[seqfan] Re: The Annoyance Sequence

[Ali Sada]

 Hi Alex and Yifan Xe,
Thank you very much for your responses. 
Alex wrote  "I wonder if it's a permutation of the integers" 
I have seen similar comments on the OEIS regarding permutations of positive integers. TBH, I don't know why we need a proof in these cases.  For example, in this sequence we started we the list of positive integers and the algorithm doesn't add or delete any numbers, so, the sequence should be a permutation of positive integers. 

Alex wrote  " And how tightly we can estimate a(n) from above (the supremum growth rate)" 
This is way above my level of education :) 

Yifan Xe wrote "For the nth turn, find the smallest number x which all the numbers behind it's position is not changed. a(x-1) numbers are annoyed"I'm sorry my explanation might not have been clear. Please see the steps below1. a(1) = 1, 2 is annoyed and it jumps 2 steps 1,3,4,2,5,6,7,8,9,10. 
2. a(2) = 3 (the first number after the 1 in the modified list). 3 now annoys three numbers- 4 moves four steps 1,3,2,5,6,7,4,8,9,10...- 2 moves two steps 1,3,5,6,2,7,4,8,9,10...- 5 moves 5 steps 1,3,6,2,7,4,8,5,9,10...   
3. Now, a(3) = 6, the first number after the previous annoying number (3). The 6 annoys six numbers  
- 2 gets annoyed and the list becomes 1,3,6,7,4,2,8,5,9,10...
- now, 7 is annoyed. It moves seven steps 1,3,6,4,2,8,5,9,10,11,7,12,13,14...
- the annoyance reaches 4, and it moves 4 steps 1,3,6,2,8,5,9, 4,10,11,7,12,13,14...
- the next in line for annoyance is 2 (again!). It moves 2 steps 1,3,6,8,5,2,9, 4,10,11,7,12,13,14...
- next comes 8. It moves 8 steps 1,3,6,5,2,9, 4,10,11,7,12,8,13,14...
- finally for this stage, 5 gets annoyed and it moves 5 steps 1,3,6,2,9, 4,10,11,5,7,12,8,13,14...
That concludes the annoyance of 6. The next number in the list is 2, so a(4) = 2 and we continue. 
As you can see, m being the smallest or the largest has nothing to do with the process.
Best,
Ali 

    On Sunday, July 16, 2023 at 10:23:48 AM GMT+1, Yifan Xie <xieyifan4013 at 163.com> wrote:  
 
 First of all, your sequence looks wrong: At the first turn, 2 is annoyed; At the second turn, 3 is not annoyed.
Maybe you can change the definition to:
For the nth turn, find the smallest number x which all the numbers behind it's position is not changed. a(x-1) numbers are annoyed.
Example:
1->
(first turn)1,3,4,2->
(second turn)1,3,4,2,6,7,8,9,10,5->1,3,4,2,7,8,9,10,5,11,6->
(third turn) 6 numbers starting from 12 are annoyed.

Best regards,
Yifan Xie

(xieyifan4013 at 163.com)




| |
xieyifan4013 at 163.com
|
|
Email:xieyifan4013 at 163.com
|




---- Replied Message ----
| From | Ali Sada via SeqFan<seqfan at list.seqfan.eu> |
| Date | 07/16/2023 09:34 |
| To | Sequence Fanatics Discussion List<seqfan at list.seqfan.eu> |
| Cc | Ali Sada<pemd70 at yahoo.com> |
| Subject | [seqfan] The Annoyance Sequence |
Hi everyone,
We start with the list of positive integers 1,2,3,4,....When it turn comes, a(n) can annoy a(n) numbers and force them to jump to the right and insert themselves. When annoyed, the number m can jump m steps.
We start with a(1) =1. The 1 can annoy only 1 number, 2. The 2 jumps 2 steps and the list becomes 1,3,4,2,5,6,7,8,9...Now, a(2) = 3. It annoys 3 numbers and force them to jump
First, the 4 jumps 4 steps 1,3,2,5,6,7,4,8,9Then the 2 jumps 2 steps 1,3,5,6,2,7,4,8,9Finally, the 5 jumps 5 steps 1,3,6,2,7,4,8,5,9At this stage, 3 finished its annoyance and a(3) = 6, which in turn annoys 6 numbers and so on.
The resulted sequence is
1, 3, 6, 2, 10, 9, 5, 8, 22, 4, 13, 15, 20, 29, 7, 12, 18, 32,59, 50, 19, 31, 14, 81, 16, 90, 17, 25, 78, 83, 21, 46, 65, 23, 41, 71, 64, 36,53, 47, 58, 44, 35, 76, 62, 43, 88, 49, 123, 116, 27, 111, 40, 11, 69, 30, 79, 102,24, 60, 159, 73, 248, 72, 55, 185, 45, 101, 38, 95, 141
I would really appreciate any help defining this sequence, if it were suitable for the OEIS. Note: Maybe the VBA/Excel program below explains the sequence better

Sub Annoyance()
For i = 1 To 1000
    Cells(1, i) = i
    Next iFor i1 = 1 To 71    k1 = Cells(1, i1)
    For i2 = i1 + 1 To i1 + k1
        k2 = Cells(1, i1 + 1)
        For i3 = i1 + 1 To i1 + k2
            Cells(1, i3) = Cells(1, i3 + 1)
        Next i3
    Cells(1, i1 + k2 + 1) = k2
    Next i2
Next i1
End Sub
Best,
Ali



--
Seqfan Mailing list - http://list.seqfan.eu/

--
Seqfan Mailing list - http://list.seqfan.eu/