Assuming that the problem of exhibiting a bijection is not considerd a frivolous pursuit, allow me to ask a question troubling me for some time now.
Let $\lambda \vdash n$ denote the fact that $\lambda$ is a partition of $n$. Denote the number of parts by $l(\lambda)$. If $T$ is a standard Young tableau (SYT), we will denote the underlying partition shape by $sh(T)$.
Given a positive even integer $2n$, let
$$ Pe_{2n}=\{ \lambda: \lambda\vdash 2n,\text{ } l(\lambda) \leq3 \text{ and all parts of } \lambda \text{ are even} \}$$
and
$$ Qe_{2n}=\{ \lambda: \lambda\vdash 2n, \lambda = (k,k,1^{2n-2k}), \text{ }k\geq 1 \}$$
Using these sets we will define two more sets whose elements are SYTs.
$$ TPe_{2n}=\{T: T \text{ an SYT, } sh(T)\in Pe_{2n} \}$$
and
$$ TQe_{2n}=\{T: T \text{ an SYT, } sh(T)\in Qe_{2n} \}$$
$\textbf{Question}$:
Is there a bijective proof exhibiting the fact that the cardinalities of $TPe_{2n}$ and $TQe_{2n}$ are equal?
The second question is very similar. Given an odd positive integer $2n+1$, let
$$ Po_{2n+1}=\{ \lambda: \lambda\vdash 2n+1,\text{ } l(\lambda)=3 \text{ and all parts of } \lambda \text{ are odd} \}$$
and
$$ Qo_{2n+1}=\{ \lambda: \lambda\vdash 2n+1, \lambda = (k,k,1^{2n+1-2k}), \text{ }k\geq 1 \}$$
Using these sets we will define two more sets whose elements are SYTs.
$$ TPo_{2n+1}=\{T: T \text{ an SYT, } sh(T)\in Po_{2n+1} \}$$
and
$$ TQo_{2n+1}=\{T: T \text{ an SYT, } sh(T)\in Qo_{2n+1} \}$$
$\textbf{Question}$:
Is there a bijective proof exhibiting the fact that the cardinalities of $TPo_{2n+1}$ and $TQo_{2n+1}$ are equal?
I tried quite a few approaches ( Motzkin path interpretations, matching diagrams etc) but did not succeed. I hope somebody here can guide me.
The relevant OEIS entry would be link text
Thanks,
Vasu
Best Answer
Not an answer, but too long for a comment. This addresses a previous comment and points out the smallest case where the requested bijection is not obvious (for the even case).
Through $n = 8$, the partition sets are in bijection via a combination of transposition and the identity map, so there is a direct correspondence for the tableaux.
$Pe_{10} = \{10, 82, 64, 622, 442\}$ and $Qe_{10} = \{55, 4411, 331^4, 221^6, 1^{10}\}$. The partitions that don't match up are 64 & 442 versus 55 & 4411.
Using the hook length formula, there are 90 standard Young tableaux of shape (6,4) and 252 of shape (4,4,2), so these contribute 342 to $TPe_{10}$. (Sorry for the change in partition notation, but "90 of shape 64" looked strange.) There are 42 tableaux of shape (5,5) and 300 of shape (4,4,1,1), also contributing 342 to $TQe_{10}$. So the conjecture does hold for $n=10$.
This example shows that the requested bijection cannot preserve the tableaux shape/underlying partition.
The next case is even more interesting because the tableaux counts match even though there are not the same number of partitions in the sets.
The unmatched partitions are $(8,4),(6,4,2),(4,4,4) \in Pe_{12}$ and $(5,5,1,1),(4,4,1,1,1,1) \in Qe_{12}$. And the sums work: Writing $\textrm{SYT}(\lambda)$ for the number of standard Young tableaux with shape $\lambda$, we have $$\textrm{SYT}((8,4))+\textrm{SYT}((6,4,2))+\textrm{SYT}((4,4,4))=275+2673+462=3410,$$ $$\textrm{SYT}((5,5,1,1))+\textrm{SYT}((4,4,1,1,1,1))=1485+1925=3410.$$
The next two cases work but don't offer anything new. Here are a few general comments.
From a formula for the number of partitions of $n$ with up to three parts, $|Pe_{2n}| = \lfloor \frac{(n+3)^2}{12} \rceil$ where $\lfloor \cdot \rceil$ denotes nearest integer.
For $n$ odd, 3 partitions match up between the two sets: $(2n)$, $(2n-2,2)$, and $(2n-4,2,2)$ in $Pe_{2n}$ with $1^{2n}$, $221^{2n-4}$, and $331^{2n-6}$ in $Qe_{2n}$ (matching by transposition).
For $n$ even, 4 partitions match because $(n,n)$ is in both sets.
Continuing along this line of reasoning would require formulas for $\textrm{SYT}(\lambda)$ for partitions from each of the sets.
Following up on the last bullet, in the 1996 JCTA note "Polygon Dissections and Standard Young Tableaux," Stanley credits O'Hara and Zelevinsky with a formula for $\textrm{SYT}(\lambda)$ for $\lambda = kk1^m$. Rewritten in terms of $k$ and $m$, it is $$\frac{1}{2k+m+1} \binom{2k+m+1}{k} \binom{m+k-1}{k-1}.$$ Perhaps Stanley's bijection between such tableaux and certain sets of nonintersecting diagonals in a polygon discussed in the note could help.
It's not hard to work out that $$\textrm{SYT}((a,b,c))=\frac{(a+b+c)!(a-b+1)(b-c+1)(a-c+2)}{(a+2)!(b+1)!c!}$$ (without regard to the parity of $a,b,c$).