• Let $a(n)$ be A309099 i.e. the number of partitions of $n$ avoiding the partition $(4,3,1)$.

• We say a partition $\alpha$ contains $\mu$ provided that one can delete rows and columns from (the Ferrers board of) $\alpha$ and then top/right justify to obtain $\mu$. If this is not possible then we say $\alpha$ avoids $\mu$. For example, the only partitions avoiding $(2,1)$ are those whose Ferrers boards are rectangles.

• Let $$ b(n) = n + \sum\limits_{i=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\sum\limits_{j=1}^{n-2i}\left\lfloor\frac{i+j-1}{j+1}\right\rfloor $$

I conjecture that $$b(n) = a(n).$$

Is there a way to prove it?

• Let $a(n)$ be A309099 i.e. the number of partitions of $n$ avoiding the partition $(4,3,1)$.

• We say a partition $\alpha$ contains $\mu$ provided that one can delete rows and columns from (the Ferrers board of) $\alpha$ and then top/right justify to obtain $\mu$. If this is not possible then we say $\alpha$ avoids $\mu$. For example, the only partitions avoiding $(2,1)$ are those whose Ferrers boards are rectangles.

• Let $$ b(n) = n + \sum\limits_{i=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\sum\limits_{j=1}^{n-2i}\left\lfloor\frac{i+j-1}{i+1}\right\rfloor $$

I conjecture that $$b(n) = a(n).$$

Is there a way to prove it?