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%I #28 Dec 17 2023 11:21:46
%S 0,5,23,61,119,213,335,509,719,997,1319,1725,2183,2741,3359,4093,4895
%N Fill an n X n array with various permutations of the integers 1, 2, 3, 4... n^2. Define the organization number of the n X n array to be the following: Start at 1, count the rectilinear steps to reach 2, then the rectilinear steps to reach 3, etc. Add them up. The array that has the maximum organization number would be the "most disorganized." This sequence is the sequence showing the most disorganized number for n X n arrays starting at 1 X 1.
%C Similar to sequence A047838.
%C My computer program worked as follows:
%C a) generate a permutation
%C b) place the permutation into the array
%C c) calculate the array position (row, column) of each integer
%C d) sort the integers into another array preserving row and column
%C e) travel the new array from 1..n^2 and summing the absolute value of the differences between the rows of consecutive integers and summing the absolute value of the differences of the columns of consecutive integers. The organization number is the sum of the two sums.
%C For instance, with the permutation 8, 3, 6, 5, 9, 1, 2, 7, 4 place the integers into a 3 X 3 array as such:
%C 8 3 6
%C 5 9 1
%C 2 7 4
%C (Notice the next integer is a knight's move away. This is not the only sequence that will give an organization number of 23, but this is why I wonder if the sequence is the same as A098499.)
%C Then sort the integers preserving their row and column:
%C number, row, column
%C 1, 2, 3
%C 2, 3, 1
%C 3, 1, 2
%C 4, 3, 3
%C 5, 2, 1
%C 6, 1, 3
%C 7, 3, 2
%C 8, 1, 1
%C 9, 2, 2
%C Traveling from 1 to 9, the differences in the row numbers are 1, 2, 2, 1, 1, 2, 2, 1 (a sum of 12) and the differences in the column numbers are 2, 1, 1, 2, 2, 1, 1, 1 (a sum of 11) therefore the organization number is 23.
%C This is basically a traveling salesman variant. - _D. S. McNeil_, Aug 26 2010
%H Sela Fried, <a href="https://arxiv.org/abs/2208.03788">On a conjecture of McNeil</a>, arXiv:2208.03788 [math.CO], 2023.
%F A possible formula: a(n) = 0 for n=1, n^3-n-1 for odd n > 1, n^3-3 for even n? - _D. S. McNeil_, Aug 26 2010
%F Let b(n) correspond to McNeil's formula. Then b(n) <= a(n) <= b(n) + 1 (see link). - _Sela Fried_, Nov 28 2023
%F Empirical G.f.: x^2*(5+13*x+10*x^2-6*x^3+x^4+x^5)/((1-x)^4*(1+x)^2). - _Colin Barker_, Mar 29 2012
%Y Cf. A047838, A098499.
%K nonn,more
%O 1,2
%A _Thomas Young_, Jun 29 2010
%E a(3) corrected and a(4)-a(17) computed by _D. S. McNeil_, Aug 26 2010. _D. S. McNeil_ also finds that a(19)=6839, a(21)=9239, a(23)=12143.
%E Edited by _N. J. A. Sloane_, Aug 26 2010
%E Typo in formula corrected by _D. S. McNeil_, Aug 26 2010
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