[seqfan] Re: Permutations with 2K-2 odd displacements, formula?

[Ron Hardin]

I suppose it's easy to prove, given the answer

Define a permutation by four choosings:

Order the even elements (factorial((n+1)/2))
Order the odd elements (factorial(n/2))
The first k-1 even elements go to the first k-1 odd elements in all ways 
(binomial(n/2,k-1))
The first k-1 odd elements go to the first k-1 even elements in all ways 
(binomial((n+1)/2,k-1))

The remaining even elements appear in unfilled even slots in their new order
The remaining odd elements appear in unfilled odd slots in their new order

Assuming this is right, the word empirical can be dropped and the formula gets 
promoted to the title, and the problem to a comment.

 rhhardin at mindspring.com
rhhardin at att.net (either)



----- Original Message ----
> From: Ron Hardin <rhhardin at att.net>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Sent: Sun, June 2, 2013 6:32:14 AM
> Subject: [seqfan] Re: Permutations with 2K-2 odd displacements, formula?
> 
> Great!  so the formula in my n,k terms checks  as
>T(n,k)=factorial(n/2)*factorial((n+1)/2)*binomial(n/2,k-1)*binomial((n+1)/2,k-1);
>;
> 
> 
> 
> with integer (floor) divides.
> 
> rhhardin at mindspring.com
> rhhardin at att.net (either)
> 
> 
> 
> ----- Original Message ----
> > From: Neil Sloane  <njasloane at gmail.com>
> > To:  Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> > Sent:  Sat, June 1, 2013 10:36:46 PM
> > Subject: [seqfan] Re: Permutations with  2K-2 odd displacements, formula?
> > 
> > In fact, after dividing by the  entries at the start of the rows,
> > we seem to  get  A124428
> 
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