'Mixing number' of permutations

[Brendan McKay]

Some more numerology:  I expanded sum(1/binomial(n,k),k=1...)
for large n and got 
  1/n + 2/n^2 + 8/n^3 + 44/n^4 + 308/n^5 + ...
where the coefficients seem to be
  http://www.research.att.com/~njas/sequences/A005649 .
Someone like to prove it?   Incidentally the numbers in 
A005649 are coefficients for the EXPONENTIAL generating
function of (2 - e^x)^(-2), which is not too clear there.

Brendan.


* Brendan McKay <bdm at cs.anu.edu.au> [060817 23:18]:
> 
> The average is  sum(1/binomial(n,k),k=1..n-1).  I don't know
> if this has a closed form or a name.
> 
> Proof:  There are clearly k! (n-k)! permutations that have a 
> break after the k-th value, so the probability of a break in
> that place is 1/binomial(n,k).
> 
> Asymptotically, the values k=1 and k=n-1 dominate, so the
> expectation is 2/n + O(1/n^2).
> 
> Brendan.
> 
> 
> * Hugo Pfoertner <all at abouthugo.de> [060817 22:30]:
> > SeqFans,
> > 
> > today Hauke Reddmann started a new thread in the NG sci.math
> > 
> > http://groups.google.com/group/sci.math/msg/b1c1164b936c6dca
> > 
> > <<
> > For easyness, I use the data of my lest chess tournament :-)
> > The finish in terms of the starting numbers was
> > 7 1 6 4 3 8 5 2|9|10|14 15 20 13 18 16 23 22 12 26 11 17 19 27 25 21
> > 24|28
> > 
> > A | marks boundaries between consecutive number subsets that permute
> > to themselves. Note that I (the 16) also permute to myself, but there
> > are number crossing from both sides and so this is no boundary.
> > 
> > Obvious question: How many boundaries occur in a random permutation?
> > Clearly a tournament is about the opposite of random, as the swap
> > numbers will be low.
> > 
> > Example n=3
> > 
> > 1|2|3
> > 1|3 2
> > 2 1|3
> > 2 3 1
> > 3 1 2
> > 3 2 1
> > -- 
> > Hauke Reddmann <:-EX8    fc3a501 at uni-hamburg.de
> > His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn
> > >>
> > 
> > Is there anything in the OEIS that can answer his question? Any other
> > ideas?
> > 
> > Can you please CC answers also to Hauke; AFAIK he is not member of the
> > Seqfan community.
> > 
> > Hugo Pfoertner