Let $F([x,x_2,...x_n],...[y,y_2,...y_n])=[z,z_2,....z_n]$ denote a function with $2$ parameters, each is a vector containing $n$ variables. Let $U_k$ be defined by the following recurrence relation:
$U_k = F([x,x_2,...x_n],U_{k-1})$ where the intitial value $U_1=[x,x_2,...x_n]$.
Let $U_{k,k_2} = F(U_k,U_{k_2})$
How can I determine if $F$ is a function such that
$U_{k,k_2} = U_{k+k_2}$
will hold for all integers $k$ and $k_2$? Furthermore, how can I construct one where the number of variables $n$ is a fixed integer?
I'm not sure of the proper terminology for such functions, but for now, I will refer to them as "completely additive" because the addition properties involving $U_{k,k_2}, U_k, U_{k_2}$ will hold.
Simple examples of completely additive functions are addition:
$F(x,y)=x+y$
$U_1 = x$, $U_2 = F(x,x) = 2x$, $U_3= F(x,2x) = 3x,...$
$U_{k}=kx$
$U_{k,k_2}=F(U_k, U_{k_2}) = F(kx, k_2x) = (k + k_2)x$
and multiplication:
$F(x,y)=x\times y$
$U_1 = x$, $U_2 = F(x,x) = x^2$, $U_3= F(x,x^2) = x^3,...$
$U_{k}=x^k$
$U_{k,k_2}=F(U_k, U_{k_2}) = F(x^k, x^{k_2}) = x^{k+k_2}$
Here is another example of a completely additive function involving four variables (this is derived from matrix multiplication):
$F([a,b,c,d],[e,f,g,h])=[ae+bg, af+bh, ce+dg, cf+dh]$
Yet, when the "dot products" are rearranged, in some cases, $F$ is no longer a completely additive function. For example,
$F([a,b,c,d],[e,f,g,h])=[ae+ch, af+bg, bf+dh, cg+de]$
is not completely additive.
I already know matrix multiplication is not commentative, so requiring that $F$ be a communitive function is non-essential for it to be a completely additive function. So then that leaves the question, what properties should $F$ have for it to be completely additive?
Expanding both sides of $U_{k,k_2} = U_{k+k_2}$ gives
$$F(U_k,U_{k_2}) = F(U_1,U_{k+k_2-1})$$
Using $k=2$ gives
$$F(U_2,U_{k_2}) = F(U_1,U_{k_2+1})$$
$$F(F(U_1,U_1),U_{k_2}) = F(U_1,F(U_1,U_{k_2}))$$
Because $U_1$ and $U_{k_2}$ can be anything, this can be seen a special form of the associative property in which the first two variables are equal.
In a similar way to $k=2$, for any $k$, the formula can be expanded on both sides to a mess of $k$ instances of $F$, $k$ instances of $U_1$, and one $U_{k_2}$ at the end. By our modified associative property, these $F$s and $U_1$s can be arranged in any order. Therefore both sides are equal. This means that the modified associative property is both necessary and sufficient to show that $F$ is completely additive.
I am not sure whether the modified version is equivalent to the original associative property.