Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31
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Parenthesised tensor words of a fixed length are counted by the Catalan numbers

Statement

For n≥1, let an be the number of parenthesised tensor words on the letters x1,…,xn with no inserted unit symbol. Then a1=1 and

an=∑k=1n−1akan−k(n≥2).

Equivalently, an=Cn−1, where the Catalan numbers are defined by C0=1 and

Cm=∑k=0m−1CkCm−1−k(m≥1).

Facts & Assumptions

Given: The recursive formation rule for parenthesised tensor words.

[L1]

A parenthesised tensor word is either one letter or a composite (u⊗v) built recursively, with the letters kept in order (Parenthesised tensor words and their evaluation functors).

Proof

technique · direct
1.1givenL1

For n=1 there is only the word x1, so a1=1.

1.2L1

If n≥2, every word has a unique outermost decomposition (u⊗v), where u uses the first k letters and v uses the remaining n−k letters for a unique k with 1≤k≤n−1. Conversely, every such pair (u,v) produces a word on n letters.

2.1step 1.2algebra

Therefore the words on n letters are partitioned by the value of k, and for fixed k there are akan−k choices. Summing over k gives an=∑k=1n−1akan−k.

3.1step 1.1step 2.1algebra∎

The recurrence in step 2.1 is exactly the Catalan recurrence after the index shift an=Cn−1, and step 1.1 matches the initial value C0=1. Hence an=Cn−1 for all n≥1.

Depends on

Used by

Dependency tree · two levels

2 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