cp's OEIS Frontend

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.

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A231348 Number of triangles after n-th stage in a cellular automaton based in isosceles triangles of two sizes (see Comments lines for precise definition).

Original entry on oeis.org

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
Offset: 0

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Author

Omar E. Pol, Dec 15 2013

Keywords

Comments

On the semi-infinite 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 n-th stage. A231349 (the first differences) gives the number of triangles added at n-th stage.
For a more complex structure see A233780.

Examples

			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

A194276 Number of distinct polygonal shapes after n-th stage in the D-toothpick structure of A194270.

Original entry on oeis.org

0, 0, 0, 0, 1, 3, 4, 5, 6, 7, 9, 10, 10, 11, 13, 13, 14
Offset: 0

Views

Author

Omar E. Pol, Aug 23 2011

Keywords

Comments

The cellular automaton of A194270 contains a large number of distinct polygonal shapes. For simplicity we also call polygonal shapes "polygons".
In order to construct this sequence we use the following rules:
- Consider only the convex polygons and the concave polygons. Self-intersecting polygons are not counted. (Note that some polygons contain in their body a toothpick or D-toothpick with an exposed endpoint; that element is not a part of the perimeter of the polygon.)
- If two polygons have the same shape but they have different size then these polygons must be counted as distinct polygonal shapes.
- The reflected shapes of asymmetric polygons, both with the same area, must be counted as distinct polygonal shapes.
For more information see A194277 and A194278.
Question: Is there a maximal record in this sequence?

Examples

			Consider toothpicks of length 2 and D-toothpicks of length sqrt(2).
.
Stage       New type  Perimeter    Area   Term       a(n)
. 0            -          -          -    a(0) =       0
. 1            -          -          -    a(1) =       0
. 2            -          -          -    a(2) =       0
. 3            -          -          -    a(3) =       0
. 4         hexagon   4*sqrt(2)+4    6    a(4) =       1
. 5   5.1   hexagon   2*sqrt(2)+8    8
.     5.2   octagon   4*sqrt(2)+8   14    a(5) = 1+2 = 3
. 6         pentagon  2*sqrt(2)+6    5    a(6) = 3+1 = 4
. 7         enneagon  6*sqrt(2)+6   13    a(7) = 4+1 = 5
		

Crossrefs

A212008 D-toothpick sequence of the second kind starting with a single toothpick.

Original entry on oeis.org

0, 1, 5, 13, 29, 51, 71, 95, 131, 171, 203, 247, 303, 397, 457, 513, 589, 661, 693, 741, 813, 925, 1057, 1197, 1333, 1501, 1613, 1745, 1885, 2123, 2271, 2391, 2547, 2683, 2715, 2763, 2835, 2947, 3079
Offset: 0

Views

Author

Omar E. Pol, Dec 15 2012

Keywords

Comments

This cellular automaton uses elements of two sizes: toothpicks of length 1 and D-toothpicks of length 2^(1/2). Toothpicks are placed in horizontal or vertical direction. D-toothpicks are placed in diagonal direction. Toothpicks and D-toothpicks are connected by their endpoints.
On the infinite square grid we start with no elements.
At stage 1, place a single toothpick on the paper, aligned with the y-axis.
The rule for adding new elements is as follows. If it is possible, each exposed endpoint of the elements of the old generation must be touched by the two endpoints of two elements of the new generation such that the angle between the old element and each new element is equal to 135 degrees, otherwise each exposed endpoint of the elements of the old generation must be touched by an endpoint of an element of the new generation such that the angle between the old element and the new element is equal to 135 degrees. Intersections and overlapping are prohibited. The sequence gives the number of toothpicks and D-toothpicks in the structure after n-th stage. The first differences (A212009) give the number of toothpicks or D-toothpicks added at n-th stage.
It appears that if n >> 1 the structure looks like an octagon. This C.A. has a fractal (or fractal-like) behavior related to powers of 2. Note that for some values of n we can see an internal growth.
The structure contains eight wedges. Each vertical wedge also contains infinitely many copies of the oblique wedges. Each oblique wedge also contains infinitely many copies of the vertical wedges. Finally, each horizontal wedge also contains infinitely many copies of the vertical wedges and of the oblique wedges.
The structure appears to be a puzzle which contains at least 50 distinct internal regions (or polygonal pieces), and possibly more. Some of them appear for first time after 200 stages. The largest known polygon is a concave 24-gon.
Also the structure contains infinitely many copies of two subsets of distinct size which are formed by five polygons: three hexagons, a 9-gon and a pentagon. The distribution of these subsets have a surprising connection with the Sierpinski triangle A047999, but here the pattern is more complex.
For another version see A220500.

Crossrefs

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