Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16
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Binary words avoiding 101 have generating function (1+x2)/(12x+x2x3)

Example

Let an be the number of binary words of length n that avoid 101. Then

n0anxn=1+x212x+x2x3.

The first values are

a0,a1,,a6=1,2,4,7,12,21,37.

Facts & Assumptions

Given: The singleton forbidden set B={101} over {0,1}.

[L1]

The prefix automaton uses the allowed proper-prefix states and the longest suffix transition (The longest-suffix prefix automaton for a finite set of forbidden factors).

[L2]

Its fixed-entry walk series are cofactors of IxA divided by det(IxA) (Transfer-matrix theorem: weighted-walk generating functions are cofactors of IxA divided by det(IxA)).

Verification

technique · transfer matrix with finite enumeration check
1.1

By [L1], the states are ε,1,10 and, in that order, their transfer matrix is A=(110011100). The missing transition from 10 on 1 is precisely the rejected factor 101.

givenL1
2.1

Avoiding words are walks from ε to any state. Summing the three corresponding entries supplied by [L2], or solving (IxA)y=1, gives (1+x2)/(12x+x2x3).

step 1.1L2algebra
3.1

Direct enumeration by length gives 1,2,4,7,12,21,37 through length six. These values also satisfy an=2an1an2+an3 for 3n6, as the denominator in step 2.1 requires.

step 2.1algebra

Depends on

Used by

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Dependency tree · next 3 levels

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