cp's OEIS Frontend

A141728 Triangle read by rows T(n,k). Triangle elements are 0 and 1. Starting with 1 in the top add below a second row of (2n-1) elements (with n=2 -> 3). Moving from left to right add 1 if the number of adjacent 1's is even or add 0 if it is odd. See example below.

%I A141728 #8 Aug 24 2012 10:50:01
%S A141728 1,0,0,0,1,0,1,0,1,0,0,1,1,0,0,0,1,0,0,1,0,0,1,0,1,0,0,0,0,0,0,0,0,1,
%T A141728 1,1,1,0,1,0,1,0,1,0,0,1,1,0,0,0,0,1,1,0,0,1,1,1,1,0,1,1,0,1,1,0,0,1,
%U A141728 0,0,0,1,1,1,0,1,0,1,0,0,0,0,0,0,0,0,0,1,1,0,0,1,0,0,1,1,1,0,1,0,1,0,1,0,1
%N A141728 Triangle read by rows T(n,k). Triangle elements are 0 and 1. Starting with 1 in the top add below a second row of (2n-1) elements (with n=2 -> 3). Moving from left to right add 1 if the number of adjacent 1's is even or add 0 if it is odd. See example below.
%C A141728 Any diagonal, read top down from right to left, expresses a periodic sequence of 0'0's and 1's Lengths of the periods are alway powers of 2. Here below the periods for the first 20 diagonals:
%C A141728 10
%C A141728 0
%C A141728 0110
%C A141728 0110
%C A141728 1000
%C A141728 0
%C A141728 01011010
%C A141728 00011110
%C A141728 11011000
%C A141728 11110000
%C A141728 11001010
%C A141728 01100000
%C A141728 01000110
%C A141728 0110
%C A141728 1011011101001000
%C A141728 0111111110000000
%C A141728 0000111101011010
%C A141728 1110000100011110
%C A141728 0100000111011000
%C A141728 1001011100001110
%H A141728 Paolo P. Lava, <a href="/A141728/a141728.pdf">Picture of Triangle A141728</a>
%e A141728 .....................................1 First Row
%e A141728 ..................................0 ... Add 0 to have an odd number of adjacent 1's
%e A141728 .....................................1 First Row
%e A141728 ...................................0.0 ... Add again 0 to have an odd number of adjacent 1's
%e A141728 ......................................1 First Row
%e A141728 ...................................0.0.0 ... Again add 0 to have an odd number of adjacent 1's
%e A141728 The second row is now complete.
%e A141728 .....................................1 First Row
%e A141728 ...................................0.0.0 Second Row
%e A141728 .................................1 ... Add 1 because there are no adjacent 1's
%e A141728 .....................................1 First Row
%e A141728 ...................................0.0.0 Second Row
%e A141728 .................................1.0 ... Add 0 because there is one adjacent 1 (third row)
%e A141728 .....................................1 First Row
%e A141728 ...................................0.0.0 Second Row
%e A141728 .................................1.0.1 ... Add 1 because there is no adjacent 1
%e A141728 .....................................1 First Row
%e A141728 ...................................0.0.0 Second Row
%e A141728 .................................1.0.1.0 ... Add 0 because there is only an 1 adjacent (third row)
%e A141728 .....................................1 First Row
%e A141728 ...................................0.0.0 Second Row
%e A141728 .................................1.0.1.0.1 ... Add 1 because there is no adjacent 1
%e A141728 The third row is now complete. Then repeat the process for the other rows.
%Y A141728 Cf. A141727, A141729-A141746.
%K A141728 easy,nonn,tabf
%O A141728 0,1
%A A141728 _Paolo P. Lava_ and _Giorgio Balzarotti_, Jul 02 2008