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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A182617 Number of toothpicks in a toothpick spiral around n cells on hexagonal net.

Original entry on oeis.org

0, 5, 9, 12, 15, 18, 21, 23, 26, 29, 31, 34, 36, 39, 41, 44, 46, 49, 51, 53, 56, 58, 61, 63, 65, 68, 70, 72, 75, 77, 79, 82, 84, 86, 89, 91, 93, 95, 98, 100
Offset: 0

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Author

Omar E. Pol, Dec 13 2010

Keywords

Comments

The toothpick spiral contains n hexagonal "ON" cells that are connected without holes. A hexagonal cell is "ON" if the hexagon has 6 vertices that are covered by the toothpicks.
Attempt of an explanation: in the hexagonal grid, we can pick any of the hexagons as a center, and then define a ring of 6 first neighbors (hexagons adjacent to the center), then define a ring of 12 second neighbors (hexagons adjacent to any of the first ring) and so on. The current sequence describes a self-avoiding walk which starts in a spiral around the center hexagon, which covers 5 edges. The walk then takes one step to reach the rim of the first ring and travels once around this ring until it reaches a point where self-avoidance stops it. It then takes one step to reach the rim of the second ring and walks around that one, etc. Imagine that on each edge we place a toothpick if it's on the path, and interrupt counting the total number of toothpicks each time one of the hexagons has six vertices covered. The total number of toothpicks after n-th stage define this sequence. Note that, except from the last sentence, this comment is a copy from R. J. Mathar's comment in A182618 (Dec 13 2010). - Omar E. Pol, Sep 15 2013

Examples

			On the infinite hexagonal grid we start at stage 0 with no toothpicks, so a(0) = 0.
At stage 1 we place 5 toothpicks on the edges of the first hexagonal cell, so a(1) = 5.
At stage 2, from the last exposed endpoint, we place 4 other toothpicks on the edges of the second hexagonal cell, so a(2) = 5 + 4 = 9 because there are 9 toothpicks in the structure.
At stage 3, from the last exposed endpoint, we place 3 other toothpicks on the edges of the third hexagonal cell, so a(3) = 9 + 3 = 12 because there are 12 toothpicks in the spiral.
From _Omar E. Pol_, Sep 14 2013: (Start)
Illustration of initial terms:
.                        _       _         _         _
.              _       _/ \    _/ \_     _/ \_     _/ \_
.  _      _   /  _    /  _    /  _  \   /  _  \   /  _  \
. / \    / \  \ / \   \ / \   \ / \ /   \ / \ /   \ / \ /
.  _/  /  _/  /  _/   /  _/   /  _/     /  _/ \   /  _/ \
.      \_/    \_/     \_/     \_/       \_/  _/   \_/  _/
.                                                    _/
.
.  5     9      12      15       18        21       23
.
(End)

Links

Crossrefs

Cf. A139250, A182618, A182619, A182632, A182840.

Formula

Conjecture: a(n) = 2*n + ceiling(sqrt(12*n - 3)), for n > 0. - Vincenzo Librandi, Sep 20 2017

A182618 Number of new grid points that are covered by the toothpicks added at n-th-stage to the toothpick spiral of A182617.

Original entry on oeis.org

6, 4, 3, 3, 3, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3
Offset: 1

Views

Author

Omar E. Pol, Dec 12 2010

Keywords

Comments

In the toothpick spiral the toothpicks are connected by their endpoints. See A182617 for more information.
Attempt at an explanation, R. J. Mathar, Dec 13 2010: (Start)
In the hexagonal grid, we can pick any of the hexagons as a center, and then define a ring of 6 first neighbors (hexagons adjacent to the center), then define a ring of 12 second neighbors (hexagons adjacent to any of the first ring) and so on. The current sequence describes a self-avoiding walk which starts in a spiral around the center hexagon, which covers 5 edges. The walk then takes one step to reach the rim of the first ring and travels once around this ring until it reaches a point where self-avoidance stops it. It then takes one step to reach the rim of the second ring and walks around that one, etc. Imagine that on each edge we place a toothpick if it's on the path, and interrupt counting the total number of toothpicks each time one of the hexagons has six vertices covered. The first differences of these intermediate totals define the sequence. (End)

Examples

			At stage 1, starting from a node on the hexagonal net, we place 5 toothpicks on 5 edges of the first hexagon, so a(1)= 6 because there are 6 grid points that are covered by the toothpicks.
At stage 2, starting from the last exposed endpoints, we place 4 toothpicks on the edges of the second hexagon, so a(2)=4 because there are new 4 grid points that are covered by the toothpicks.
At stage 3, starting from the last exposed endpoints we place 3 toothpicks on the edges of the third hexagon, so a(3)=3 because there are new 3 grid points covered. Etc.
If written as a triangle, begins:
6,
4,3,3,3,3,2,
3,3,2,3,2,3,2,3,2,3,2,2,
3,2,3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,2,
3,2,2,3,2,2,2,3,2,2,2,3,2,2,2,3,2,2,2,3,2,2,2,2,
3,2,2,2,3,2,2,2,2,3,2,2,2,2,3,2,2,2,2,3,2,2,2,2,3,2,2,2,2,2

Links

Crossrefs

Cf. A121149, A139250, A139251, A182617, A182619, A182632, A182840.
Row n has A008458(n-1) terms. Row sums give A017593.

A250300 Number of ON states after n generations of cellular automaton based on triangles (compare A161644).

Original entry on oeis.org

0, 3, 6, 12, 24, 36, 42, 54, 72, 90, 108, 126, 162, 198, 210, 234, 264, 282, 300, 324, 366, 420, 462, 498, 558, 624, 678, 726, 816, 906, 936, 990, 1044, 1062, 1080, 1104, 1146, 1200, 1242, 1284, 1350, 1428, 1506, 1584, 1698, 1848, 1950, 2022, 2130
Offset: 0

Views

Author

Omar E. Pol, Jan 15 2015

Keywords

Comments

The same rules as A161644 but here we start with three ON cells which share only one vertex.

Links

Crossrefs

Cf. A139250, A147562, A151723, A160120, A161644, A182632, A253770.

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.

Links

Crossrefs

Cf. A139250, A153006, A182632, A182634, A182837, A182840.

A360512 Total number of edges after n generations in hexagonal graph constructed in first quadrant (see Comments in A360501 for precise definition).

Original entry on oeis.org

0, 1, 2, 4, 8, 13, 19, 26, 34, 44, 54, 66, 78, 93, 107, 124, 140, 160, 178, 200, 220, 245, 267, 294, 318, 348, 374, 406, 434, 469, 499, 536, 568, 608, 642, 684, 720, 765, 803, 850, 890, 940, 982, 1034, 1078, 1133, 1179, 1236, 1284, 1344, 1394, 1456, 1508, 1573, 1627, 1694, 1750
Offset: 0

Views

Author

N. J. A. Sloane, Feb 09 2023

Keywords

Links

Crossrefs

Partial sums of A360501.
Cf. A182632, A182838, A182840.

Formula

G.f. = x*(1+x+x^2+3*x^3+2*x^4+x^5+x^6-x^7)/((1-x)*(1-x^2)*(1-x^4)).
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