$\DeclareMathOperator\sym{sym}$Let $i$, $j$, $k$ be the units of quaternions, in particular $i^2=j^2=k^2=-1$, $ijk=-1$.


We will use non commutative variables $x$, $y$, $z$. Define $\sym\_{a,b,c}$ to be the polynomial made of the sum of monomials which are all possible products of $a$ variables $x$, $b$ variables $y$ and $c$ variables $z$.


For example $\sym\_{2,1,0}(i,j,k)=i^2 j+i j i+j i^2$.


Considering the symmetric definition of $\sym\_{a,b,c}$ and the non commutativity of quaternions I would expect a lot of simplifications on $\sym\_{a,b,c}(i,j,k)$, in fact $\sym\_{a,0,0}(i,j,k)=i^a$, and $\sym\_{1,1,1}(i,j,k)=0$ but $\sym\_{2,2,0}(i,j,k)=i^2 j^2+j^2 i^2=2$.


Is there a simple formula to determine the value of $\sym\_{a,b,c}(i,j,k)$ in terms of $a$, $b$, $c$?




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May/29/2019
By the way I introduced those functions we have:
$e^{(i+j+k)/\sqrt3}=\sum\_n (\sqrt{3}^n/n!)\sum\_{a,b,c} \sym\_{a,b,c}(i,j,k) =(i+j+k)\sin(90)=i+j+k$.



Possible values of symmetric functions evaluated on quaternions
Careful counting gives the following formula:
$$ \operatorname{sym}\_{a,b,c}(i,j,k) = 
{\left\lfloor \frac{a}{2}\right\rfloor +\left\lfloor \frac{b}{2}\right\rfloor +\left\lfloor \frac{c}{2}\right\rfloor \choose \left\lfloor \frac{a}{2}\right\rfloor ,\left\lfloor \frac{b}{2}\right\rfloor ,\left\lfloor \frac{c}{2}\right\rfloor }\cdot \delta(a,b,c) \cdot i^a j^b k^c $$
where
$$ \delta(a,b,c) = \left(1-(a\%2)\cdot(b\%2)-(b\%2)\cdot(c\%2)-(c\%2)\cdot(a\%2)+2(a\%2)(b\%2)(c\%2)\right) $$
is $0$ when at least two of $a$, $b$, $c$ are odd, and $1$ otherwise.
One might further evaluate using
$$ i^aj^bk^c = (-1)^{bc+\frac{c(c-1)}{2}+\left\lfloor \frac{a+c}{2}\right\rfloor +\left\lfloor \frac{b+c}{2}\right\rfloor }\cdot i^{(a+c)\%2}j^{(b+c)\%2}
$$
to actually get a multiple of one of the basis elements.


