# Bijective proof problems by Stanley R. By Stanley R.

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For instance, when n = 2 we get the three oscillating tableaux (∅, 1, ∅, 1, ∅), (∅, 1, 2, 1, ∅), and (∅, 1, 11, 1, ∅). The number of oscillating tableaux of length 2n and shape ∅ is equal to 1 · 3 · 5 · · · (2n − 1) (the number of partitions of [2n] into n 2-element blocks). 217. [*] The major index maj(T ) of an SYT is defined to be the sum of all entries i of T for which i + 1 appears in a lower row than i. Fix n ∈ P and λ ⊢ n, and let m ∈ Z. Then the number of SYT T of shape λ satisfying maj(T ) ≡ m (mod n) depends only on λ and gcd(m, n).

3] The number of SYT with n entries and at most four rows is equal to C⌊(n+1)/2⌋ C⌈(n+1)/2⌉ . Note. There is a similar, though somewhat more complicated, formula for the case of five rows. For six and more rows, no “reasonable” formula is known. 207.  The number of pairs (P, Q) of SYT of the same shape with n entries each and at most two rows is the Catalan number Cn . 208.  The number of pairs (P, Q) of SYT of the same shape with n entries each and at most three rows is given by 1 (n + 1)2 (n + 2) n k=0 2k k n+1 k+1 n+2 .

2 2 0 0  2 1 0 0 2 2 0 0 A descending plane partition is an array of positive integers satisfying: (i) Each row after the first contains fewer elements than the row above, (ii) each row is indented one space to the right from the row above, (iii) the entries weakly decrease in each row, (iv) the entries strictly decrease in each column, (v) the first entry in each row (except the first) does not exceed the number of entries in the preceding row, and (vi) the first entry in each row is greater than the number of entries in its own row.