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 1-3 of 3 results.

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)
		

Crossrefs

Formula

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

A182619 Number of vertices that are connected to two edges in a spiral without holes constructed with n hexagons.

Original entry on oeis.org

6, 8, 9, 10, 11, 12, 12, 13, 14, 14, 15, 15, 16, 16, 17, 17, 18, 18, 18, 19, 19, 20, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 23, 24, 24, 24, 24
Offset: 1

Views

Author

Omar E. Pol, Dec 13 2010

Keywords

Comments

The structure shows a hexagonal growth as in A182618.
a(n) is the number of vertices of the convex parts of the perimeter of the structure.
This sequence can be constructed geometrically in the following manner: Construct a gapless array of n equal circles with the rule of always choosing an arrangement with the maximum number of completely enclosed inner circles. Then, a(n) equals the number of circles required to create a kissing perimeter around the original array. Examples: a(1) = 6 because it takes 6 circles to create a kissing perimeter around 1 circle. a(7) = 12 because it takes 12 circles to create a kissing perimeter around 7 circles, which are arranged with 1 circle in center surrounded by 6 kissing circles. One could describe this as the "kissing numbers of kissing circles" sequence. - Peter Woodward, Apr 25 2015
a(n) is also the size of the smallest hexagonal polyomino that admits a hole of size n (Cf. A257594). - Luca Petrone, Feb 28 2017

Examples

			For n=1 there is 1 hexagon, so a(1)= 6 because there are 6 vertices that are connected to two edges.
For n=2 there are 2 connected hexagons, so a(2)= 8 because there are 8 vertices that are connected to two edges.
For n=3 there are 3 connected hexagons, so a(3)= 9 because there are 9 vertices that are connected to two edges.
If written as a triangle, begins:
6,
8,9,10,11,12,12,
13,14,14,15,15,16,16,17,17,18,18,18,
19,19,20,20,20,21,21,21,22,22,22,23,23,23,24,24,24,24
		

Crossrefs

Row n has A008458(n-1) terms.

A194811 Number of grid points that are covered after n-th stage of A139250 version "Tree", starting with a(0) = 1 and assuming the toothpicks have length 4, 3, and 2.

Original entry on oeis.org

1, 5, 13, 25, 37, 53, 81, 113, 133, 149, 177, 213, 253, 313, 401, 481, 517, 533, 561, 597, 637, 697, 785, 869, 925, 985, 1077, 1189, 1329, 1537, 1793, 1985, 2053, 2069, 2097, 2133, 2173, 2233, 2321, 2405, 2461, 2521, 2613, 2725, 2865, 3073, 3329, 3525, 3613
Offset: 0

Views

Author

Omar E. Pol, Oct 24 2011

Keywords

Comments

The first differences give A147646.

Crossrefs

Formula

a(n) = 1 + 4*A153006(n) = 1 + A159795(n).
Showing 1-3 of 3 results.