# [seqfan] Re: a permutation of the naturals

Benoît Jubin benoit.jubin at gmail.com
Wed Oct 28 20:35:52 CET 2009

```On Wed, Oct 28, 2009 at 7:50 PM, Andrew Weimholt
<andrew.weimholt at gmail.com> wrote:
> I will submit the sequences tonight, along with one more related sequence.
> Each number (assuming there is no number catapulted an infinite number
> of times)

Actually, it would be nice to prove this assumption since we don't
know yet if your sequence is a permutation of the naturals, and hence
if the "inverse permutation" is everywhere defined.  Anyone can come
with a proof?  I tried a little bit, in vain.

Also, the sequence "number of times catapulted" would change to
b(0)=0, b(1)=1, b(2)=3, ... (because 1 is catapulted one time by 0
position)

and another sequence worth submitting would be "total offset by which
n is catapulted": c(0)=0, c(1)=0, c(2)=1+3+4=8, ...

Benoit

will eventually catapult one and only one other number, so
> we can define another sequence in which a(n) is the number that n
> catapults in the generation of the first sequence.
>
> I will also incorporate Benoit's suggestion of beginning with 0, which
> does not affect the
> rest of the sequence.
>
> One interesting observation, when collecting more terms: 29 is
> catapulted 17 times and ends up in a position offset by more than
> 150000 from its original position.
> Also, 0,1,54 are the only numbers I found so far with a(n)=n.
>
> Andrew
>
> On 10/28/09, Benoît Jubin <benoit.jubin at gmail.com> wrote:
>> Interesting sequence, I'm looking forward to seeing its graph.
>>
>>  I suggest you begin the sequence at 0 since it does not require any
>>  special treatment.
>>
>>  Benoit
>>
>>
>>
>>  On Wed, Oct 28, 2009 at 10:10 AM, Andrew Weimholt
>>  <andrew.weimholt at gmail.com> wrote:
>>  > This idea was inspired by one of Eric Angelini's recent posts (angry numbers).
>>  >
>>  > We start with the natural numbers in their normal positions,
>>  > and then the number in position 1 (which happens to be 1), catapults
>>  > the number to its right to a position 1 further to the right.
>>  > So after the first step, we have 1,3,2,4,5,6,7,8...
>>  > Then the number now in position 2, (which is 3), catapults the number
>>  > to its right (which is 2) to a position 3 further to the right
>>  > Now we have, 1,3,4,5,6,2,7,8...
>>  > In the nth step, the number now in the nth position (which will be
>>  > a(n)) catapults the number to its right to a position a(n) further to
>>  > the right.
>>  >
>>  > The sequence, beginning at n=1 is...
>>  >
>>  > 1, 3, 4, 6, 7, 5, 10, 2, 13, 12, 14, 16, 18, 19, 21,
>>  > 23, 8, 25, 15, 28, 17, 24, 32, 33, 20, 36, 22, 38, 40, 41, 42,
>>  > 44, 45, 47, 31, 35, 50, 52, 27, 55, 11, 58, 59, 61, 63, 64, 66,
>>  >
>>  > The inverse permutation is...
>>  >
>>  > 1, 8, 2, 3, 6, 4, 5, 17, 152, 7, 41, 10, 9, 11, 19,
>>  > 12, 21, 13, 14, 25, 15, 27, 16, 22, 18, 57, 39, 20,
>>  >
>>  > The following sequence gives the number of times n is catapulted
>>  >
>>  > 0, 3, 0, 0, 1, 0, 0, 2, 6, 0, 3, 1, 0, 0, 1,
>>  > 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 2, 2, 0,
>>  >
>>  > Not sure these are worth submitting, but thought I'd at least share
>>  > them the seqfan list
>>  >
>>  > Andrew
>>  >
>>  >
>>
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