Minimizing the sum(permutation*inverse)

[franktaw at netscape.net]

These aren't finite sequences; just the first 9 terms of two infinite sequences.  Permutations are of numbers, not digits.
 
The second one is A000330.
 
Franklin T. Adams-Watters
 
 
-----Original Message-----
From: zak seidov zakseidov at yahoo.com


Here's table of min and max for n=1...9 (first entry)

{1,1,1}
{2,5,5}
{3,11,14}
{4,20,30}
{5,35,55}
{6,57,91}
{7,85,140}
{8,120,204}
{9,165,285}
which gives two full fini seqs:

1, 5, 11, 20, 35, 57, 85, 120, 165, 
1, 5, 14, 30, 55, 91, 140, 204, 285

Zak

--- Leroy Quet <qq-quet at mindspring.com> wrote:

> Regarding the sci.math thread:
> 
>
http://groups.google.com/group/sci.math/browse_thread/thread/6cf265b0d0e2f1
> 57
> 
> Let {b(k)} be a permutation of {1,2,3,...,n}. 
> Let {c(k)} be the inverse-permutation of {b(k)}. 
> (ie. b(c(j)) =j, for every j.) 
> 
> What is the minimum possible sum:
> sum{k=1 to n}  b(k) * c(k)  ? 
> 
> For example, if b is: 
> [1,2,3], 
> [2,1,3], 
> [3,2,1], 
> each give a sum of 14 (since each of these
> permutations is its own 
> inverse). 
> 
> But 
> [2,3,1], 
> [3,1,2], 
> (which are inverses of each other) 
> both give a sum of 11.
> 
> I get (possibly erroneously) that the sequences of
> minimum sums begins:
> 1, 5, 11, 20, 35,...
> 
> Could someone please calculate/submit this sequence
> (unless it is already 
> in the EIS, of course).
> 
> thanks,
> Leroy Quet
> 


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