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North–east–west walks without immediate horizontal reversal satisfy $a_{n} = 2 a_{n - 1} + a_{n - 2}$

Example

Let $a_{n}$ be the number of length- $n$ words over ${N, E, W}$ in which neither $E W$ nor $W E$ occurs. Then

$a_{0} = 1, a_{1} = 3, a_{n} = 2 a_{n - 1} + a_{n - 2} (n \geq 2),$

and

$\sum_{n \geq 0} a_{n} x^{n} = \frac{1 + x}{1 - 2 x - x^{2}} .$

For nonempty words, classification by the last letter gives the transfer matrix

$A = (\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array})$

in the state order $N, E, W$ .

Facts & Assumptions

Given: Words over ${N, E, W}$ with the adjacent factors $E W$ and $W E$ forbidden.

[L1]

An eventual recurrence is equivalent to rationality of the ordinary formal generating function, with the numerator determined by the initial coefficients (A coefficient sequence is eventually linearly recurrent if and only if its formal generating function is rational).

[L2]

Finite-state walk series are rational cofactor quotients of their transfer matrix (Transfer-matrix theorem: weighted-walk generating functions are cofactors of $I - x A$ divided by $\det (I - x A)$ ).

Verification

technique · state elimination and transfer check

1.1givenalgebra

Let $N_{n}, E_{n}, W_{n}$ count valid nonempty words by their last letter. Appending $N$ is always allowed, whereas $E$ may not follow $W$ and $W$ may not follow $E$ ; this gives the displayed matrix and $N_{n} = a_{n - 1}$ for $n \geq 1$ .

2.1step 1.1algebra

For $n \geq 2$ , the state equations give $E_{n} + W_{n} = 2 N_{n - 1} + E_{n - 1} + W_{n - 1} = a_{n - 1} + a_{n - 2}$ . Together with $N_{n} = a_{n - 1}$ this yields $a_{n} = 2 a_{n - 1} + a_{n - 2}$ .

3.1step 2.1L1algebra

The empty word and the three one-letter words give $a_{0} = 1, a_{1} = 3$ . Multiplying the coefficient series by $1 - 2 x - x^{2}$ and using step 2.1 leaves $1 + x$ , so [L1] gives the displayed generating function.

4.1step 1.1L2algebra∎

As a consistency check, [L2] applied to the displayed matrix makes $1 + x 1^{T} (I - x A)^{- 1} 1$ equal to the same quotient; the leading $1$ counts the empty word.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

R. P. Stanley, Enumerative Combinatorics, vol. 1, 2nd ed., Example 4.1.3 (standard reference, not scraped)

North–east–west walks without immediate horizontal reversal satisfy an=2an−1+an−2an​=2an−1​+an−2​