[seqfan] Fwd: [math-fun] Re: Nice (nontrivial) sequence idea

[Fred Lunnon]

---------- Forwarded message ---------
From: Fred Lunnon <fred.lunnon at gmail.com>
Date: Thu, Mar 2, 2023 at 2:06 PM
Subject: Re: [math-fun] Re: Nice (nontrivial) sequence idea
To: math-fun <math-fun at mailman.xmission.com>


  Over on  seqfan , Christian Sievers reports the following 96 stable terms
for  n < 2*10^9 :

[ 1, 4, 11, 29, 47, 18, 61, 165, 434, 703, 1675, 972, 2213, 10093, 17973,
25853,
59586, 33733, 7880, 21427, 56401, 204177, 147776, 91375, 217724, 126349,
414021,
287672, 161323, 34974, 48521, 13547, 5667, 9121, 12575, 28604, 16029, 3454,
1241, 269, 1180, 3271, 2091, 911, 2464, 11409, 20354, 90361, 250729,
160368,
551111, 1492965, 941854, 390743, 1011861, 621118, 2714847, 15667964,
44289045,
161488216, 278687387, 117199171, 72910126, 174441333, 450413873, 275972540,
101531207, 28621081, 12953117, 36144504, 23191387, 33429657, 10238270,
7523423,
19855422, 52042843, 32187421, 44519420, 12331999, 4808576, 21328033,
59175523,
156198536, 253221549, 350244562, 97023013, 231893516, 134870503, 37847490,
16519457, 61269252, 167288299, 273307346, 106019047, 362806936, 1345208697
];

Remark that there are no more Fibonacci numbers or powers of 2 so far,
after the first two terms.

WFL


On Thu, Mar 2, 2023 at 7:34 AM Andy Latto <andy.latto at pobox.com> wrote:

> On Wed, Mar 1, 2023 at 9:45 PM Dan Asimov <dasimov at earthlink.net> wrote:
>
> >
> >
> > -----
> > The idea is to insert all integers n = 1, 2, 3, ... in a list, between
> > the pair of consecutive elements whose sum equals n and whose absolute
> > difference is the smallest possible, or at the end of the list if
> > there's no such pair.
> > -----
> >
> > Question:
> > ---------
> > Is it possible that there could be *more than one* pair of
> > consecutive elements whose sum equals n and whose absolute difference
> > is the smallest possible? And if so, what does the algorithm dictate?
> >
>
> Each integer only occurs once in the list. You can't have two distinct
> pairs of integers that have the same sum and the same absolute difference.
>
> Andy
>
>
> >
> > —Dan
> >
> > Allan Wechsler a écrit:
> > -----
> > I confess that I don't share Maximilian's intuition that all numbers will
> > eventually find a home. It feels likely that 2 won't. Every time 2 gets
> > pushed to the right, new, bigger sums appear to its left. Another
> sequence,
> > giving the number of pairs to the left of two whose sum is greater than n
> > after the nth step, might be instructive. Is it growing? Shrinking?
> Bopping
> > around randomly?
> > ...
> > ...
> > > On Wed, Mar 1, 2023 at 8:17 PM M F Hasler wrote:
> > >
> > > > Hello math-fun,
> > > > I encountered a nice idea on reddit today.
> > > > The idea is to insert all integers n = 1, 2, 3, ... in a list,
> between
> > > the
> > > > pair of consecutive elements whose sum equals n and whose absolute
> > > > difference is the smallest possible,
> > > > or at the end of the list of there's no such pair :
> > > >
> > > > For n = 1 and n = 2, there is no pair in the sequence that sums to n,
> > so
> > > > both of these are appended to the list, which then is [1, 2].
> > > > Now n = 3 is the sum of 1 & 2 so it is inserted between these
> two,
> > the
> > > list
> > > > becomes [1, 3, 2]
> > > > Similarly, n = 4 is inserted between 1 & 3 and n = 5 is inserted
> > between
> > > 3
> > > > & 2, to get [1, 4, 3, 5, 2].
> > > > Now n = 6 isn't the sum of any two adjacent terms, so it is appended:
> > [1,
> > > > 4, 3, 5, 2, 6].
> > > > Then n = 7 is the sum of 4+3 as well as 5+2, one has to compare the
> > > > differences |4-3| and |5-2| to find that it must be inserted between
> 4
> > > and
> > > > 3. And so on.
> > > >
> > > > I wonder this is a permutation of the positive integers, i.e., any n
> > will
> > > > eventually reach its final position in the sequence.
> > > > This is the case when it is no more preceded by a pair of consecutive
> > > terms
> > > > with a sum larger than the current length of the sequence. It appears
> > > that
> > > > sufficiently many terms are not equal to a sum of a consecutive pair
> > and
> > > > therefore appended to the end of the sequence, so that any n reaches
> a
> > > > stable position, sooner or later.
> > > >
> > > > But I have absolutely no proof for this, and I don't even know at
> which
> > > > position the number 2 will eventually remain:
> > > > It's at position 106 when we've got 13 terms for sure (requiring 3185
> > > terms
> > > > to be computed), at position 176 when we've got 14 good terms (using
> > more
> > > > than 12K terms) and at position 234 when we've got 15 good terms
> (using
> > > > about 28K terms),
> > > > (1, 4, 11, 29, 47, 18, 61, 165, 434, 703, 1675, 972, 2213, 10093,
> > 17973,
> > > > ...)
> > > >
> > > > Although the number 2 goes away 60 more places each time I get one
> more
> > > > term for sure, I don't give up hope that it might eventually become
> > > > "stabilized"...
> > > > (BTW, the whole group (3, 8, 13, 5, 2, 6) moves away together, so all
> > of
> > > > them will be "stabilized" at once (if they are) ; the number 10 also
> > > occurs
> > > > about 15 indices earlier, and 7 roughly half way there (around index
> > 120
> > > > when they are around position 240).)
> > > >
> > > > Can someone try to compute their final indices or get some estimate,
> > > maybe
> > > > proving that they (and/or any n) will reach its final destination
> > sooner
> > > > or later?
> >
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> >
>
>
> --
>                  Andy.Latto at pobox.com
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