> of the table given below.
>
> Not sure exactly what you mean by "regularity."
> >
> >
> Something like what is present in these https://oeis.org/A213709 &
> https://oeis.org/A226060 & https://oeis.org/A218543
> Some regularity (e.g. partial scale-invariance) mixed with some "random" or
> not-so-easily computable aspects.
>
>
> > Nice ideas, Antti.
> >
> >
> Not wholly my idea. I just tried to find a variant with a more
> "number-theoretical taste to it" than Carl White's original "binary
> beanstalk" sequence https://oeis.org/A179016 (to which the above three
> seq-links are related to) and other examples of such sequences that I have
> submitted in other bases.
> However, while it is easy to guarantee the existence of "choking points"
> (through which all the iterated paths will come through) in base-related
> beanstalk sequences, it seems not that easy when trying to construct such
> phenomena from sequences related to the traditional number theory.
> But what I glean from Pari-code at https://oeis.org/A002828
> it seems actually to be a some kind of "crypto-base" sequence.
> (Or at least involving modulus some power of 2. I have to re-read my
> Burton.)
>
>
> -- Antti
>
> PS: If anybody comments on this, please add CC: to my gmail-account, so I
> will get the reply even before the mailing list digest is ready.
>
> >
> > n Seq of trajectories for 0 <= n <= 48
> > --- -----------------------------------------
> > 0 0
> > 1 0
> > 2 0
> > 3 0
> > 4 3,0
> > 5 3,0
> > 6 3,0
> > 7 3,0
> > 8 6,3,0
> > 9 8,6,3,0
> > 10 8,6,3,0
> > 11 8,6,3,0
> > 12 9,8,6,3,0
> > 13 11,8,6,3,0
> > 14 11,8,6,3,0
> > 15 11,8,6,3,0
> > 16 15,11,8,6,3,0
> > 17 15,11,8,6,3,0
> > 18 16,15,11,8,6,3,0
> > 19 16,15,11,8,6,3,0
> > 20 18,16,15,11,8,6,3,0
> > 21 18,16,15,11,8,6,3,0
> > 22 19,16,15,11,8,6,3,0
> > 23 19,16,15,11,8,6,3,0
> > 24 21,18,16,15,11,8,6,3,0
> > 25 24,21,18,16,15,11,8,6,3,0
> > 26 24,21,18,16,15,11,8,6,3,0
> > 27 24,21,18,16,15,11,8,6,3,0
> > 28 24,21,18,16,15,11,8,6,3,0
> > 29 27,24,21,18,16,15,11,8,6,3,0
> > 30 27,24,21,18,16,15,11,8,6,3,0
> > 31 27,24,21,18,16,15,11,8,6,3,0
> > 32 30,27,24,21,18,16,15,11,8,6,3,0
> > 33 30,27,24,21,18,16,15,11,8,6,3,0
> > 34 32,30,27,24,21,18,16,15,11,8,6,3,0
> > 35 32,30,27,24,21,18,16,15,11,8,6,3,0
> > 36 35,32,30,27,24,21,18,16,15,11,8,6,3,0
> > 37 35,32,30,27,24,21,18,16,15,11,8,6,3,0
> > 38 35,32,30,27,24,21,18,16,15,11,8,6,3,0
> > 39 35,32,30,27,24,21,18,16,15,11,8,6,3,0
> > 40 38,35,32,30,27,24,21,18,16,15,11,8,6,3,0
> > 41 38,35,32,30,27,24,21,18,16,15,11,8,6,3,0
> > 42 39,35,32,30,27,24,21,18,16,15,11,8,6,3,0
> > 43 40,38,35,32,30,27,24,21,18,16,15,11,8,6,3,0
> > 44 41,38,35,32,30,27,24,21,18,16,15,11,8,6,3,0
> > 45 43,40,38,35,32,30,27,24,21,18,16,15,11,8,6,3,0
> > 46 43,40,38,35,32,30,27,24,21,18,16,15,11,8,6,3,0
> > 47 43,40,38,35,32,30,27,24,21,18,16,15,11,8,6,3,0
> > 48 45,43,40,38,35,32,30,27,24,21,18,16,15,11,8,6,3,0
> >
> > n n^2-1 Trajectories for n^2 - 1, 1 <= n <= 7
> > --- ------- -----------------------------------------
> > 1 0 0
> > 2 3 0
> > 3 8 6,3,0
> > 4 15 11,8,6,3,0
> > 5 24 21,18,16,15,11,8,6,3,0
> > 6 35 32,30,27,24,21,18,16,15,11,8,6,3,0
> > 7 48 45,43,40,38,35,32,30,27,24,21,18,16,15,11,8,6,3,0
> >
> >
>
> > Ed Jeffery
> >
> > -------------------------------------------------------------
> >
> AK wrote earlier:
>
> > Just wondering:
> >
> > If we iterate the function a(n) = n - A002828(n) (
> > https://oeis.org/A002828 ) from any starting value k, will the
> trajectory
> > visit every square-1 (A005563) < k before it reaches zero?
> > Which would imply that there would be only one infinite sequence a_n such
> > that a(n-1) = a(n) - the least number of squares that add up to a(n).
> >
> > Then, would there be any regularity in the derived sequence counting the
> > number of iterations needed to hop from A005563(n+1) to A005563(n) ?
> >
> >
> >
> > Cheers,
> >
> > Antti
> >
> >
> >
>
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