This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A141743 #12 Jan 01 2013 02:29:52 %S A141743 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 A141743 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 A141743 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 A141743 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 A141743 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 A141743 01 %C A141743 0 %C A141743 1100 %C A141743 1100 %C A141743 1110 %C A141743 0 %C A141743 00011110 %C A141743 10010110 %C A141743 00101000 %C A141743 00011110 %C A141743 10000100 %C A141743 01001000 %C A141743 00011010 %C A141743 1110 %C A141743 0101011110101000 %C A141743 1011000101001110 %C A141743 0111111011010100 %C A141743 1000100101110110 %C A141743 0111011101000100 %C A141743 1011010110000110 %H A141743 Paolo P. Lava, <a href="/A141743/a141743.pdf">Picture of Triangle A141743</a> %e A141743 ...............................0 First Row %e A141743 ............................1.... Add 1 to have an odd number of adjacent 1's %e A141743 ..............................0 First Row %e A141743 ............................1.0.. Add 0 because there is an odd number of adjacent 1's (second row) %e A141743 ..............................0 First Row %e A141743 ............................1.0.1 Again add 1 because there is a null number of adjacent 1's. %e A141743 The second row is now complete. %e A141743 ..............................0 First Row %e A141743 ............................1.0.1 Second Row %e A141743 ..........................0... Add 0 because there is only an 1 adjacent (second row) %e A141743 ..............................1 First Row %e A141743 ............................1.0.1 Second Row %e A141743 ..........................0.0.... Add 0 because there is only an 1 adjacent (second row) %e A141743 .............................0 First Row %e A141743 ...........................1.0.1 Second Row %e A141743 .........................0.0.1 Add 1 because there are two 1's adjacent (second row) %e A141743 .............................0 First Row %e A141743 ...........................1.0.1 Second Row %e A141743 .........................0.0.1.1 Add 1 because there are two 1's adjacent (second and third row) %e A141743 .............................0 First Row %e A141743 ...........................1.0.1 Second Row %e A141743 .........................0.0.1.1.1 Add 1 because there are two 1's adjacent (second and third row) %e A141743 The third row is now complete. Then repeat the process for the other rows. %K A141743 easy,nonn,tabf %O A141743 0,1 %A A141743 _Paolo P. Lava_ and _Giorgio Balzarotti_, Jul 07 2008