For a partition $\lambda=( \lambda_1,\dots,\lambda_n)\vdash n$ with $\lambda_1\ge\dots\ge\lambda_n\ge0$ and any function $f:\mathbb Z\to\mathbb C$, define a Hankel-like $n\times n$ matrix $$M_f(\lambda):=\begin{bmatrix} f(\lambda_1)&f(\lambda_1+1)&\cdots&f(\lambda_1+n-1)\\\ f(\lambda_2-1)&f(\lambda_2)&\cdots&f(\lambda_2+n-2)\\\ \vdots&\ddots&\ddots&\vdots\\\ f(\lambda_n-(n-1))&\cdots&f(\lambda_n-1)&f(\lambda_n)\qquad \end{bmatrix}.$$

(Note that the parts $\lambda_k$ are the arguments on the main diagonal; the partition is padded with zeros to get length $n$.)

Then an interesting identity is implied in this question: if $g(z):=\dfrac1{\Gamma(z+1)}=\dfrac1{z!}$, then we have $$\boxed{\sum\limits_{\lambda\vdash n}\det[M_{fg}(\lambda)\cdot M_{g}(\lambda)] =\dfrac {f(1)^n}{n!}=f(1)^ng(n)},$$ meaning that all other $f(k)$ for $k\ne1$ are canceled out in the summing.
The order of the $\lambda_k$'s is crucial here. Even if all partitions are ordered "backwards", i.e. increasingly, nothing cancels out.

What are references for this striking identity?


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.