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%I #4 Jan 02 2013 11:43:17
%S 1,1,2,3,4,3,4,4,6,4,4,8,7,7,5,6,10,11,10,10,6,7,12,14,14,13,12,7,7,
%T 12,17,18,19,16,13,8,9,16,19,23,24,24,19,16,9,10,18,24,28,28,30,28,22,
%U 18,10,10,20,27,32,32,34,35,31,25,19,11,12,20,30,32,40,40,42,40,34,28,22,12,13
%N 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 antidiagonal neighbors in a random 0..1 nXk array
%C Table starts
%C ..1..1..3..4..4..6..7..7...9..10..10..12.13.13..15.16.16.18.19.19.21
%C ..2..4..4..8.10.12.12.16..18..20..20..24.26.28..28.32.34.36.36.40
%C ..3..6..7.11.14.17.19.24..27..30..30..36.39.42..43.47.50.53.55
%C ..4..7.10.14.18.23.28.32..32..40..44..47.50.54..58.63.68.72
%C ..5.10.13.19.24.28.32.40..45..50..53..60.65.70..72.78.84
%C ..6.12.16.24.30.34.40.48..52..58..66..72.76.84..90.96
%C ..7.13.19.28.35.42.47.56..61..68..76..82.89.98.103
%C ..8.16.22.31.40.48.54.62..72..80..88..96.98
%C ..9.18.25.34.44.54.61.70..77..88..97.108
%C .10.19.28.39.48.58.68.80..90.100.108
%C .11.22.31.44.54.66.75.87..97.108
%C .12.24.34.48.60.71.82.96.106
%H R. H. Hardin, <a href="/A221108/b221108.txt">Table of n, a(n) for n = 1..217</a>
%F Empirical for column k:
%F k=1: a(n) = 2*a(n-1) -a(n-2) increment period 1: 1
%F k=2: a(n) = a(n-1) +a(n-3) -a(n-4) increment period 3: 3 2 1
%F k=3: a(n) = 2*a(n-1) -a(n-2) for n>3 increment period of length 1: 3
%F k=4: a(n) = a(n-1) +a(n-5) -a(n-6) increment period 5: 4 3 3 5 5
%F k=5: a(n) = 2*a(n-1) -2*a(n-2) +2*a(n-3) -2*a(n-4) +2*a(n-5) -a(n-6) for n>7 increment period of length 6: 4 4 6 6 5 5
%F k=6: a(n) = a(n-1) +a(n-9) -a(n-10) for n>11 increment period of length 9: 5 6 5 6 8 6 6 4 8
%F k=7: a(n) = 2*a(n-1) -a(n-2) for n>7 increment period of length 1: 7
%F Empirical for row n:
%F n=1: a(n) = a(n-1) +a(n-3) -a(n-4) increment period 3: 0 2 1
%F n=2: a(n) = a(n-1) +a(n-4) -a(n-5) increment period 4: 2 0 4 2
%F n=3: a(n) = a(n-1) +a(n-12) -a(n-13) increment period 12: 3 1 4 3 3 2 5 3 3 0 6 3
%F n=4: a(n) = a(n-1) +a(n-10) -a(n-11) increment period 10: 3 3 4 4 5 5 4 0 8 4
%F n=5: a(n) = a(n-1) +a(n-24) -a(n-25) increment period 24: 5 3 6 5 4 4 8 5 5 3 7 5 5 2 6 6 5 4 7 5 5 0 10 5
%F n=6: a(n) = 2*a(n-1) -a(n-2) -a(n-3) +2*a(n-4) -a(n-5) -a(n-6) +2*a(n-7) -a(n-8) -a(n-9) +2*a(n-10) -a(n-11) -a(n-12) +2*a(n-13) -a(n-14) -a(n-15) +2*a(n-16) -a(n-17) increment period 18: 6 4 8 6 4 6 8 4 6 8 6 4 8 6 6 0 12 6
%e Some solutions for n=3 k=4
%e ..1..1..1..0....1..1..1..1....0..1..0..0....0..0..0..1....0..0..0..0
%e ..0..1..0..0....1..1..0..1....0..0..1..1....1..1..1..1....0..0..0..0
%e ..0..0..0..1....0..0..0..1....0..0..1..0....1..0..1..1....1..0..1..0
%K nonn,tabl
%O 1,3
%A _R. H. Hardin_ Jan 02 2013
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