[seqfan] Re: Should A000201 and A026352 be cross-referenced?
Ali Sada
pemd70 at yahoo.com
Thu Jun 15 02:31:34 CEST 2023
Thank you, Allan. I will add the comment below to A026352 and cross-refrenced it to A000201.
"Consider the state of a system consists of an infinite sequence of integers, and it is initialized to 1,2,3,4,5,... The transition rule is to remove the leading
element k, and increment the next k entries. So, after one step, the state is 3,3,4,5,6,...; after two steps it is 4,5,6,6,7,8,...; after three steps it is 6,7,7,8,8,9,10,... We then read off either the largest or the smallest entry that was incremented by that step. The resulting sequences are A026352 and A000201 respectively." [From Allan Wechsler via Seqfan]
Best,
Ali
On Wednesday, June 14, 2023 at 08:32:38 AM GMT+1, Allan Wechsler <acwacw at gmail.com> wrote:
I think I understand the system being suggested by Ali Sada. The "state" of
the system consists of an infinite sequence of integers, and it is
initialized to 1,2,3,4,5,.... The transition rule is to remove the leading
element k, and increment the next k entries. So, after one step, the state
is 3,3,4,5,6,...; after two steps it is 4,5,6,6,7,8,...; after three steps
it is 6,7,7,8,8,9,10,... He then "reads off" either the largest or the
smallest entry that was incremented by that step, and conjectures that the
resulting sequences are A026352 and A000201 respectively.
To me that certainly warrants a cross-reference, assuming the
interpretation is really true of A026352. There is already a comment on
A000201 from Roland Schroeder, dated June 19, 2010, which describes the
model Aii Sada proposes.
On Tue, Jun 13, 2023 at 12:43 PM jean-paul allouche <
jean-paul.allouche at imj-prg.fr> wrote:
> I guess "divided" = "equally distributed" by addition
> jp
>
>
> Le 13/06/2023 à 18:38, Michel Marcus a écrit :
> > I do not understand this part: The 3 gets divided on the next three
> numbers
> > (3,4,5) and they become (4,5,6)
> >
> > Le mar. 13 juin 2023 à 18:03, Ali Sada via SeqFan <seqfan at list.seqfan.eu>
> a
> > écrit :
> >
> >> Hi everyone,
> >> I was trying the following algorithm:
> >> Start with 1. We add 1 to 2 and get a(1) = 3. The 3 gets divided on the
> >> next three numbers (3,4,5) and they become (4,5,6). The largest is 6,
> so,
> >> a(2) = 6. Then, we take the 4 and we divide it on the next four numbers
> >> (5,6,6,7) and they become (6,7,7,8). The largest is 8, so a(3) = 8, and
> so
> >> on.
> >> We get 3, 6, 8, 11, 14, 16, 19, 21, 24, 27, 29, 32, 35,37, 40, 42, 45,
> 48,
> >> 50, 53, 55, 58, 61, 63, 66, 69, 71, 74.
> >> This seems to be A026352.
> >> If we choose the smallest number after each operation we get
> >>
> >> 1, 3, 4, 6, 8, 9, 11, 12, 14, 16,17, 19, 21, 22, 24, 25, 27, 29, 30, 32,
> >> 33, 35, 37, 38, 40, 42, 43, 45, 46, 48,50
> >> Which seems to be A000201
> >> Best,
> >> Ali
> >>
> >> --
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> >>
> > --
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>
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