The binomial Sheffer sequence of Bell / Touchard / exponential polynomials $\phi\_n (x) $, whose coefficients are the Stirling numbers of the second kind, have the representation 


$(RL)^n=\phi\_n (:RL:) $


where $R $ and $L $ are the raising and lowering operators of any sequence of Sheffer polynomials and $:RL:^n=R^n \; L^n$ by definition (a notational convenience).


Are there other polynomial functions of the ladder operators $K(L,R)$ such that 


$ K^n(L,R) = u\_n(B.(L,R))$ 


for $u\_n(x)$ a Sheffer sequence and $(B.(L,R))^n = B\_n(L,R)$ a sequence of operator polynomials?


(Other than for the trivial case $u\_n(x) = x^n$.)


Note that $R^nL^n= (RL)\_n = (RL)!/(RL-n)!$, the falling factorial polynomials, whose coefficients are the signed Stirling numbers of the first kind.



A uniqueness of the Stirling numbers?
Well, upon further reflection, the answer is that any pair of Sheffer sequences that are an umbral inverse pair suffice to construct such an arrangement. Let $u\_n (x)$ and $v\_n (x)$ be such a pair, i.e., 


$u\_n (v.(x)) = x^n=v\_n(u.(x))$.


Then


$K^n = u\_n(v.(K))$ and $B\_n(K) = v\_n (K) $.


So, the Stirling pair is not unique.


