extending a partition of a number to get new partition

Question

Let $\lambda = (k_1^{m_1}\,k_2^{m_2})$ where $0<k_1<k_2$ be a partition of $n$ in the power notation.

Let $\mu = p_0^{r_0}\,p_1^{r_1}\,\cdots\, p_t^{r_t} \,(k_1^{m_1}\,k_2^{m_2})\,q_0^{s_0}\,q_1^{s_1}\,\cdots\, q_u^{s_u}$ where $0<p_0<p_1<\cdots<p_t<k_1<k_2<q_0<q_1<\cdots<q_u$ be a partition of some positive integer $m>n$.

Pictorially, I am adding new rows of length different from $r$ and $s$ on top and bottom of the Ferrer diagram of $\lambda$.

My question is, Given an $m > n$, how many such $\mu$'s are there which are also partition of $m$?

Also, what is the relation between $\lambda$ and $\mu$.

More precisely are they comparable in any natural partial order defined on partitions and set of all such $\mu$'a form the ideal generated by lambda?

Kindly share your thoughts.

Thank you.

Answer 1

Here's a way to count the number of $\mu \vdash m$ determined as you specify by a given $\lambda \vdash n$ that is a modification of the standard generating function for integer partitions.

Writing $p(n)$ for the number of partitions of $n$, Euler gave us $$\sum_{n=0}^\infty p(n) x^n = \prod_{i=1}^\infty \frac{1}{1-x^i}.$$ Your $\mu$ are partitions that "contain" $\lambda = (k_1^{m_1}, k_2^{m_2})$ with the additional restriction that other parts be either less than $k_1$ or greater than $k_2$. Writing $r(m)$ for the number of partitions of $m$ with this restriction, we have \begin{align} \sum_{m=0}^\infty r(m) x^m & = (x^{k_1})^{m_1} (x^{k_2})^{m_2} \left(\prod_{i=1}^{k_1-1} \frac{1}{1-x^i}\right) \left(\prod_{i=k_2+1}^\infty \frac{1}{1-x^i}\right) \\ & = x^n (1-x^{k_1})\cdots(1-x^{k_2})\prod_{i=1}^\infty \frac{1}{1-x^i} \\ & = x^n \left(\prod_{i=k_1}^{k_2}(1-x^i)\right) \left(\sum_{n=0}^\infty p(n) x^n\right). \end{align}

As an example, consider $\lambda = (2,6,6)$. The polynomial giving counts for $\mu$ begins $$x^{14}+x^{15}+x^{16}+x^{17}+x^{18}+x^{19}+x^{20}+2x^{21}$$ where the 2 partitions of 21 are $(1^7,2,6,6)$ and $(2,6,6,7)$.

I chose $\lambda = (2,6,6)$ to highlight that this approach applies to $\lambda$ with more than 2 distinct parts. Using $(2,2,4,6)$ or $(2,3,3,6)$ for $\lambda$ would give the same number of $\mu$ as using $(2,6,6)$. All that matters are the smallest part, largest part, and sum of $\lambda$.