[seqfan] Equals sum of neighbors mod 2

Ron Hardin rhhardin at att.net
Sun Nov 11 14:12:54 CET 2012


Take a random nXk binary array
Mark the elements equal to the sum of their horizontal and vertical neighbors 
modulo 2.
Form an nXk binary array of these marks.
How many of the latter arrays are there?

It turns out empirically to always be a power of 2 with (so far) periodic 
increments.  Theory?

T(n,k)=Sum of neighbor maps: log base 2 of the number of nXk binary arrays 
indicating the locations of corresponding elements equal to the sum mod 2 of 
their horizontal and vertical neighbors in a random 0..1 nXk array

Table starts
..1..1..3..4..4..6..7..7..9.10.10.12.13.13.15.16
..1..4..4..8..9.12.12.16.17.20.20.24.25.28.28...
..3..4..9.12.12.18.21.22.27.30.30.36.39.40......
..4..8.12.12.20.24.28.32.32.40.44.48.52.........
..4..9.12.20.23.30.31.39.44.50.51.60............
..6.12.18.24.30.36.42.42.54.60.66...............
..7.12.21.28.31.42.49.54.63.70..................
..7.16.22.32.39.42.54.64........................
..9.17.27.32.44.54.63...........................
.10.20.30.40.50.60..............................
.10.20.30.44.51.................................
.12.24.36.48....................................

column 1
Empirical: a(n)=a(n-1)+a(n-3)-a(n-4)
Apparent periodic increment sequence of length 3: 0 2 1

column 2
Empirical: a(n)=a(n-1)+a(n-4)-a(n-5)
Apparent periodic increment sequence of length 4: 3 0 4 1

column 3
Empirical: a(n)=a(n-1)+a(n-6)-a(n-7)
Apparent periodic increment sequence of length 6: 1 5 3 0 6 3

column 4
Empirical: a(n)=a(n-1)+a(n-5)-a(n-6)
Apparent periodic increment sequence of length 5: 4 4 0 8 4

column 5
Empirical: a(n)=a(n-3)+a(n-8)-a(n-11)
Apparent periodic increment sequence of length 24: 5 3 8 3 7 1 8 5 6 1 9 4 5 2 9 
3 7 2 7 5 6 0 10 4

column 6
Empirical: a(n)=a(n-1)+a(n-9)-a(n-10)
Apparent periodic increment sequence of length 9: 6 6 6 6 6 6 0 12 6

column 7
Empirical: a(n)=a(n-1)+a(n-12)-a(n-13)
Apparent periodic increment sequence of length 12: 5 9 7 3 11 7 5 9 7 0 14 7


 rhhardin at mindspring.com
rhhardin at att.net (either)



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