[seqfan] Re: a permutation of the naturals
Benoît Jubin
benoit.jubin at gmail.com
Wed Oct 28 15:14:48 CET 2009
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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