Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • 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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Small Kostka numbers

Example

Write a filling of a diagram with at most two rows by its rows, so that 11/2 denotes the tableau whose first row is 1,1 and whose second row is 2. For n=3 the small Kostka numbers are K(2,1),(2,1)=1,K(2,1),(1,1,1)=2,K(3),(2,1)=1,K(1,1,1),(2,1)=0. The witnesses are: the single filling 11/2 for shape (2,1) and content (2,1); the two fillings 12/3 and 13/2 for shape (2,1) and content (1,1,1); and the single filling 112 for shape (3) and content (2,1). There is no semistandard filling of the column (1,1,1) with content (2,1).

Facts & Assumptions

Given: The partitions (2,1), (3), (1,1,1) and (2,1) of 3, and the filling notation of the Example section.

[F1]

A semistandard tableau of shape λ and content μ is a filling of [λ] such that the entry i occurs exactly μi times, entries weakly increase along each row and strictly increase down each column; Kλ,μ is the number of such fillings (Semistandard tableaux and Kostka numbers).

[F2]

A filling of content (1n) is semistandard exactly when it is standard, so Kλ,(1n)=fλ (Semistandard tableaux and Kostka numbers).

Verification

technique · direct
1.1

For shape (2,1) and content (2,1) the filling carries two 1's and one 2, so the single entry 2 occupies one of the three boxes; placing it in the first box of the top row gives 21/1, whose top row violates weak increase since 2>1; placing it in the second box of the top row gives 12/1, whose column has entries 1,1 and violates strict increase; placing it in the bottom box gives 11/2, whose rows are weakly increasing and whose column entries 1<2 strictly increase, and no other filling is available, so K(2,1),(2,1)=1, realized by 11/2.

givenF1
1.2

For shape (2,1) and content (1,1,1) the fillings are the bijections of the three boxes onto {1,2,3}, six in all; each of 12/3 and 13/2 is semistandard, since its rows are weakly increasing and its column entries are 1<3 and 1<2, while 21/3, 31/2 and 32/1 have a top row that is not weakly increasing and 23/1 has column entries 2,1, which are not strictly increasing, and hence K(2,1),(1,1,1)=2, in agreement with Kλ,(1n)=fλ and f(2,1)=2.

F1F2
1.3

For shape (3) the diagram is a single row, whose weak increase forces the entries to be sorted, so the filling of content (2,1) must be 1,1,2, that is 112, and this filling is semistandard because a one-row diagram has no column condition; hence K(3),(2,1)=1.

givenF1
1.4

For shape (1,1,1) the diagram is a single column of three boxes, and strict increase down the column forces the three entries to be pairwise distinct; the content (2,1) supplies only the two distinct labels 1 and 2, with 1 repeated twice, so no filling of that content is semistandard and K(1,1,1),(2,1)=0.

givenF1
2.1

Steps 1.1, 1.2, 1.3 and 1.4 compute the four displayed numbers 1,2,1,0, by checking all three content fillings in step 1.1 and all six in step 1.2, and by using the row and column conditions in steps 1.3 and 1.4 to leave respectively one and no semistandard fillings. Each of the latter two shapes has three fillings of content (2,1) before imposing those conditions. ∎

step 1.1step 1.2step 1.3step 1.4F1F2

Depends on

Used by

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Dependency tree · two levels

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Sources