'Mixing number' of permutations

[Henry Gould]

My memory made a small made a mistake. Here is a correct statement which 
I just picked up on the web at

<http://f2.org/maths/resnet/>

"When I read the rec.puzzles FAQ there was [November 2004: & still is] 
no general solution to the resistance between opposite corners of an 
arbitrary hypercube of 1-ohm resistors; solutions were only given for 3D 
and 4D hypercubes. This problem seemed quite tractable - just 
combinatorics. The same reasoning about symmetry for the cube case can 
be applied to the arbitrary d-dimensional case: joining the 
equipotential points results in d layers containing d(d-1 choose i) 
resistors in parallel in the i'th layer, giving an overall resistance of 
(sum from i=0 to d-1 of 1/(d-1 choose i)) / d. The values for the first 
few d are:

d 	2 	3 	4 	5 	6 	7 	8 	9 	10 	11
resistance 	1 	5/6 	2/3 	8/15 	13/30 	151/420 	32/105 	83/315 	73/315 
1433/6930

As d gets larger, the resistance tends toward 2/d [more accurately, 2/d 
+ 2/(d-1)^2] ; the dense layers in the middle become relatively perfect 
conductors compared to the outermost layers."

So what I wanted to recall is (1/n)*sum(1/binomial(n-1,k),k=0..n-1). Se 
were asked in a calculus class at University of Virginia in 1948 yo find 
the resistance of a rhree dimensional cube and we extended it to n by 
this formula, which you can get by using Kirchhoff's laws. The fractions 
abovve may be adjusted or normalized and entered in the OEIS if not 
already there.

Regards,

Henry Gould


Brendan McKay wrote:
> 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
>>     
>
>