This is a question inspired by T. Amdeberhan's recent question, as well as another previos MO question.

For an integer partition $\lambda$, and $k\in \mathbb{N}\cup\{\infty\}$, let $|\lambda|_k$ denote the sum of the parts of $\lambda$, but where we only count each number at most $k$ times. E.g., $|\lambda|_1$ is the sum of the parts of $\lambda$ after removing repeated parts, and $|\lambda|_{\infty}=|\lambda|$ is the usual size of the partition.

Define the coefficients $a_k(n)$ by $$ \frac{\sum_{\lambda} |\lambda|_k \cdot q^{|\lambda|}}{\sum_{\lambda} q^{|\lambda|}} = \left( \sum_{\lambda} |\lambda|_k \cdot q^{\lambda} \right) \cdot \prod_{i=1}^{n} (1-q^i) = \sum_{n\geq 0} a_k(n) q^{n}.$$

From the above-linked questions, we see that $a_1(n) = n$, while $a_{\infty}(n) = \sigma(n) = \sum_{d\mid n} d$, the sum of divisors of $n$. So $a_k(n)$ give a sequence of numbers which "interpolate" between $n$ and $\sigma(n)$ in some sense.

Question: What are these $a_k(n)$ for arbitrary $k$? Do they have any nice expression in general?


1 Answer 1


The bijection described by Mark Wildon in the second linked question can be adapted to your generalization. Indeed, $|\lambda|_k$ counts the number of ways of selecting a box $B$ of $\lambda$ such that, if the row containing $B$ has length $m$, then there are at most $k-1$ other rows of length $m$ above it. By erasing the rows of length $m$ that contain box $B$ or are above the row that contains $B$, we obtain the partition $\mu$. Let $1\le c\le m$ be the column containing $B$. Then for each $m$ we have a bijection $(\lambda, B)\leftrightarrow (\mu, c)$. Which gives $$\sum_{\lambda \vdash n}|\lambda|_k=\sum_{m=1}^n\sum_{r=1}^k p(n-rm)m$$ With a little calculation we see that this statement is equivalent to $$a_k(n)=n\left(\sum_{d\le k \,\text{and}\, d|n}\frac{1}{d}\right).$$ In this form, it is easier to see how $a_k(n)$ interpolates between $n$ and $\sigma(n)$.

  • $\begingroup$ Excellent! This is indeed as nice an interpolation as one could ask for. $\endgroup$ Nov 11, 2021 at 3:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.