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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

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1 Answer 1

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

Ferrer diagram

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.

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  • $\begingroup$ So, what you are suggesting is that that they might even be equal? $\endgroup$
    – Ayan Shah
    Feb 21, 2019 at 16:05
  • $\begingroup$ 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. $\endgroup$
    – Dubs
    Feb 21, 2019 at 16:07
  • $\begingroup$ So, how do i prove that F(n) and G(n) are always equal? $\endgroup$
    – Ayan Shah
    Feb 21, 2019 at 16:08
  • $\begingroup$ 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 $1$. $\endgroup$
    – Dubs
    Feb 21, 2019 at 16:11
  • $\begingroup$ Cann you elaborate a bit on this "the conjugate will never have consecutive parts and will never be 1." $\endgroup$
    – Ayan Shah
    Feb 21, 2019 at 16:17

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