Can the following expression be related to the multinomial formula? The following formula
$$(n,p,q,r\ \text{odd})\quad \sum_{\genfrac{}{}{0pt}{1}{p\leq q \leq r}{p+q+r =n}} \frac{n!}{p!\, q!\, r!} \times \begin{cases} 1 & \text{if}\ p< q <r\\
 \frac{1}{2} & \text{if exactly two indices are equal}\\
 \frac{1}{3!}& \text{if}\ p=q=r= \frac{n}{3} \end{cases}$$
reminds me very much of the "multinomial formula"
$$ \left( x_1 + x_2 + \cdots + x_d \right)^n = \sum_{\left|\boldsymbol{\alpha}\right|=n} \genfrac{(}{)}{0pt}{0}{\left|\boldsymbol{\alpha}\right|}{\boldsymbol{\alpha}}\, \mathbf{x}^{\boldsymbol{\alpha}}
$$
where $\boldsymbol{\alpha}\in \mathbb{N}^d,\ \left|\boldsymbol{\alpha}\right| = \alpha_1 + \alpha_2 + \cdots + \alpha_d $  and
$$ \begin{split}
 \genfrac{(}{)}{0pt}{0}{\left|\boldsymbol{\alpha}\right|}{\boldsymbol{\alpha}} & = \genfrac{(}{)}{0pt}{0}{n}{\alpha_1} \genfrac{(}{)}{0pt}{0}{n-\alpha_1}{\alpha_2}\, \cdots \, \genfrac{(}{)}{0pt}{0}{n-\alpha_1 -\alpha_2 - \cdots - \alpha_{d-1}}{\alpha_d}= \frac{n!}{\alpha_1 !\, (n-\alpha_1)!} \frac{(n-\alpha_1)!}{\alpha_2 !\, (n-\alpha_1-\alpha_2)!}\, \cdots\, \frac{(n-\alpha_1 -\alpha_2 - \cdots - \alpha_{d-1})!}{\alpha_d !\, 0 !} \\
 & = \frac{n!}{\alpha_1 !\, \alpha_2 ! \, \cdots \, \alpha_d!}
 \end{split}
$$
is the number of times the facteur $x^{\boldsymbol{\alpha}}:= x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_d^{\alpha_d}$ appears in the expansion of $\left( x_1 + x_2 + \cdots + x_d \right)^n$.
In our case $d=3,\ \boldsymbol{\alpha}= (p,q,r)$.
There could be other possibilities but I'm thinking of taking $(x_1,x_2,x_3)= \left(1, \frac{1}{2}, \frac{1}{3} \right)$ then I need to find a "simple" function of $\boldsymbol{\alpha}$, $f:\mathbb{N}^3 \to \mathbb{N}^3 $  which would take value $(1,0,0)$ if $p<q<r$ (i.e. if $\boldsymbol{\alpha}$ view as a function $1\mapsto \alpha_1=p,\ 2\mapsto \alpha_2=q,\ 3\mapsto \alpha_3=r$ is injective), take value $(0,1,0)$ if exactly two indices are equal (or if $\boldsymbol{\alpha}$ view as a the previous function takes only two values) and $(0,1,1)$ if $p=q=r$.
If we find such a function then the first sum can be rewritten
$$\sum_{\genfrac{}{}{0pt}{1}{p\leq q \leq r}{p+q+r =n}} \frac{n!}{p!\, q!\, r!} \mathbf{x}^{f(\boldsymbol{\alpha})}$$
The final step, if possible would be to relate $\mathbf{x}^{f(\boldsymbol{\alpha})}$ to $\mathbf{x}^{\boldsymbol{\alpha}}$ by integration or derivation or maybe something different.
 A: Yes, they are related. The coefficients in the case bracket account for the fact that you require $p\leqslant q \leqslant r$, whereas the usual formula for the multinomial theorem makes no such assumption about $\alpha_1$, $\alpha_2$ and $\alpha_3$.
If $p$, $q$ and $r$ are pairwise distinct, the coefficient $\binom{n}{p,q,r}$ would appear $3! = 6$ times in the multinomial formula, corresponding to the ways of assigning the values of $p$, $q$ and $r$ to $\alpha_1$, $\alpha_2$ and $\alpha_3$.
Similarly, if $p=q\neq r$, the coefficient $\binom{n}{p,q,r}$ would appear $3$ times in the multinomial formula, corresponding to the ways of assigning the values of $p$, $q$ and $r$ to $\alpha_1$, $\alpha_2$ and $\alpha_3$.
Finally, if they are all the same, the coefficient appears only once, as there is a signle way to assign the values.
So, aside from the restriction that $r$ be odd, $3!$ times your expression would be another representation of the expression you get from the multinomial theorem for $(1+1+1)^n$.
Put another way, your expression equals
$$
\frac1{3!}\sum_{\genfrac{}{}{0pt}{1}{p+q+r =n}{\max\{p,q,r\} \text{ is odd}}}
\binom{n}{p,q,r}
$$
A: I leave here my first answer but there is in fact a formula for the sum, cf. the end...

by Fimpellizieri's answer, one can get rid of the $p\leq q\leq r$ condition in the summation.
In addition, the condition $p+q+r=n$ (each quantity being odd and $n\geq 3$) is equivalent to $p'+q'+r'=n'$ with $p=2\, p' +1,\ q=2\, q' +1, r=2\, r' +1$ and $n'= \frac{n-3}{2}$.
Formally (or algorithmically), one should transform each summand of the following sum as:
$$(1+1+1)^{n'}= \sum_{p'+q'+r'=n'} \frac{n'!}{p'!\, q'!\, r'!} \quad \longrightarrow \sum_{p'+q'+r'=n'}\frac{(2\,n'+3)!}{(2\,p'+1)!\, (2\,q'+1)!\, (2\,r'+1)!}= ??$$
This looks like applying a linear form (on suitable spaces of functions of $(x,y,z)$ or polynomials in $(X,Y,Z)$). Let us start with $\mathcal{A}:=\mathbb{R}[X]$ the algebra of polynomials in $X$ (one variable). The transformations I mentionned in the question e.g. evaluation at $X:=1$ or multiple derivations and then evaluation, or integration from $a$ to $b\in \mathbb{R}$ are all linear maps from $\mathbb{R}[X]$ to $\mathbb{R}$. We want to define a map (let us denote it $L_X$) which to a basis element $X^{p'}\mapsto \frac{p'!}{(2\,p'+1)} $. And in fact repeat it for $\mathcal{A}:=\mathbb{R}[X,Y,Z]$. We are thus looking for
$$ \frac{(2\, n' + 3)!}{n'!}\ L_X \circ L_Y \circ L_Z \left( (X+Y+Z)^{n'}\right)$$

Comments/Parenthesis: One may wonder how a general linear map looks like. For $\mathcal{A}:=\mathcal{C}(X),\ X$ compact, the positive linear forms are given by so-called positive radon measure. For any other $\mathcal{A}$ containing $\mathbb{R}[X]$, the question of whether a linear form is completely defined by its value on $X^p$ is known as the moment problem. For $\mathcal{A}:=\mathcal{C}^{\infty}_c(\Omega)$ with suitable topology, the continuous linear maps are given by distributions
Edit: in fact the idea of considering linear forms on spaces of functions (specifically polynomials) is known as umbral calculus!!!


*

*Let us give an (almost non-trivial) example of a linear map from $\mathbb{R}[X]$ to itself (one can always evaluate at some point to have a map to $\mathbb{R}$): translation
$$P(X) \longmapsto P(X+a) = \left. e^{a \frac{d}{dt}} P(X+t) \right)|_{t:=0}$$
can also be defined on each basis elements by
$$ X^p \longmapsto (X+a)^p = X^p + p\, X^{p-1} a + {p \choose 2}\, X^{p-2}\, a^2 + \cdots$$
An explicit answer to the initial question would be provided by an expression of $L_X$ analogous to translation, or even in the form $e^{a \frac{d}{dt}}$ instead of an algorithm for each $X^p$.

*Using another hammer tool to solve this little problem... Our map which could be written
$$ L_X: P(X) \longmapsto \sum_{k=0}^{\infty} \frac{1}{(2k+1)!} \left.\frac{d^k P(X)}{dX^k} \right|_{X:=0}$$
commutes with translation and can thus be written as a convolution product, i.e. there exists some function $F$ (or maybe some distribution?) such that
$$ L_X\Big( P(X) \Big) = \left. \int_{\mathbb{R}} F(x-u)\, P(u)\, du \right|_{x=??}$$


Let us call $S_{\text{odd}}(n,3)$ the original sum. By an analogue the recurrence relation for Stirling numbers of the second kind, one finds for $n\geq 3$ odd
$$ S_{\text{odd}}(n+2,3) = 3\, S_{\text{odd}}(n,3) + 2\big( S(n,3)+ S(n,2)  -S_{\text{odd}}(n,3) \big) + 1 =  S_{\text{odd}}(n,3) +\, 3^{n-1} $$
$$\Longrightarrow\quad S_{\text{odd}}(2 p +1,3)= \frac{3^{2p}-1}{8}$$
