cp's OEIS Frontend

A141727 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.

%I A141727 #19 Nov 25 2015 22:58:41
%S A141727 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 A141727 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 A141727 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 A141727 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 A141727 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 A141727 1
%C A141727 0
%C A141727 10
%C A141727 10
%C A141727 0110
%C A141727 0
%C A141727 0100
%C A141727 1000
%C A141727 11110000
%C A141727 1110
%C A141727 01001110
%C A141727 00101000
%C A141727 01011100
%C A141727 1000
%C A141727 11100000
%C A141727 11001110
%C A141727 0111000110001110
%C A141727 01101000
%C A141727 0011011010011100
%C A141727 0010001010001000
%C A141727 If we draw a large number of rows we obtain an interesting figure with several large islands of zeros.
%H A141727 Paolo P. Lava, <a href="/A141727/a141727.pdf">Picture of Triangle A141727</a>
%e A141727 .....................................1 First Row
%e A141727 ...................................1 ... Add 1 to have an even number of adjacent 1's (2)
%e A141727 .....................................1 First Row
%e A141727 ...................................1.0 ... Add 0 because there are two adjacent 1's (in the first and second rows)
%e A141727 ......................................1 First Row
%e A141727 ....................................1.0.1 ... Again add 1 to have an even number of adjacent 1's (2)
%e A141727 The second row is now complete.
%e A141727 .....................................1 First Row
%e A141727 ...................................1.0.1 Second Row
%e A141727 .................................1 ... Add 1 because there is only an 1 adjacent (second row)
%e A141727 .....................................1 First Row
%e A141727 ...................................1.0.1 Second Row
%e A141727 .................................1.0 ... Add 0 because there are two 1's adjacent (second and third row)
%e A141727 .....................................1 First Row
%e A141727 ...................................1.0.1 Second Row
%e A141727 .................................1.0.0 ... Again add 0 because there are two 1's adjacent (second row)
%e A141727 .....................................1 First Row
%e A141727 ...................................1.0.1 Second Row
%e A141727 .................................1.0.0.1 ... Add 1 because there is only an 1 adjacent (second row)
%e A141727 .....................................1 First Row
%e A141727 ...................................1.0.1 Second Row
%e A141727 .................................1.0.0.1.0 ... Add 0 because there are two 1's adjacent (second and third row)
%e A141727 The third row is now complete. Then repeat the process for the other rows.
%e A141727 The triangle begins:
%e A141727 ...........................1
%e A141727 ........................1..0..1
%e A141727 .....................1..0..0..1..0
%e A141727 ..................1..0..1..0..1..0..0
%e A141727 ...............1..0..0..1..1..0..1..1..1
%e A141727 ............1..0..1..0..0..0..0..0..1..1..0
%e A141727 .........1..0..0..1..0..0..0..0..1..1..1..0..0
%e A141727 ......1..0..1..0..1..0..0..0..1..1..0..0..1..1..1
%e A141727 ...1..0..0..1..1..0..1..1..0..0..0..1..0..0..1..1..0
%e A141727 1..0..1..0..0..0..0..0..0..1..1..0..1..0..1..1..1..0..0
%Y A141727 Cf. A141728-A141746.
%K A141727 easy,nonn,tabf
%O A141727 0,1
%A A141727 _Paolo P. Lava_ & _Giorgio Balzarotti_, Jul 02 2008
%E A141727 Minor edits by _N. J. A. Sloane_, Sep 10 2012