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

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

Author

Omar E. Pol, Dec 13 2010

Keywords

• nonn
• more

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

• David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
• N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
• Index entries for sequences related to toothpick sequences

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