Consider the ring of polynomials $R:=\mathbb{Z}[x\_1,x\_2,x\_3]$. Define the operators $E, I:R\rightarrow R$ by $Ef(x\_1,x\_2,x\_3)=f(x\_1-1,x\_2,x\_3)$ and the identity $If=f$.


Let $\mathcal{L}:R\rightarrow R$ be the operator given by
$$\mathcal{L}f=[(x\_1+x\_2)(x\_1+x\_3)E-x\_1^2I]f.$$


Let $1$ stand for the constant function $f(x\_1,x\_2,x\_3)=1$ and $\mathcal{L}^2f=\mathcal{L}(\mathcal{L}f)$, etc.



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> **CLAIM.** Experiments suggest that $\mathcal{L}^n1$ is always a symmetric polynomial in $R$. Any proof?
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**EDIT.** This has found a resolution (see Pietro Majer's answer).


For example, $\mathcal{L} 1=e\_2$ and $\mathcal{L}^21=e\_2^2-e\_1e\_2+e\_3$ where $e\_1=x\_1+x\_2+x\_3, e\_2=x\_1x\_2+x\_1x\_3+x\_2x\_3, e\_3=x\_1x\_2x\_3$ are the standard elementary symmetric polynomials.



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> **QUESTIONS.** (EDIT) These did not find a definitive answer (apart from Brendan McKay's evidence and argument).
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> (1) Are there other orbits of symmetric polynomials under $\mathcal{L}$?
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> (2) Are there other non-trivial operators with similar property over $R$?
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> (3) What about over rings of many more variables?
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An orbit of symmetric polynomials
Suppose $f(x,y,z)$ is symmetric (in the following, *symmetric* tout court always means "symmetric w.r.to the three variables $(x,y,z)$") . Then $\mathcal{L}f(x,y,z):=(x+y)(x+z)f(x-1,y,z)-x^2f(x,y,z)$ is already symmetric w.r.to $(y,z)$, so it is symmetric if and only if it is symmetric w.r.to $(x,y)$, that is, after simplifications, if and only if $f$ satisfies the functional equation
$$(x-y)f(x,y,z)-(x+z)f(x-1,y,z)+(y+z)f(x,y-1,z)=0.$$


*Claim*: if both $f$ and $\mathcal{L}f$ are symmetric, so is $\mathcal{L}^2f$. In other words, the space of all symmetric solutions to the above functional equation is $\mathcal{L}$-invariant. 


*Proof:* According to the observation above, in order to prove the claim, we need to check that the following is identically zero:
$$(x-y)\mathcal{L}f(x,y,z)-(x+z)\mathcal{L}f(x-1,y,z)+(y+z)\mathcal{L}f(x,y-1,z) $$
and we exploit the symmetry of $\mathcal{L}f$ writing it
$$(x-y)\mathcal{L}f(x,y,z)-(x+z)\mathcal{L}f(y,x-1,z)+(y+z)\mathcal{L}f(x,y-1,z) $$
namely
$$(x-y)\big[(x+y)(x+z)f(x-1,y,z)-x^2f(x,y,z)\big]$$$$
-(x+z)\big[(y-1+x)(y+z)f(y-1,x-1,z)-y^2f(y,x-1,z)\big]$$$$
+(y+z)\big[(x+y-1)(x+z)f(x-1,y-1,z)-x^2f(x,y-1,z)\big]$$


$$=\big[(x-y) (x+y)(x+z) +(x+z) y^2 \big]f(x-1,y,z)$$$$-(x-y)x^2f(x,y,z) 
-(y+z) x^2f(x,y-1,z) =$$


$$=-x^2\big[(x-y)f(x,y,z)-(x+z)f(x-1,y,z)+(y+z)f(x,y-1,z)\big]$$


which is indeed zero, according to the initial remark, since $\mathcal{L}f$ was assumed symmetric. 


