# Counting the number of semistandard Young Tableaux with maximum entry $n.$ Reference/Formula request

Question: If $k \leq n$ let $\lambda_k$ be a Young diagram with square $k \times k$ shape. I write $\#_{\lambda_{k}^n}$ to count the number of semistandard Young tableaux with shape $\lambda_k$ and maximum entry $n$. For example if $n=4$ and $k=3$ then we draw $\lambda_{3}= \begin{matrix} \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet \\ \bullet & \bullet & \bullet \\ \end{matrix}$ and $\#_{\lambda_{3}^4}=20.$ Is there a known formula to compute $\#_{\lambda_{k}^n}$ ? "Googling" picks up the "content hook formula" - but I am not sure.

• Yes, you're on the right track! See if this answer helps: math.stackexchange.com/a/25943/1242. Oct 30, 2017 at 16:06
• That answers it ! It is Stanley's hook content formula. Are there any "concrete formulas", in the sense of explicit closed products involving binomials if the diagram is known to be square ? Oct 30, 2017 at 16:12
• It's certainly possible to write it down in the form of a product, but I don't know if it simplifies to anything nice. Oct 30, 2017 at 16:15

By the hook content formula we have $$\prod_{i=0}^{k-1} \prod_{j=0}^{k-1} \frac{N+i-j}{k+i-j}$$ We can rewrite that as any of $$\prod_{i=0}^{k-1} \frac{(N+i)^\underline{k}}{(k+i)^\underline{k}} = \prod_{i=0}^{k-1} \frac{(N+i)!}{(N-k+i)!} \frac{i!}{(k+i)!} = \prod_{i=0}^{k-1} \binom{N+i}{k} \binom{k+i}{k}^{-1}$$ The closest to a "closed product" is probably to write it as $$\frac{G(N+k+1)\;G(N-k+1)\;G(k+1)^2}{G(N+1)^2\;G(2k+1)}$$ where $G$ is the Barnes G-function or superfactorial (although "supergamma" would perhaps be a better name given the offset).