# Showing two partitions to be equal

I was posed with the following problem " Let $$F(n)$$ denote the number of partitions of $$n$$ with every part appearing at least twice and $$G(n)$$ denote the number of partitions of $$n$$ into parts larger than 1 such that no two parts are consecutive integers. Use conjugate partitions to prove that $$F(n)=G(n).$$"

What i have so far:

I realize that for G(n), the statement that no two parts are consecutive implies that distance between two is at least two, which gives a similarity with F(n). I think using ferrer diagrams will help a bit, but i could not see how to get started. Please help

Here's the Ferrers diagram for $$n=6$$.

I realize that for $$G(n)$$, the statement that no two parts are consecutive implies that distance between two is at least two,

This might be causing a problem. No two parts are consecutive implies the difference between two parts is not $$1$$ instead of at least two as you have stated.

• So, what you are suggesting is that that they might even be equal? Feb 21, 2019 at 16:05
• That's correct. And you can see from my Ferrer diagram that $3+3$ in the $F(n)$ count is the same as $2+2+2$ in the $G(n)$ count.
– Dubs
Feb 21, 2019 at 16:07
• So, how do i prove that F(n) and G(n) are always equal? Feb 21, 2019 at 16:08
• Try something along this line: given a Ferrer diagram of a partition in $F(n)$, the conjugate is in $G(n)$. It should be fairly straight forward using the property that each row is shows up at least twice due to $F(n)$ property, the conjugate will never have consecutive parts and will never be $1$.
– Dubs
Feb 21, 2019 at 16:11
• Cann you elaborate a bit on this "the conjugate will never have consecutive parts and will never be 1." Feb 21, 2019 at 16:17