We know the Arithmetic-Geometric mean:
$$a_0=a, \qquad b_0=b$$
$$a_n=\frac{a_{n-1}+b_{n-1}}{2}, \qquad b_n=\sqrt{a_{n-1} b_{n-1}}$$
$$\text{agm} (a,b)=\lim_{n \to \infty}a_n=\lim_{n \to \infty}b_n$$
Now we introduce a family of modified 'iterated means':
$$a_n=\frac{a_{n-1}+b_{n-1}}{2}\left(1-\frac{(a_{n-1}-b_{n-1})^2}{(a_{n-1}+b_{n-1})^2} \right)^p$$
$$b_n=\sqrt{a_{n-1} b_{n-1}}\left(1-\frac{(a_{n-1}-b_{n-1})^2}{(a_{n-1}+b_{n-1})^2} \right)^q$$
$$M_{pq} (a,b)=\lim_{n \to \infty}a_n=\lim_{n \to \infty}b_n$$
I obtained the following numerical results:
$$M_{10} (a,b)=\frac{ab}{\text{agm}(a,b)}$$
(This is actually the Harmonic-Geometric mean, and this result is known and easy to prove).
$$M_{11} (a,b)=\frac{a^2b^2}{\text{agm}^3(a,b)}$$
$$M_{21} (a,b)=\frac{a^3b^3}{\text{agm}^5(a,b)}$$
$$M_{22} (a,b)=\frac{a^4b^4}{\text{agm}^7(a,b)}$$
$$M_{32} (a,b)=\frac{a^5b^5}{\text{agm}^9(a,b)}$$
I think the pattern is easy to see here. We increase $p$ by one and leave $q$ the same. Then we increase $q$ by one as well.
For negative $p,q$ we have a corresponding situation:
$$M_{0-1} (a,b)=\frac{\text{agm}^3(a,b)}{ab}$$
$$M_{-1-1} (a,b)=\frac{\text{agm}^5(a,b)}{a^2b^2}$$
And so on.
How can we prove this observation? Is there some way to prove it for the general case, because I find it unlikely that we can directly prove something like:
$$M_{100,100} (a,b)=\frac{a^{200}b^{200}}{\text{agm}^{399}(a,b)}$$
In particular cases, the 'modification' introduced here turns one mean into another.
$$\frac{a+b}{2}\left(1-\frac{(a-b)^2}{(a+b)^2} \right)=\frac{a+b}{2} \frac{4 ab}{(a+b)^2}=\frac{2 ab}{a+b}$$
$$\frac{a+b}{2}\left(1-\frac{(a-b)^2}{(a+b)^2} \right)^{1/2}=\frac{a+b}{2} \frac{2 \sqrt{ab}}{a+b}=\sqrt{ab}$$
$$\sqrt{ab} \left(1-\frac{(a-b)^2}{(a+b)^2} \right)^{1/2}=\frac{2a b }{a+b}$$
$$\sqrt{ab} \left(1-\frac{(a-b)^2}{(a+b)^2} \right)^{-1/2}=\frac{a+b}{2}$$
etc. In the general case it probably doesn't always result in a mean.
And the 'iterated means' above certainly aren't means in the general case, since the limits for large positive or negative $p,q$ can go to zero or infinity.
I'll show (skipping some parts of the proof) $$M_p(a,b):=M_{p,p}(a,b)=(ab)^{2p} \mathrm{agm}(a,b)^{1-4p},$$ in the same way one can show $$M_{p,p-1}(a,b)=(ab)^{2p-1} \mathrm{agm}(a,b)^{3-4p}.$$ Let us write $S(a,b)=\frac12 (a+b)$ and $P(a,b) = \sqrt{ab}$. One can rewrite the "$p$ iterated means" as follows: \begin{align} a_n & = Q_1(a_{n-1},b_{n-1}) := S(a_{n-1},b_{n-1})^{1-2p} P(a_{n-1},b_{n-1})^{2p},\\ b_n & = Q_2(a_{n-1},b_{n-1}) := S(a_{n-1},b_{n-1})^{-2p} P(a_{n-1},b_{n-1})^{2p+1}. \end{align} Let us also define $F_p(a,b) = (ab)^{2p} \mathrm{agm}(a,b)^{1-4p}$, we want to prove $M_p(a,b) = F_p(a,b)$. The key observation is that we have (this follows immediately from the definition), $$\mathrm{agm}(a,b) = \mathrm{agm}(S(a,b),P(a,b)) \label{1}\tag{1}.$$ Suppose we can prove $$ F_p(a,b) = F_p(Q_1(a,b),Q_2(a,b)) = F_p(a_1,b_1) \label{2} \tag{2}.$$ Iterating the same equality (and using the continuity of $F_p$) we find \begin{align} F_p(a,b) & = F_p(a_1,b_1) = F_p(a_2,b_2) = \dots = F_p(a_n,b_n) = \dots = F_p(M_p(a,b),M_p(a,b))\\ & = (M_p(a,b) \cdot M_p(a,b))^{2p} \cdot \mathrm{agm} (M_p(a,b),M_p(a,b))^{1-4p} = M_p(a,b), \end{align} where we used $\mathrm{agm}(\lambda, \lambda) = \lambda$.
To prove $\eqref{2}$ we use property $\eqref{1}$. We have \begin{align} F_p(a_1,b_1) & = F_p(Q_1(a,b),Q_2(a,b)) = (Q_1(a,b) \cdot Q_2(a,b))^{2p} \cdot \mathrm{agm}(Q_1(a,b), Q_2(a,b))^{1-4p}\\ & = \left( S(a,b)^{1-2p} P(a,b)^{2p} \cdot S(a,b)^{-2p} P(a,b)^{2p+1} \right)^{2p} \cdot \\ & \quad \mathrm{agm}\left( S(a,b)^{1-2p} P(a,b)^{2p} , S(a,b)^{-2p} P(a,b)^{2p+1} \right)^{1-4p}\\ & = S(a,b)^{2p(1-4p)} P(a,b)^{2p(4p+1)} \cdot \left( S(a,b)^{-2p} P(a,b)^{2p} \mathrm{agm}\left( S(a,b), P(a,b) \right)\right)^{1-4p}\\ & = P(a,b)^{4p} \mathrm{agm}\left( a, b \right)^{1-4p}\\ & = F_p(a,b). \end{align}