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.

Showing 1-8 of 8 results.

A182635 Number of toothpicks added at n-th stage to the toothpick structure of A182634.

Original entry on oeis.org

0, 1, 2, 4, 4, 4, 8, 12, 8, 4, 8, 16, 20, 16, 20, 28, 16, 4, 8, 16, 20, 20, 28, 44, 44, 24, 20, 40, 56, 48, 52, 64, 32, 4, 8, 16, 20, 20, 28, 44, 44, 28, 28, 52, 76, 76, 76
Offset: 0

Views

Author

Omar E. Pol, Dec 08 2010

Keywords

Comments

First differences of A182634.
First differs from A139251 at a(11).

Examples

			Contribution from _Omar E. Pol_, Dec 06 2012 (Start):
When written as an irregular triangle begins:
0;
1;
2;
4,4;
4,8,12,8;
4,8,16,20,16,20,28,16;
4,8,16,20,20,28,44,44,24,20,40,56,48,52,64,32;
4,8,16,20,20,28,44,44,28,28,52,76,76,76,...
(End)
		

Crossrefs

Formula

a(n) = A182633(n)/3.

A182840 Toothpick sequence on hexagonal net.

Original entry on oeis.org

0, 1, 5, 13, 27, 43, 57, 81, 119, 151, 165, 189, 235, 299, 353, 409, 495, 559, 573, 597, 643, 707, 769, 849, 975, 1119, 1205, 1261, 1371, 1539, 1697, 1841, 2039, 2167, 2181, 2205, 2251, 2315, 2377, 2457, 2583, 2727, 2821, 2901, 3043, 3267, 3505, 3729, 4015
Offset: 0

Views

Author

Omar E. Pol, Dec 09 2010

Keywords

Comments

Rules:
- Each new toothpick must lie on the hexagonal net such that the toothpick endpoints coincide with two consecutive nodes.
- Each exposed endpoint of the toothpicks of the old generation must be touched by the endpoints of two toothpicks of new generation.
The sequence gives the number of toothpicks after n stages. A182841 (the first differences) gives the number added at the n-th stage.
The toothpick structure has polygons in which there are uncovered grid points, the same as A160120 and A161206. For more information see A139250.
Has a behavior similar to A151723, A182632. - Omar E. Pol, Feb 28 2013
From Omar E. Pol, Feb 17 2023: (Start)
Assume that every triangular cell has area 1.
It appears that the structure contains only three types of polygons:
- Regular hexagons of area 6.
- Concave decagons (or concave 10-gons) of area 12.
- Concave dodecagons (or concave 12-gons) of area 18.
There are infinitely many of these polygons.
The structure contains concentric hexagonal rings formed by hexagons and also contains concentric hexagonal rings formed by alternating decagons and dodecagons.
For an animation see the movie version in the Links section.
The animation shows the fractal-like behavior the same as in other members of the family of toothpick cellular automata.
The structure has internal growth.
For another version starting from a node see A182632.
For a version of the structure in the first quadrant but on the square grid see A182838. (End)

Examples

			We start at stage 0 with no toothpicks.
At stage 1 we place a toothpick anywhere in the plane (For example, in vertical position). There are two exposed endpoints, so a(1)=1.
At stage 2 we place 4 toothpicks. Two new toothpicks touching each exposed endpoint. So a(2)=1+4=5. There are 4 exposed endpoints.
At stage 3 we place 8 toothpicks. a(3)=5+8=13. The structure has 8 exposed endpoints.
At stage 4 we place 14 toothpicks (Not 16) because there are 4 endpoints that are touched by new 8 toothpicks but there are 4 endpoints that are touched by only 6 new toothpicks (not 8), so a(4)=13+14=27.
After 4 stages the toothpick structure has 4 hexagons and 8 exposed endpoints.
		

Crossrefs

Extensions

More terms from Olaf Voß, Dec 24 2010
Wiki link added by Olaf Voß, Jan 14 2011

A182632 Toothpick sequence on the hexagonal net starting from a node.

Original entry on oeis.org

0, 3, 9, 21, 33, 45, 69, 105, 129, 141, 165, 213, 273, 321, 381, 465, 513, 525, 549, 597, 657, 717, 801, 933, 1065, 1137, 1197, 1317, 1485, 1629, 1785, 1977, 2073, 2085, 2109, 2157, 2217, 2277, 2361, 2493, 2625, 2709, 2793, 2949, 3177, 3405, 3633
Offset: 0

Views

Author

Omar E. Pol, Dec 07 2010

Keywords

Comments

A connected network of toothpicks is constructed by the following iterative procedure. At stage 1, place three toothpicks each of length 1 on a hexagonal net, as a propeller, joined at a node. At each subsequent stage, add two toothpicks (which could be called a single V-toothpick with a 120-degree corner) adjacent to each node which is the endpoint of a single toothpick.
The exposed endpoints of the toothpicks of the old generation are touched by the endpoints of the toothpicks of the new generation. In the graph, the edges of the hexagons become edges of the graph, and the graph grows such that the nodes that were 1-connected in the old generation are 3-connected in the new generation.
It turns out heuristically that this growth does not show frustration, i.e., a free edge is never claimed by two adjacent exposed endpoints at the same stage; the rule of growing the network does apparently not need specifications to address such cases.
The sequence gives the number of toothpicks in the toothpick structure after n-th stage. A182633 (the first differences) gives the number of toothpicks added at n-th stage.
a(n) is also the number of components after n-th stage in a toothpick structure starting with a single Y-toothpick in stage 1 and adding only V-toothpicks in stages >= 2. For example: consider that in A161644 a V-toothpick is also a polytoothpick with two components or toothpicks and a Y-toothpick is also a polytoothpick with three components or toothpicks. For more information about this comment see A161206, A160120 and A161644.
Has a behavior similar to A151723, A182840. - Omar E. Pol, Mar 07 2013
From Omar E. Pol, Feb 17 2023: (Start)
Assume that every triangular cell has area 1.
It appears that the structure contains only three types of polygons:
- Regular hexagons of area 6.
- Concave decagons (or concave 10-gons) of area 12.
- Concave dodecagons (or concave 12-gons) of area 18.
There are infinitely many of these polygons.
The structure contains concentric hexagonal rings formed by hexagons and also contains concentric hexagonal rings formed by alternating decagons and dodecagons.
The structure has internal growth.
For an animation see the movie version in the Links section.
The animation shows the fractal-like behavior the same as in other members of the family of toothpick cellular automata.
For another version starting with a simple toothpick see A182840.
For a version of the structure in the first quadrant but on the square grid see A182838. (End)

Examples

			a(0)=0. At stage 1 we place 3 toothpicks connected to the initial grid point of the structure. Note that there are 3 exposed endpoints. At stage 2 we place 6 toothpicks, so a(2)=3+6=9, etc.
		

Crossrefs

Formula

a(n) = 3*A182634(n).
a(n) = 1 + 2*A161644(n), n >= 1. - Omar E. Pol, Mar 07 2013

A182838 H-toothpick sequence in the first quadrant starting with a D-toothpick placed on the diagonal [(0,1), (1,2)] (see Comments for precise definition).

Original entry on oeis.org

0, 1, 3, 7, 11, 15, 21, 31, 39, 43, 49, 61, 77, 91, 105, 127, 143, 147, 153, 165, 181, 197, 217, 249, 285, 307, 321, 349, 391, 431, 467, 517, 549, 553, 559, 571, 587, 603, 623, 655, 691, 715
Offset: 0

Views

Author

Omar E. Pol, Dec 12 2010

Keywords

Comments

An H-toothpick sequence is a toothpick sequence on a square grid that resembles a partial honeycomb of hexagons.
The structure has two types of elements: the classic toothpicks with length 1 and the "D-toothpicks" with length sqrt(2).
Classic toothpicks are placed in the vertical direction and D-toothpicks are placed in a diagonal direction.
Each hexagon has area = 4.
The network looks like an elongated hexagonal lattice placed on the square grid so that all nodes of the hexagonal net coincide with some of the grid points of the square grid. Each node in the hexagonal network is represented with coordinates x,y.
The sequence gives the number of toothpicks and D-toothpicks after n steps. A182839 (first differences) gives the number added at the n-th stage.
[It appears that for this sequence a classic toothpick is a line segment of length 1 that is parallel to the y-axis. A D-toothpick is a line segment of length sqrt(2) with slope +-1. D stands for diagonal. It also appears that classic toothpicks are not placed on the y-axis. - N. J. A. Sloane, Feb 06 2023]
From Omar E. Pol, Feb 17 2023: (Start)
This cellular automaton appears to be a version on the square grid of the first quadrant of the structure of A182840.
The rules are as follows:
- The elements (toothpicks and D-toothpicks) are connected at their ends.
- At each free end of the elements of the old generation two elements of the new generation must be connected.
- The toothpicks of length 1 must always be placed vertically, i.e. parallel to the Y-axis.
- The angle between a toothpick of length 1 and a D-toothpick of length sqrt(2) that share the same node must be 135 degrees, therefore the angle between two D-toothpicks that share the same node is 90 degrees.
As a result of these rules we can see that in the odd-indexed rows of the structure are placed only the toothpicks of length 1 and in the even-indexed rows of the structure are placed the D-toothpicks of length sqrt(2).
Apart from the trapezoids, pentagons and heptagons that are adjacent to the axes of the first quadrant it appears that there are only three types of polygons:
- Regular hexagons of area 4.
- Concave decagons (or concave 10-gons) of area 8.
- Concave dodecagons (or concave 12-gons) of area 12.
There are infinitely many of these polygons.
The structure shows a fractal-like behavior as we can see in other members of the family of toothpick cellular automata.
The structure has internal growth as some members of the mentioned family. (End)

Examples

			We start at stage 0 with no toothpicks.
At stage 1 we place a D-toothpick [(0,1),(1,2)], so a(1)=1.
At stage 2 we place a toothpick [(1,2),(1,3)] and a D-toothpick [(1,2),(2,1)], so a(2)=1+2=3.
At stage 3 we place 4 elements: a D-toothpick [(1,3),(0,4)], a D-toothpick [(1,3),(2,4)], a D-toothpick [(2,1),(3,2)] and a toothpick [(2,1),(2,0)], so a(3)=3+4=7. Etc.
The first hexagon appears in the structure after 4 stages.
		

Crossrefs

See A360501 and A360512 for another hexagonal net built on the square grid. - N. J. A. Sloane, Feb 09 2023

Formula

Conjecture: a(n) = (A182840(n+1) + A267458(n+1) - 2)/4. - Omar E. Pol, Feb 10 2023

Extensions

Partially edited by N. J. A. Sloane, Feb 06 2023
a(19)-a(41) from Omar E. Pol, Feb 06 2023

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

Views

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

A233970 Toothpick sequence on hexagonal net starting from the vertex of a 60-degree wedge (see Comments lines for precise definition).

Original entry on oeis.org

0, 1, 3, 5, 9, 11, 15, 21, 29, 31, 35, 41, 51, 61, 69, 83, 99, 101, 105, 111, 121, 131, 141, 159, 183, 205, 213, 227, 249, 275, 291, 321, 353, 355, 359, 365, 375, 385, 395, 413, 437, 459, 469, 487, 515, 553, 581, 627, 683, 737, 745, 759, 781, 807
Offset: 0

Views

Author

Omar E. Pol, Dec 18 2013

Keywords

Comments

Toothpicks are connected by their endpoints. The toothpicks placed in north direction are prohibited. The sequence gives the number of toothpicks after n-th stage in the structure. A233971 (the first differences) give the number of toothpicks added at n-th stage.
First differs from A169780 at a(24).
First differs from both A233764 and A233780 at a(25).

Crossrefs

A182836 Toothpick sequence starting at the vertex of the outside corner of an infinite 120-degree wedge on hexagonal net.

Original entry on oeis.org

0, 1, 3, 7, 15, 27, 39, 51, 71, 91, 107
Offset: 0

Views

Author

Omar E. Pol, Dec 12 2010

Keywords

Comments

Corner sequence for the toothpick structure on hexagonal net.
The sequence gives the number of toothpicks after n stages. A182837 (the first differences) gives the number added at the n-th stage. For more information see A182632 and A153006.

Examples

			We start at stage 0 with no toothpicks.
At stage 1 we place a single toothpick touching a vertex of the infinite hexagon, in direction to the center of the hexagon, but on the outside corner, so a(1)=1.
At stage 2 we place 2 toothpicks touching the exposed endpoint of the initial toothpick, so a(2)=1+2=3.
At stage 3 we place 4 toothpicks, so a(3)=3+4=7.
At stage 4 we place 8 toothpicks, so a(4)=7+8=15.
At stage 5 we place 12 toothpicks, so a(5)=15+12=27.
After 5 stages the toothpick structure has 5 hexagons and 6 exposed endpoints.
		

Crossrefs

A342271 A161645(n)/3.

Original entry on oeis.org

1, 2, 2, 2, 4, 6, 4, 2, 4, 8, 10, 8, 10, 14, 8, 2, 4, 8, 10, 10, 14, 22, 22, 12, 10, 20, 28, 24, 26, 32, 16, 2, 4, 8, 10, 10, 14, 22, 22, 14, 14, 26, 38, 38, 38, 50, 46, 20, 10, 20, 28, 30, 38, 58, 66, 44, 30, 48, 70, 64, 64, 70, 32, 2, 4, 8
Offset: 2

Views

Author

N. J. A. Sloane, Mar 11 2021

Keywords

References

  • S. Ulam, On some mathematical problems connected with patterns of growth of figures, pp. 215-224 of R. E. Bellman, ed., Mathematical Problems in the Biological Sciences, Proc. Sympos. Applied Math., Vol. 14, Amer. Math. Soc., 1962. See Example 3.

Crossrefs

Showing 1-8 of 8 results.