Binomial transforms, floretions, and a divisors property

[Creighton Dement]

Sorry for the mistake, should read: 

>
>
>Conjecture: If a' = (1, a(0), a(1), ) = (1, tes[x], tes[x^2], ) is an
>integer sequence then  b' = (1, b(0), b(1), b(2), ) = (1, tes[x+e],
>tes[(x+e)^2, ) is the binomial transform of a'.
>
>Proof. tes is a linear mapping. Moreover, x and the unit e commute.
>Thus,  
>SUM C(n,k)a'(k), k=0..n = SUM C(n,k)tes[x^(k)], k=0..n
>= tes[SUM C(n,k)x^k], k=0..n

= tes[(x+e)^n]
= b'(n), q.e.d.