[seqfan] Re: A179094

[Benoît Jubin]

On Fri, Aug 27, 2010 at 5:37 AM, Douglas McNeil <mcneil at hku.hk> wrote:
> [Aside: since 1 is odd and not even, I should've written the
> conjectural formula as 'A179094(n) = 0 for n=1, n^3-n-1 for odd n > 1,
> n^3-3 for even n.']
>
> 2010/8/27 Benoît Jubin:
>
>> What would your program give when you also add the distance between
>> the labels n^2 and 1 ? (0, 6, >=24, ...)
>
> If I understand you then this is the round-trip variant, right?

Yes, that's it: consider the labels to be integers modulo n^2.

> Assuming I didn't break anything, I find
>
> [   0    6   24   62  120  214  336  510  720  998 1320 1726 2184 2742 3360
>  4094 4896 5830 6840]
>
> which looks like a(1) = 0, a(n) = n^3-n for odd n > 1, a(n) = n^3-2 for even n.

Very nice: it's exactly A179094(n)+1, as I suspected.  One of the
reasons I asked was to include the case n=1 in your conjectured
formula (it's not a special case anymore).  This makes the sequence
more natural, to me.  Also, there should be more tie winning
configurations.

The same applies to A047838(n)=Floor(n^2/2)-1: for the associated
"cyclic" version, we would have a(n)=Floor(n^2/2), and we could begin
at the index 1, and not 2.

>> Also, can your program output the "winning configurations" for small n ?
>
> Just to be clear, it's only my program in the sense that it's the
> program that I used: namely, Concorde with glpk as the linear solver,
> both of which can be freely downloaded, so the two groups get all the
> props.  The methods are constructive so they'll output winning paths,
> which can be converted into configurations and thus they provide
> concrete evidence of the >= part, at least. :^)  Not sure what the
> best way to store auxiliary data like that on the wiki will be.

Plain text as in the current oeis entry for this sequence seems best.
Looking at all the winning configurations for n=3, 4 might hint at a
proof strategy.

Benoit

>
>
> Doug
>
> --
> Department of Earth Sciences
> University of Hong Kong
>
>
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