Let φ(BBn) be the set of permutations under Corteel and Nadeau’s bijection apply-ing on the set BBn.
proof of Theorem 2(3). Let T ∈ BBn with columns c1, c2, . . . , cs, where c1 > c2 >
· · · > cs. Since the first row of T contains no 0’s, row 1 is an unrestricted row.
Moreover, (1, cl) are topmost 1’s for all 1 ≤ l ≤ s. The first step of φ is to write down the unrestricted rows in increasing order, thus the number ‘1’ is the first label in the list. Then insert Rlclto the list. Since (1, cl) are topmost 1’s for all 1 ≤ l ≤ s, we obtain σ = R1c1R2c2· · · RscsR0. Note that, R0 is the unrestricted rows of T in increasing order, thus there is no descent after ‘1’ in σ. Furthermore, the descents are decreasing, hence σ avoids patterns 31-42 and 32-41.
Conversely, suppose that there exists a permutation σ ∈ Sn(31-42, 32-41) and there is no descent after the number ‘1’ such that φ−1(σ) = T /∈ BBn; i.e., there exists a column cl such that the cell rl 6= 1. If rl is an unrestricted row, then we have · · · 1 · · · Rlclrl· · · in Listl. Thus cl will be a descent after the number ‘1’, a contradiction. If rl is a restricted row. Let rl ∈ Rw for some cw > cl. Thus we have
· · · rl· · · cwrw· · · in Listl−1. Hence · · · clrl· · · cwrw· · · in Listl. This pair of descents cl, cw forms either a 31-42 or a 32-41 pattern in σ, a contradiction.
We also obtain a bijection between set partitions and the set of Sn(31-42,32-41) in which there is no descent after the number ‘1’.
Theorem 8. There is a bijection between set partitions of [n] and the set of permu-tations in Sn(31-42, 32-41) in which there is no descent after the number ‘1’.
Proof. For every block in a set partition of [n], we arrange the elements in it in increasing order. Then we arrange the blocks that do not contain 1 in increasing order by its largest number. Finally, we put the block that contains 1 in the last.
Thus after deleting the blocks, we get a permutation whose descents are increasing and in which there is no descent after the number ‘1’.
The reverse is easy. We form a set partition by cutting a permutation in Sn(31-42, 32-41) into blocks after each descent.
By Theorems 2(3) and 8, we know that the B-Bell tableaux are counted by the Bell numbers.
3 p, q-Statistics on Bell tableaux
3.1 q-statistics on L-Bell and R-Bell Tableaux
We called the bijection in Proposition 1 f and let T be a L-Bell tableau whose sum of the number of columns and the number of zero rows is k. Then f (T ) is a set partition of [n] with k blocks formed by the following:
1. For every column j of T , construct a block of f (T ) that contains j and row i whenever its leftmost 1 is in column j,
2. Every zero row forms a block of size 1.
Note that the largest element in each block of f (T ) is a column of T .
For a L-Bell tableau T , let hinv(T ) be the sum of the number of the restricted 0’s, the number of 0’s for which there is a topmost 1 to the left of it and the number of the superfluous 1’s for which there is a topmost 1 to the left of it (stars in Figure 8).
For example, the L-Bell tableau T in Figure 8 has hinv(T ) = 6. The image of T under f is π = 178/25/346. The complement of π, πc = 128/356/47, also has hinv(πc) = 6.
Now we are ready to prove our first result in this section:
Proof of Theorem 4 (1). Suppose that T is a L-Bell tableau of length n with n − k nonzero rows, then f (T ) is a set partition of [n] with k blocks. As we have mentioned
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Figure 8: A L-Bell tableau of length 8 and its hinv.
in section 1.3, for a set partition π,
hinv(π in standard form) = lcb(π in standard form)
= los(πc in max-decreasing form).
We prove this theorem by showing that
hinv(T ) = los(f (T ) in max-decreasing form), thus hinv(T ) = lcb(f (T )c in standard form)
= hinv(f (T )c in standard form).
By applying Theorem 3 we complete the proof.
Now we show that hinv(T ) = los(f (T ) in the max-decreasing form). We first classify the hinv(T ) into four types:
1. For a superfluous 1’s in cell (i, j) that has topmost 1’s to its left, it contributes once to hinv(T ). For example, (3, 5) in Figure 8. Let the cell (i0, j) be the topmost 1 above it and (i, j0) be a topmost 1 to its left, where i0 < i < j < j0. Since T is a L-Bell tableau, they are also leftmost 1’s. Thus in f (T ), i and j0 are in the same block. Besides, i and j0 are the opener and closer since (i, j0) is a topmost 1. Hence, we have the block containing i, j0 as its opener and closer is to the left of the block containing i0, j as its opener and closer in f (T ).
Therefore the opener i contributes once to losj(f (T )), where j is a closer.
2. For a 0 in cell (i, j) that has a topmost 1 to its left, it contributes once to hinv(T ). For example, (1, 6) and (1, 5) in Figure 8. Let the cell (i, j0) be
a topmost 1 to its left and (i0, j) be the topmost 1 in column j, where i <
i0 < j < j0. By the similar discussion in previous case, we have the block containing i, j0 as its opener and closer is to the left of the block containing i0, j as its opener and closer in f (T ). Therefore the opener i contributes once to losj(f (T )), where j is a closer.
3. For a restricted 0’s in cell (i, j) that has topmost 1’s to its right, it contributes once to hinv(T ). For example, (2, 8) and (3, 8) in Figure 8. Let the cell (i0, j) be the topmost 1 above it and (i, j0) be a topmost 1 to its right, where i0 < i < j0 < j. Similarly, we have the block containing i0, j as its opener and closer is to the left of the block containing i, j0 as its opener and closer in f (T ). We regard the contribution made by (i, j) in T as the contribution made by the opener i0 to losi(f (T )). Therefore the opener i0 contributes once to losi(f (T )), where i is an opener.
4. For a restricted 0’s in cell (i, j) that has is no topmost 1 to its right, it con-tributes once to hinv(T ). For example, (4, 8) in Figure 8. Let the cell (i0, j) be the topmost 1 above it and (i, j0) be a leftmost 1 in row i, where i0 < i < j0 < j.
Similarly, we have the block containing i0, j as its opener and closer is to the left of the block containing i, j0, where i is neither an opener nor a closer and j0 is a closer in f (T ). We regard the contribution made by (i, j) in T as the contribution made by the opener i0 to losi(f (T )). Therefore the opener i0 contributes once to losi(f (T )), where i is neither an opener nor a closer.
Thus hinv(T ) ≤ los(f (T ) in the max-decreasing form). Besides, it is easy to see from above that every pair of (i, j), where i contributes to losj(f (T )), is counted exactly once.
Hence, we have hinv(T ) = los(f (T ) in the max-decreasing form).
Now we turn to the q-statistic on R-Bell tableaux. For a R-Bell tableau T , let hinv(T ) be the sum of the followings:
1. the number of the restricted 0’s (diamonds in Figure 9), and
2. twice the number of the superfluous 1’s for which there is a topmost 1 to the right of it (stars in Figure 9).
For example the R-Bell tableau T in Figure 9 has hinv(T ) = 6. Note that the 1 in cell (2, 8) contributes twice hinv(T ) as well as the cell (3, 6) does.
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Figure 9: A R-Bell tableau of size 8.
We prove the distribution of statistic hinv on R-Bell tableaux by providing a bijection between LBn,k and RBn,k. Although Corteel and Nadeau already gave a bijection between LBn,k and RBn,kin [5], the statistic hinv between the correspond-ing tableaux vary and the bijection is tedious and use PT-words as its medium. The one we give here is easier than Corteel and Nadeau’s and is a statistic-preserving map for hinv.
Now we introduce the bijection between LBn,k and RBn,k, say θ.
From L-Bell tableaux to R-Bell tableaux. The shapes of a L-Bell tableau T and its corresponding R-Bell tableau T0 are the same, by [5]. In each row, the topmost 1’s in T remain topmost 1’s in T0 (but the positions may change). We do the bijection row by row from the bottom to the top. For each row, we fill the cells in T0 with 0’s whenever there is a topmost 1 below it in the same column. Then
1. if there is no topmost 1 in the row, we fill the empty cells in T0 from the left to the right with the restricted 0’s and the superfluous 1’s in the same order in T ;
2. otherwise, we write down the restricted 0’s and the superfluous 1’s in T to form a list, and we transform 0’s to 1’s and 1’s to 0’s in it, respectively; then
we fill the empty cells in T0 from the left to the right with elements in reversing order of the list. Finally, we fill the remaining empty cell in T0 with topmost 1’s to complete this row.
See Figure 10 as an example.
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Figure 10: A bijection from L-Bell tableau to R-Bell tableau.
Note that, we put the restricted 0’s to the left of 1’s in each row operation, hence a topmost 1 in T becomes a rightmost 1 in the resulting tableau T0. Thus T0 is a R-Bell tableau.
From R-Bell tableaux to L-Bell tableaux. We apply the inverse operation in each row operation. Similarly, the topmost 1’s in R-Bell tableaux T0 remain topmost 1’s in L-Bell tableaux T in each row (but the positions may change). We apply the inverse operation row by row from the bottom to the top. For each row, we fill the cells in T with 0’s whenever there is a topmost 1 below it in the same column. Then 1. if there is no topmost 1 in the row, we fill the empty cells in T from the left to the right with the restricted 0’s and the superfluous 1’s in the same order in T0;
2. otherwise, we write down the restricted 0’s and the superfluous 1’s in the reversing order in T0 to form a list, and we transform 0’s to 1’s and 1’s to 0’s in it, respectively; then we insert the topmost 1 to the list between 0’s and 1’s.
Finally, we fill the empty cells in T from the left to the right with elements in the order of the list.
proof of Theorem 4 (2). Now we prove that the statistics hinv(T ) and hinv(T0) are equal. If we regard a contribution made by a 0 of type(2) as a contribution made by a 0 of type(3) in the proof of 4(1), then hinv(T ) of a L-Bell tableau T can be classified into three types:
1. For a superfluous 1 which has topmost 1’s to the left of it contributes once to hinv(T ).
2. For a restricted 0 which has topmost 1’s to the right of it contributes twice to hinv(T ).
3. For a restricted 0 which has no topmost 1 to its right contributes once to hinv(T ).
On the other hand, for the statistics hinv(T0) of the corresponding R-Bell tableau T0, we also classify them into three types:
1. For a restricted 0 which has topmost 1’s to the right of it contributes once to hinv(T0).
2. For a superfluous 1 which has topmost 1’s to the right of it contributes twice to hinv(T0).
3. For a restricted 0 which has no topmost 1 to its right contributes once to hinv(T0).
For example, (3, 8), (2, 8), and (4, 8) are of type(1), (2), and (3), respectively in Figure 9.
It is easy to see that the bijection θ maps elements of types (i) in T to type (i) in T0, where i = 1, 2, 3, hence the hinv is invariant under θ.