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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?

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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 $\sigma(n)$.

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

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