

A231348


Number of triangles after nth stage in a cellular automaton based in isosceles triangles of two sizes (see Comments lines for precise definition).


5



0, 1, 3, 7, 11, 15, 23, 33, 41, 45, 53, 65, 81, 91, 111, 133, 149, 153, 161, 173, 189, 201, 225, 253, 285, 295, 315, 343, 383, 405, 449, 495, 527, 531, 539, 551, 567, 579, 603, 631, 663, 675, 699, 731, 779, 807, 863, 923, 987, 997, 1017, 1045, 1085, 1113, 1169, 1233, 1313, 1335, 1379, 1439, 1527, 1573, 1665, 1759, 1823
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,3


COMMENTS

On the semiinfinite square grid the structure of this C.A. contains "black" triangles and "gray" triangles (see the Links section). Both types of triangles have two sides of length 5^(1/2). Every black triangle has a base of length 2 hence its height is 2 and its area is 2. Every gray triangle has a base of length 2^(1/2) hence its height is 3/(2^(1/2)) and its area is 3/2. Both types of triangles are arranged in the same way as the triangles of Sierpinski gasket (see A047999 and A006046). The black triangles are arranged in vertical direction. On the other hand the gray triangles are arranged in diagonal direction in the holes of the structure formed by the black triangles. Note that the vertices of all triangles coincide with the grid points.
The sequence gives the total number of triangles (black and gray) in the structure after nth stage. A231349 (the first differences) gives the number of triangles added at nth stage.
For a more complex structure see A233780.


LINKS

Table of n, a(n) for n=0..64.
Omar E. Pol, Illustration of the structure after 16 stages
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS


EXAMPLE

We start at stage 0 with no triangles, so a(0) = 0.
At stage 1 we add a black triangle, so a(1) = 1.
At stage 2 we add two black triangles, so a(2) = 1+2 = 3.
At stage 3 we add two black triangles and two gray triangles from the vertices of the master triangle, so a(3) = 3+2+2 = 7.
At stage 4 we add four black triangles, so a(4) = 7+4 = 11.
At stage 5 we add two black triangles and two gray triangles from the vertices of the master triangle, so a(5) = 11+2+2 = 15.
At stage 6 we add four black triangles and four gray triangles, so a(6) = 15+4+4 = 23.
At stage 7 we add four black triangles and six gray triangles, so a(7) = 23+4+6 = 33.
At stage 8 we add eight black triangles, so a(8) = 33+8 = 41.
And so on.
Note that always we add both black triangles and gray triangles except if n is a power of 2. In this case at stage 2^k we add only 2^k black triangles, for k >= 0.


CROSSREFS

Cf. A047999, A006046, A139250, A151566, A160406, A173530, A182634, A194444, A220524, A220478, A230981, A231349, A233780.
Sequence in context: A279106 A172306 A112714 * A194444 A220524 A169626
Adjacent sequences: A231345 A231346 A231347 * A231349 A231350 A231351


KEYWORD

nonn


AUTHOR

Omar E. Pol, Dec 15 2013


STATUS

approved



