A141727 - OEIS

%I #19 Nov 25 2015 22:58:41

%S 1,1,0,1,1,0,0,1,0,1,0,1,0,1,0,0,1,0,0,1,1,0,1,1,1,1,0,1,0,0,0,0,0,1,

%T 1,0,1,0,0,1,0,0,0,0,1,1,1,0,0,1,0,1,0,1,0,0,0,1,1,0,0,1,1,1,1,0,0,1,

%U 1,0,1,1,0,0,0,1,0,0,1,1,0,1,0,1,0,0,0,0,0,0,1,1,0,1,0,1,1,1,0,0,1,0,0,1,0

%N Triangle T(n,k) read by rows. Entries are 0 and 1. Start with 1 in the top row, add 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 even or add 1 if it is odd.

%C Any diagonal, read top down from right to left, is a periodic sequence of 0's and 1's. The lengths of the periods are always powers of 2. Here are the periods for the first 20 diagonals:

%C 1

%C 0

%C 10

%C 10

%C 0110

%C 0

%C 0100

%C 1000

%C 11110000

%C 1110

%C 01001110

%C 00101000

%C 01011100

%C 1000

%C 11100000

%C 11001110

%C 0111000110001110

%C 01101000

%C 0011011010011100

%C 0010001010001000

%C If we draw a large number of rows we obtain an interesting figure with several large islands of zeros.

%H Paolo P. Lava, <a href="/A141727/a141727.pdf">Picture of Triangle A141727</a>

%e .....................................1 First Row

%e ...................................1 ... Add 1 to have an even number of adjacent 1's (2)

%e .....................................1 First Row

%e ...................................1.0 ... Add 0 because there are two adjacent 1's (in the first and second rows)

%e ......................................1 First Row

%e ....................................1.0.1 ... Again add 1 to have an even number of adjacent 1's (2)

%e The second row is now complete.

%e .....................................1 First Row

%e ...................................1.0.1 Second Row

%e .................................1 ... Add 1 because there is only an 1 adjacent (second row)

%e .....................................1 First Row

%e ...................................1.0.1 Second Row

%e .................................1.0 ... Add 0 because there are two 1's adjacent (second and third row)

%e .....................................1 First Row

%e ...................................1.0.1 Second Row

%e .................................1.0.0 ... Again add 0 because there are two 1's adjacent (second row)

%e .....................................1 First Row

%e ...................................1.0.1 Second Row

%e .................................1.0.0.1 ... Add 1 because there is only an 1 adjacent (second row)

%e .....................................1 First Row

%e ...................................1.0.1 Second Row

%e .................................1.0.0.1.0 ... Add 0 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.

%e The triangle begins:

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

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

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

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

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

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

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

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

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

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

%Y Cf. A141728-A141746.

%K easy,nonn,tabf

%O 0,1

%A _Paolo P. Lava_ & _Giorgio Balzarotti_, Jul 02 2008

%E Minor edits by _N. J. A. Sloane_, Sep 10 2012