Given an integer partition $\lambda$, introduce the following quantities:
\begin{align\*}
c(\lambda)&=\sum\_{i\geq1}\left\lceil\frac{\lambda\_i}2\right\rceil, \qquad c\_o(\lambda)=\sum\_{i\geq1}\left\lceil\frac{\lambda\_{2i-1}}2\right\rceil, \qquad
c\_e(\lambda)=\sum\_{i\geq1}\left\lceil\frac{\lambda\_{2i}}2\right\rceil, \\
f(\lambda)&=\sum\_{i\geq1}\left\lfloor\frac{\lambda\_i}2\right\rfloor, \qquad f\_o(\lambda)=\sum\_{i\geq1}\left\lfloor\frac{\lambda\_{2i-1}}2\right\rfloor, \qquad
f\_e(\lambda)=\sum\_{i\geq1}\left\lfloor\frac{\lambda\_{2i}}2\right\rfloor, \\
a\_o(\lambda)&=\sum\_{i\geq1}\lambda\_{2i-1}, \qquad
a\_e(\lambda)=\sum\_{i\geq1}\lambda\_{2i}.
\end{align\*}
With these in mind, we have the following identities:



> 
> **QUESTIONS.** All sums run through all partitions of the integer $n$. Are these true?
> 
> 
> 


(1) $\sum\_{\lambda\vdash n}f\_o(\lambda)=\sum\_{\lambda\vdash n}c\_e(\lambda)$;


(2) $\sum\_{\lambda\vdash n}a\_o(\lambda)=\sum\_{\lambda\vdash n}c(\lambda)$;


(3) $\sum\_{\lambda\vdash n}a\_e(\lambda)=\sum\_{\lambda\vdash n}f(\lambda)$;


(4) $\sum\_{\lambda\vdash n}(c\_o(\lambda)-f\_e(\lambda))
=\sum\_{\lambda\vdash n}(a\_o(\lambda)-a\_e(\lambda))$.


**Example.** These are $\{(5), (4,1), (3,2), (3,1,1), (2,2,1), (2,1,1,1), (1,1,1,1,1)\}$ the partitions of $n=5$. The relevant vectors for the identities in question are:


$f\_o$ in $\{(2), (2), (1), (1,0), (1,0), (1,0), (0,0,0)\}$


$c\_e$ in $\{(0), (1), (1), (1), (1), (1,1), (1,1)\}$


$a\_o$ in $\{(5), (4), (3), (4), (3), (3), (3)\}$


$c$ in $\{(3), (3), (3), (4), (3), (4), (5)\}$


$a\_e$ in $\{(0), (1), (2), (1), (2), (2), (2)\}$


$f$ in $\{(2), (2), (2), (1), (2), (1), (0)\}$


$c\_o$ in $\{(3), (2), (2), (3), (2), (2), (3)\}$


$f\_e$ in $\{(0), (0), (1), (0), (1), (0), (0)\}$.


**Remark.** I can provide a generating function proof of (4), maybe others too. So, I only wish to see a combinatorial justification for each assertion.



Seeking for a combinatorial argument for partition identities
It's easy to check that all four identities are true when the sum is restricted to two terms indexed by a partition $\lambda$ and its conjugate $\lambda'$. (If $\lambda=\lambda'$ then the two terms are equal, so we only need one term.) From this (1)-(4) follow by grouping the terms in conjugate pairs (or just one term when $\lambda=\lambda'$).


Here is an illustrative example for the pairing of $\lambda$ and its conjugate $\lambda'$. **Just for identity (1)**.


Let $\lambda=(5,4,2)$ so that $\lambda'=(3,3,2,2,1)$. Fill in the odd-indexed rows by alternating $a$'s and $b$'s; fill in the even-indexed rows by alternating $c$'s and $d$'s (see [this paper](https://arxiv.org/pdf/math/0308012.pdf), pages 1-2). In the present case, these look like:


$$\lambda=\begin{array} 1 a&{\mathbf{\color{blue}b}}&a&{\mathbf{\color{blue}b}}&a \\ {\mathbf{\color{red}c}}&d&{\mathbf{\color{red}c}}&d \\ a&{\mathbf{\color{blue}b}}\end{array} \qquad
\lambda'=\begin{array} 1 a&{\mathbf{\color{blue}b}}&a \\ {\mathbf{\color{red}c}}&d&{\mathbf{\color{red}c}} \\ a&{\mathbf{\color{blue}b}} \\ {\mathbf{\color{red}c}}&d \\ a \end{array}
$$
We notice that $f\_o(\mu)=\#$ of ${\mathbf{\color{blue}b}}$'s while $c\_e(\mu)=\#$ of ${\mathbf{\color{red}c}}$'s in the respective diagrams. Therefore,
$$f\_o(\lambda)+f\_o(\lambda')=3+2=2+3=c\_e(\lambda)+c\_e(\lambda').$$


