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