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.

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, 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, 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

Offset: 0

Author

Paolo P. Lava & Giorgio Balzarotti, Jul 02 2008

Keywords

• easy
• nonn
• tabf

Comments

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:

1

0

10

10

0110

0

0100

1000

11110000

1110

01001110

00101000

01011100

1000

11100000

11001110

0111000110001110

01101000

0011011010011100

0010001010001000

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

Examples

			.....................................1 First Row
...................................1 ... Add 1 to have an even number of adjacent 1's (2)
.....................................1 First Row
...................................1.0 ... Add 0 because there are two adjacent 1's (in the first and second rows)
......................................1 First Row
....................................1.0.1 ... Again add 1 to have an even number of adjacent 1's (2)
The second row is now complete.
.....................................1 First Row
...................................1.0.1 Second Row
.................................1 ... Add 1 because there is only an 1 adjacent (second row)
.....................................1 First Row
...................................1.0.1 Second Row
.................................1.0 ... Add 0 because there are two 1's adjacent (second and third row)
.....................................1 First Row
...................................1.0.1 Second Row
.................................1.0.0 ... Again add 0 because there are two 1's adjacent (second row)
.....................................1 First Row
...................................1.0.1 Second Row
.................................1.0.0.1 ... Add 1 because there is only an 1 adjacent (second row)
.....................................1 First Row
...................................1.0.1 Second Row
.................................1.0.0.1.0 ... Add 0 because there are two 1's adjacent (second and third row)
The third row is now complete. Then repeat the process for the other rows.
The triangle begins:
...........................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..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..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
		

Links

• Paolo P. Lava, Picture of Triangle A141727

Crossrefs

Cf. A141728-A141746.

Extensions

Minor edits by N. J. A. Sloane, Sep 10 2012