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CorollaryStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedjudge pass (gpt-6.1-sol)
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The sum of squares of the standard tableau numbers

Statement

For every n≥0, ∑λ⊢n(fλ)2=n!, where fλ is the number of standard λ-tableaux and f∅=1, 0!=1.

Facts & Assumptions

Given: An integer n≥0, the set Xn of words (w1,…,wn) of pairwise distinct real numbers with {w1,…,wn}={1,…,n}, and for each partition λ⊢n the number fλ of standard λ-tableaux.

[L1]

The Robinson-Schensted map w↦(P(w),Q(w)) is a bijection from Xn onto the set of pairs (P,Q) of standard tableaux of the same shape λ⊢n (The Robinson-Schensted correspondence).

[F1]

A standard λ-tableau is a filling of the Young diagram of λ by 1,…,n, each once, increasing along rows and columns; fλ is the number of such tableaux, and f∅=1 is the number of fillings of the empty diagram (Tableaux and standard tableaux).

[F2]

A word of Xn is determined by the function i↦wi, which is a bijection of {1,…,n}; conversely every such bijection gives a word in Xn, and X0 consists of the empty word alone (The Robinson-Schensted correspondence).

Proof

technique · direct
1.1F1algebra

The shapes λ⊢n are pairwise distinct as subsets of the plane, so the sets of pairs of standard tableaux of shape λ are pairwise disjoint over λ⊢n.

1.2F1algebra

For fixed λ⊢n the pairs (P,Q) of standard λ-tableaux are exactly the choices of a standard λ-tableau P followed by an independent choice of a standard λ-tableau Q, so there are fλ⋅fλ=(fλ)2 of them.

2.1L1F2step 1.1step 1.2algebra

By [L1] the map w↦(P(w),Q(w)) is a bijection from Xn onto the disjoint union over λ⊢n of the sets counted in step 1.2; comparing cardinalities and using that the bijections of {1,…,n} are n!-in-number (with 0!=1) gives n!=∣Xn∣=∑λ⊢n(fλ)2.

3.1F1F2given∎

At n=0 the only partition is ∅ and the sum is the single term (f∅)2=12=1=0!, so the identity holds at the boundary.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources