login
T(n,k)=Half the number of nXk binary arrays with the number of 1-1 adjacencies equal to the number of 0-0 adjacencies
7

%I #5 Mar 31 2012 12:35:50

%S 1,1,1,1,3,1,2,6,6,2,3,18,25,18,3,6,60,164,164,60,6,10,210,1092,2101,

%T 1092,210,10,20,756,7954,29100,29100,7954,756,20,35,2772,57519,417190,

%U 807261,417190,57519,2772,35,70,10296,429936,6104168,23226996,23226996

%N T(n,k)=Half the number of nXk binary arrays with the number of 1-1 adjacencies equal to the number of 0-0 adjacencies

%C Table starts

%C ..1.....1........1...........2..............3.................6

%C ..1.....3........6..........18.............60...............210

%C ..1.....6.......25.........164...........1092..............7954

%C ..2....18......164........2101..........29100............417190

%C ..3....60.....1092.......29100.........807261..........23226996

%C ..6...210.....7954......417190.......23226996........1333831052

%C .10...756....57519.....6104168......679637064.......78105333988

%C .20..2772...429936....90554742....20179029954.....4635786110412

%C .35.10296..3205504..1356933444...604725622404...277890240523800

%C .70.38610.24288784.20489622108.18266061977928.16785123061670988

%H R. H. Hardin, <a href="/A183253/b183253.txt">Table of n, a(n) for n = 1..3000</a>

%e Some solutions with a(1,1)=0 for 4X6

%e ..0..0..0..0..0..1....0..0..0..0..1..1....0..0..0..0..1..0....0..0..0..1..0..1

%e ..1..1..1..0..1..0....1..1..0..1..1..1....0..0..0..1..1..1....0..0..1..1..1..1

%e ..0..1..1..1..1..1....1..1..0..0..0..0....1..0..1..0..0..1....0..1..1..0..1..1

%e ..0..0..0..0..0..1....1..1..0..0..1..0....1..1..1..1..1..0....0..1..0..0..0..0

%Y Column 1 is A001405(n-2)

%Y Column 2 is 3*A000984(n-2)

%K nonn,tabl

%O 1,5

%A _R. H. Hardin_ Jan 03 2011