A308628 A Van Eck-type sequence on the triangular lattice.

0, 0, 1, 0, 2, 0, 1, 4, 0, 2, 3, 0, 2, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Author

Charlie Neder, Jun 11 2019

Keywords

• nonn
• easy

Comments

For a given lattice, the Van Eck sequence over that lattice is the unique sequence of nonnegative integers such that, if all equal terms are connected by "bridges" that travel between adjacent faces, then each term is the length of the bridge connecting the previous term to a term with lower index, or 0 if no such bridge exists. Generally, the Van Eck sequence of a given lattice is not unique since it depends on the path that the sequence takes through the lattice. This sequence uses a spiral, as in A308625 and A308626, and appears as follows, starting at the cell in parentheses facing upward and traveling clockwise:

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\ / \ / \ 3 / \ 1 / \ / \

.\ / \ / 3 \ / 1 \ / 1 \ / \

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./ \ / \ 2 / \ 1 / \ 1 / \ /

/ \ / 0 \ / 0 \ / 0 \ / 1 \ /

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\ / \ 3 / \(0)/ \ 2 / \ 1 / \

.\ / \ / 2 \ / 0 \ / 1 \ / \

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./ \ / \ 0 / \ 1 / \ 1 / \ /

/ \ / \ / 4 \ / 1 \ / \ /

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Note: This sequence uses the definition that two cells are adjacent if they share an edge. Allowing vertex adjacency makes a very boring sequence: 0, 0, 1, and 0, followed by an infinite string of 1's.

a(n) = 1 for all n >= 17, since the previous 1 will always be adjacent to another 1. The Van Eck-type sequences for the square and hexagonal lattices end similarly.

Examples

			a(7) = 1, and the only other 1 to appear so far is 4 cells away (not 2, since we only consider edge adjacency), so a(8) = 4.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (1).

Crossrefs

Cf. A181391, A308625, and A308626 for Van Eck-type sequences over the 1D "lattice" and the 2D hexagonal and square lattices, respectively.

Formula

G.f.: x^3*(1 - x + 2*x^2 - 2*x^3 + x^4 + 3*x^5 - 4*x^6 + 2*x^7 + x^8 - 3*x^9 + 2*x^10 + x^11 - 2*x^13)/(1 - x). - Elmo R. Oliveira, Aug 03 2024