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Define the following statistic as follows:

\(D(a_1,a_2,\ldots, a_n)\):

 

Start at \(a_n\). Go back from \(a_n\) until you get to the nearest element of the list smaller than it. Suppose that the first element smaller than it is \(a_k\). We break the list there (right after \(a_k\)).

 

Then, repeat from \(a_k\)... go back until you get nearest element in the list smaller than \(a_k\).

 

By the end, you will have split the list a certain number of times, x, creating x+1 sub-lists.

 

Now, consider the expression:

 

G(n) = \(\frac{D(1,2,3, \ldots , (n-1) ,n) + D(1,2,3, \ldots , n ,( n-1) , \ldots + D(n , (n-1) , \ldots , 2 , 1)}{n!}\) where the numerator is \(D\) evaluated at all \(n!\)  permutations of \(1,2,3,4,\ldots,n\).

 

For example, for n=3, we have:

\(G(3) = \frac{D(1,2,3)+D(1,3,2)+D(2,1,3)+D(2,3,1)+D(3,1,2)+D(3,2,1)}{3!}\), which can be shown to equal \(\frac{11}{6}.\)

 

Prove that \(G(n)\) is the \(n\)th harmonic number!!!!!!!!!!!!!!!!!

 

Thank you so much in advance.

 Mar 14, 2022
 #1
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You asked me to clarify my last answer.

I did that at length.

I am still waiting on a response from you.

 Mar 14, 2022
 #2
avatar+72 
+1

Have responded. As I said, I will work on characterizing all numbers that serve as counter-examples.

 

I'm curious to know an approach to this post too.

 

jsaddern

jsaddern  Mar 14, 2022
 #3
avatar+117830 
0

Thanks, :)

Unfortunately I do not think this question will get my attention.

It looks very time consuming and it doesn't look like my kind of question. 

I hope someone else can help you here. :)

Melody  Mar 14, 2022

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