Example of operation that is tree associative, but not generally associative In a lot of algorithms using trees, we need the property that when folding $2^n$ elements with some operator $+$, we can do the first half of $2^{n-1}$ elements and the second half independently. That is
$$((x_1+x_2)+x_3)+x_4 = (x_1+x_2)+(x_3+x_4)$$
which then implies by repeated application
$$
\begin{align*}
((((((x_1+x_2)+x_3)+x_4)+x_5)+x_6)+x_7)+x_8 
&= (((((x_1+x_2)+x_3)+x_4)+x_5)+x_6)+(x_7+x_8)
\\&= ((((x_1+x_2)+x_3)+x_4)+(x_5+x_6))+(x_7+x_8)
\\&= ((x_1+x_2)+(x_3+x_4))+((x_5+x_6)+(x_7+x_8))
\end{align*}
$$, and so on for larger $n$.
Clearly, associativity is sufficient, but I wonder if you can think of any structures with an operator that only satisfies this weaker condition? (Notice that it is only defined for folds with length $2^n$ for some $n$). I can't think of any, but I don't see why not?
Update:
I guess for any group the two notions are the same, since we can write
$$(x_1+x_2)+x_3 = ((0+x_1)+x_2)+x_3 = (0+x_1)+(x_2+x_3) = x_1+(x_2+x_3)$$, so maybe the answer is negative.
 A: I'm supposing that $x_1+\ldots+x_k$ in $(S,+)$ without parenthesis means $$(x_1+(\cdots+(x_{k-1}+x_k)\cdots),$$ a "left-handed" association. WLOG I'll use $n=3$. Suppose first that there is a right neutral element $0$ for $+$, so that $x+0=x$ for every $x\in S$.  Then $(S,+)$ is associative, for
$$(a+b)+c=((a+b)+(c+0))+((0+0)+(0+0))=a+(b+(c+(0+(0+(0+(0+0)))))) = a+(b+c).$$
Now, if $(S,+)$ has no right neutral element, chances are that we can just add it to $S$: the defining "folding" property $F$ of $+$ does not forbid it, and then $(S,+)\subseteq(S\cup\{0\},+)$ would inherit associativity.
So, for $(S,+)$ to be nonassociative, it must satisfy some forbidding extra condition $f(x_1,\ldots,x_r)$, such that some substitution $x_i=0$ renders an incompatibility. For example, $f:x+x=a$ with $a\neq0$ would do, and it doesn't look like this property along with folding are enough to achieve associativity, so some little free algebra in the universal sense with identities $F,f$ (and derived ones), but with enough elements, possibly does the trick.
