extending a partition of a number to get new partition

Let $$\lambda = (k_1^{m_1}\,k_2^{m_2})$$ where $$0 be a partition of $$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 $$0 be a partition of some positive integer $$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.

1 Answer

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 $$\lambda$$.