$f(x+y) = f(x) + f(y) + f^k(x) f(y) + f^k(y)f(x) $ Let $k>1$ be an integer.
Consider equations of type 
$$f_k(x+y) = f_k(x) + f_k(y) + f_k^k(x) f_k(y) + f_k^k(y) f_k(x) , f(-z) = f(z), $$
where $f_k$ is the $k$ th function and $f_k^k $ is the $k$ th power of $f_k$.
Valid for all real $x,y,z$ and $f$ is real-analytic.
Consider the nonconstant solutions.
When considering $f_2$ I came to the conclusion that $f_2(z) = 0 $ is the only solution.
( this was achieved by contradictions while investigating---and equating---$f(-1),f(0),f(1),f(2),f(3), f(2) = f(1+1) = f(3-1) $ by using algebra )
As a side note I considered the julia sets of $f_k\left(2 \operatorname{arc}f_k(z)\right) = 2 z + 2 z^{k+1} $ , more specific if they are connected or not. For $k>4$ they are totally disconnected and otherwise they are totally connected. For $k = 2$ it is a circle shape. Not sure if it relates or not?
Notice $k$ needs to be even to have $f(-z) = - f(z) $.
Although odd $k$ might be intresting too if we drop that condition.
So I wonder about general case; for what even $k$ do we get a solution and when do we fail like The case $f_2$?
How does $f_k$ behave? 
Probably like a double exponential ... but a bit more precise understanding would be nice. 
Is there uniqueness?
To be a bit more concrete I particularly wonder about 
$$f(x+y) = f(x) + f(y) + f^4(x) f(y) + f^4(y)f(x) $$
and 
$$f(x+y) = f(x) + f(y) + f^6(x) f(y) + f^6(y)f(x). $$
Do any of those 2 equations have a nonconstant solution?
Note that those are around The switch from connected to disconnected associated Julia sets.
Another remark : 
$$g(x + y) = g(x) + g(y) + g(x)g(y) $$
has the following solution $g(z)=\exp(C z) - 1$.
Maybe this implies we should study instead
$$f(x+y) = f(x) + f(y) + \frac12\left( f^k(x) f(y) + f^k(y)f(x) \right)?? $$
The related Julia sets are similar btw.
Are there representations known for these functions or their functional inverses , like integral representations , differential equations , infinite sums , analogues to eulers addition theorem or gauss integral for The agm , continued fractions , ...
 have “ very “similar things been studied before ?
( not Somos )
 A: Fix a non-negative integer $k$ and $t\in\Bbb R$.  Let $f:\Bbb R\to \Bbb R$ be a function such that$$(1)\ \ \ \ \ f(x+y)=f(x)+f(y)-\frac{t}{2}\left(f^k(x)f(y)+f(x)f^k(y)\right)$$
for all $x,y\in \Bbb R$.  (Here we interpret $0^0$ as $1$.)    Plugging in $x,y=0$ in $(1)$, we have
$$f(0)\big(1-tf^k(0)\big)=0.$$
This shows that 
$(i)$ $f(0)=0$,
$(ii)$ $t\neq 0$, $k$ is odd, and $f(0)=\frac{1}{\sqrt[k]{t}}$,
$(iii)$ $t> 0$, $k>0$ is even, and $f(0)=\frac{1}{\sqrt[k]{t}}$, or
$(iv)$ $t=1$ and $k=0$.

We first deal with $(ii)$ and $(iii)$.  Plugging in $y=0$ in $(1)$ yields 
$$\big(\sqrt[k]{t}f(x)\big)^k+\big(\sqrt[k]{t}f(x)\big)-2=0$$
for all $x\in \Bbb R$.  By Descartes' rule of signs, the polynomial $u^k+u-2=0$ has exactly one real root $u=1$ for odd $k$, and this polynomial has two real roots $u=1$ and $u=-\epsilon_k$ for even $k$ (where $\epsilon_k>1$ is the only positive root of $u^k-u-2=0$).  For even $k>0$ and for $t>0$, suppose that there exists $c\in \Bbb R$ s.t. $f(a)=-\frac{\epsilon_k}{\sqrt[k]{t}}$.  Plugging in $c$ for both $x$ and $y$ gives
$$1<\epsilon_k^2=\epsilon_k(\epsilon_k^k-2)\in\{1,-\epsilon_k\}.$$
This is absurd, so $c$ does not exist.  Therefore, in both cases $(ii)$ and $(iii)$, the only solution is 
$$(2)\ \ \ \ \ f(x)=\frac1{\sqrt[k]{t}}\ \forall x$$

We now deal with the case $k=0$ separately (and this will complete the case $(iv)$ automatically).  From $(1)$, we see that
$$f(x+y)=\left(1-\frac{t}{2}\right)\big(f(x)+f(y)\big).$$
If $t=0$, we have Cauchy's functional equation, whose solutions are well-known and the only analytic solutions are the functions $f$ of the form $$(3)\ \ \ \ \ f(x)=m x$$ for some constant $m$.  If $t=1$, taking $y=0$ we have the constant solution $$(4)\ \ \ \ \ f(x)=f(0)\ \forall x.$$   If $t\notin\{0,1\}$, then $f(0)=0$.  By taking $y=0$, $$(5)\ \ \ \ \ f(x)=0\ \forall x.$$

For the remaining part of this answer, we assume that $k>0$ and $f(0)=0$.  If $t=0$, we get once again Cauchy's functional equation, and we already know the solutions.  Suppose from now on that $t\ne0 $. 
If $k=1$, we let $g(x)=1-tf(x)$.  Then $$g(x+y)=1-tf(x+y)=1-tf(x)-tf(y)+t^2f(x)f(y)=g(x)g(y).$$
This is another well-known functional equation.  Its solution is $g(x)=0$ for all $x$, or $g(x)=\exp\big(\gamma(x)\big)$ where $\gamma$ is a solution to Cauchy's functional equation.  In particular, this means that all analytic solutions are
$$(6)\ \ \ \ \ f(x)=\frac1t\ \forall x\ \wedge\ f(x)=\frac{1-\exp(\lambda x)}{t}\ \forall x.$$

Suppose now that $k>1$.  At this point I gave up on finding general solutions to $(1)$, but it is possible to show that the only analytic solution to $(1)$ is the zero function $f(x)=0$.  To see this, take derivative of $(1)$ wrt $y$ to get
$$(7)\ \ \ \ \ f'(x+y)=f'(y)-\frac{t}{2}f(x)f'(y)\left(f^{k-1}(x)+kf^{k-1}(y)\right).$$
Putting $y=0$ we get
$$(8)\ \ \ \ \ f'(x)=f'(0)\left(1-\frac{t}{2}f^{k}(x)\right).$$
Using $(8)$, $(7)$ becomes
$$f'(0)\left(1-\frac{t}{2}f^k(x+y)\right)=f'(0)\left(1-\frac{t}{2}f^{k}(y)\right)\left(1-\frac{t}{2}f(x)\left(f^{k-1}(x)+kf^{k-1}(y)\right)\right).$$
By symmetry,
$$f'(0)\left(1-\frac{t}{2}f^k(x+y)\right)=f'(0)\left(1-\frac{t}{2}f^{k}(x)\right)\left(1-\frac{t}{2}f(y)\left(f^{k-1}(y)+kf^{k-1}(x)\right)\right).$$
If $f'(0)\ne 0$, then
\begin{align}\left(1-\frac{t}{2}f^{k}(y)\right)&\left(1-\frac{t}{2}f(x)\left(f^{k-1}(x)+kf^{k-1}(y)\right)\right)\\&=\left(1-\frac{t}{2}f^{k}(x)\right)\left(1-\frac{t}{2}f(y)\left(f^{k-1}(y)+kf^{k-1}(x)\right)\right)\end{align}
and so
$$f(x)f(y)\Big(2f^{k-2}(x)-2f^{k-2}(y)-tf^{2k-2}(x)+tf^{2k-2}(y)\Big)=0.$$
If $f$ is non-zero, then 
$$2f^{k-2}(x)-2f^{k-2}(y)-tf^{2k-2}(x)+tf^{2k-2}(y)=0$$
for all $x,y$.  Differentiating the previous equation wrt $y$ yields
$$2(k-2)f^{k-3}(x)f'(x)-(2k-2)tf^{2k-3}(x)f'(x)=0.$$
Observe that $f'$ is also non-zero.  Therefore 
$$2(k-2)-(2k-2)tf^{k}(x)=0$$
or
$$f^k(x)=\frac{k-2}{(k-1)t}$$
so $f$ is constant, but this contradicts the fact that $f'$ is non-zero.

In conclusion, an analytic solution to $(1)$ is $f(x)=0$.  There are exceptional solutions, depending on the parameters $t$ and $k$.  These (analytic) solutions are listed below.
$(a)$ $t=0$: $f(x)=m x$.
$(b)$ $t\neq 0$ and $k$ is odd: $f(x)=\frac{1}{\sqrt[k]{t}}$.
$(c)$ $t>0$ and $k>0$ is even: $f(x)=\frac{1}{\sqrt[k]{t}}$.
$(d)$ $t=1$ and $k=0$: $f(x)=\kappa$.
$(e)$ $t\ne 0$ and $k=1$: $f(x)=\frac1t$.
$(f)$ $t\ne 0$ and $k=1$: $f(x)=\frac{1-\exp(\lambda x)}{t}$.
There are non-analytic solutions at least in the case $t=0$ and the case $t\ne 0$ with $k=1$.  There may be non-analytic solutions in the case $t\ne 0$, $k>1$, and $f(0)=0$, but I have no idea.
