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%I #12 Jan 01 2013 02:29:52
%S 0,1,0,1,0,0,1,1,1,1,0,0,1,1,0,0,0,0,0,0,1,0,0,1,0,1,0,1,0,0,0,0,0,0,
%T 0,1,0,0,1,1,1,0,1,0,1,0,0,0,0,1,0,0,1,1,0,1,1,0,0,0,1,0,1,0,0,0,0,0,
%U 1,0,1,0,1,1,0,0,0,1,1,1,0,1,0,1,0,0,0,1,1,0,1,0,0,1,1,0,0,1,1,0,0,0,1,1,1
%N Triangle read by rows T(n,k). Triangle elements are 0 and 1. Starting with 0 in the top add below a second row of (2n-1) elements (with n=2 -> 3). Moving from left to right add 0 if the number of adjacent 1's is odd or add 1 if it is even or null. See example below.
%C Any diagonal, read top down from right to left, expresses a periodic sequence of 0's and 1's. Lengths of the periods are always powers of 2. Here below the periods for the first 20 diagonals:
%C 01
%C 0
%C 1100
%C 1100
%C 1110
%C 0
%C 00011110
%C 10010110
%C 00101000
%C 00011110
%C 10000100
%C 01001000
%C 00011010
%C 1110
%C 0101011110101000
%C 1011000101001110
%C 0111111011010100
%C 1000100101110110
%C 0111011101000100
%C 1011010110000110
%H Paolo P. Lava, <a href="/A141743/a141743.pdf">Picture of Triangle A141743</a>
%e ...............................0 First Row
%e ............................1.... Add 1 to have an odd number of adjacent 1's
%e ..............................0 First Row
%e ............................1.0.. Add 0 because there is an odd number of adjacent 1's (second row)
%e ..............................0 First Row
%e ............................1.0.1 Again add 1 because there is a null number of adjacent 1's.
%e The second row is now complete.
%e ..............................0 First Row
%e ............................1.0.1 Second Row
%e ..........................0... Add 0 because there is only an 1 adjacent (second row)
%e ..............................1 First Row
%e ............................1.0.1 Second Row
%e ..........................0.0.... Add 0 because there is only an 1 adjacent (second row)
%e .............................0 First Row
%e ...........................1.0.1 Second Row
%e .........................0.0.1 Add 1 because there are two 1's adjacent (second row)
%e .............................0 First Row
%e ...........................1.0.1 Second Row
%e .........................0.0.1.1 Add 1 because there are two 1's adjacent (second and third row)
%e .............................0 First Row
%e ...........................1.0.1 Second Row
%e .........................0.0.1.1.1 Add 1 because there are two 1's adjacent (second and third row)
%e The third row is now complete. Then repeat the process for the other rows.
%K easy,nonn,tabf
%O 0,1
%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Jul 07 2008
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