This is essentially a request for counterexamples, since I know so few $n$-buds (or as some might say, formal group law $n$-chunks). One notices that the only $1$-bud of maximal degree 1 is the additive one, $x+y$, which also happens to be a formal group law. Also, the only 2-buds that I know of over a ring $R$ are of the form $x+y+cxy$ for some $x\in R$, which also defines a formal group law (here I'm assuming $R$ is commutative and unital). Presumably this pattern does not continue infinitely? That is to say, there is a finite polynomial $f=x+y+h.o.t.$ of total degree $n$ such that $f(f(x,y),z)-f(x,f(y,z))$ is equivalent to zero modulo degree $n+1$ terms (i.e. it defines an $n$-bud) but is not equal to zero?



Is every (one dimensional) n-bud of total degree n also a formal group law?
If $f(x, y)$ is a formal group law then so is $g(f(g^{-1}(x), g^{-1}(y))$ where $g$ is an invertible (under composition) formal power series. This suggests a strategy for writing down $n$-buds, namely pick a polynomial $g$, a group law $f$, and a polynomial approximation $h$ to $g^{-1}$ and then compute $g(f(h(x), h(y))$. If $h$ agrees with $g^{-1}$ modulo degree $n+1$ terms then this gives an $n$-bud. 


For $g$ let's pick $g(x) = x + x^2 + x^3$ and for $f$ let's pick $f(x, y) = x + y$. For $h$ let's pick $h(x) = x - x^2 + x^3$, which agrees with the compositional inverse modulo degree $4$ terms, so $g(h(x) + h(y))$ is a $3$-bud. Expanding, this is (modulo degree $4$ terms)


$$x + y + 2xy + x^2 y + y^2 x.$$


But this shouldn't give a formal group law. (I tried to check this in Sage but it's not happy with expanding the associator here.) 


