# Is there a known formula for the number of SSYT of given shape with partition type?

Let $s_{\lambda}$ and $m_{\lambda}$ be the Schur and monomial symmetric functions indexed by an integer partition $\lambda$ ($\ell(\lambda)$ is the number of parts of $\lambda$ and $m_i(\lambda)$ is the multiplicity of part $i$). By the hook-content formula we have: $$s_{\lambda}(1^n) = \prod_{u\in \lambda} \frac{n+c(u)}{h(u)},$$ where $c(u)$ and $h(u)$ are the content and hook length of the cell $u\in \lambda$.

Using $s_{\lambda} = \sum_{\mu} K_{\lambda \mu} m_{\mu}$ where $K_{\lambda \mu}$ is the Kostka number, the number of semistandard Young tableaux of shape $\lambda$ and type $\mu$. Then we get $\sum_{\mu} K_{\lambda \mu} m_{\mu}(1^n)=\prod_{u\in \lambda} \frac{n+c(u)}{h(u)}$. This counts semistandard Young tableaux of shape $\lambda$ and any type.

Does the sum $\sum_{\mu}K_{\lambda,\mu}$ have a known formula for $\ell(\lambda)\geq 2$? This would be the number of semistandard Young tableaux of shape $\lambda$ with partition type.

Let $k(\lambda)=\sum_\mu K_{\lambda\mu}$. Then we have the generating function $\sum_\lambda k(\lambda)s_\lambda = \prod_{n\geq 1} (1-h_n)^{-1}$.
• Is there a known formula for $k(\lambda)$ expressed in terms of the entries $\lambda_i$? – Per Alexandersson Sep 1 '12 at 9:57
Why would it be? From rep. theoretic point of view, your sum is the multiplicity of an irreducible $S_n$-rep $\pi_\lambda$ in the sum of induced rep's $\rho_\mu$. But the latter is an ugly rep as you can easily see. For example, when $n=4$ you get $5$ partitions, and its dimension is: $$\sum_\mu dim(\rho_\mu) = dim (\rho_{4}) + dim (\rho_{31}) + dim (\rho_{22}) + dim (\rho_{21^2}) + dim (\rho_{1^4}) = 1 +4 + 2 + 12 + 24 = 43,$$ which is not very promising. Also, as you probably noticed yourself, the multiplicity of a trivial rep $\pi_{m}$ is the number of partitions of $n$, a nice number with no product formula (unless you start taking g.f.'s which is a completely different story - you can't really do that for general $\lambda$, except by adding squares in the first row, which is not what you want I think). In conclusion, it is unlikely a nice formula exists.