Here $x_1>0$ is the initial condition and $x_{n+1}$ is defined by

$$x_{n+1}=\Big[\frac{1}{n}\sum_{k=1}^n x_k^2 -\frac{1}{n^2}\Big(\sum_{k=1}^n x_k\Big)^2 \Big]^{1/2}. $$ Clearly, $x_n=\lambda_n \cdot x_1$ where $\lambda_1,\lambda_2,\dots$ is a sequence of positive real numbers not depending on the initial condition $x_1$. For instance, $\lambda_1=1,\lambda_2=0$. It seems, based on empirical evidence, that $\lambda_n\sim \gamma \cdot n^{-\alpha}$ for some constants $\gamma\approx 0.54$ and $\alpha\approx 0.35$.

Substituting $x_k$ by its asymptotic expansion in the first formula defining the sequence $(x_n)$, one gets the following, after trivial computations and rearrangements: $\alpha^2=(1-2\alpha)(1-\alpha)^2$. Let $\tau=(13+3\sqrt{33})^{1/3}$. Then we have, according to Mathematica:

$$\alpha=\frac{1}{3}\Big(2+\frac{2^{5/3}}{\tau}-\frac{\tau}{2^{2/3}}\Big)\approx 0.3522011.$$

Is this correct? Also, I could not find how to determine the value of $\gamma$. Could you express $\gamma$ as a solution of some equation (as I did for $\alpha$) or using a series or anything else?

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    $\begingroup$ Put $\lambda_n=\gamma\,n^{-\alpha}\,(1+\mu_n)$, then empirically $\mu_n=O(1/n)$; this implies $$0=\zeta(\alpha)+\sum_{n=1}^\infty n^{-\alpha}\mu_n=\zeta(2\alpha)+\sum_{n=1}^\infty n^{-2\alpha}(2\mu_n+\mu_n^2)$$ (even under weaker assumptions), and $\mu_n=\beta/n+o(1/n)$ results in $$\beta=\frac\alpha2-\frac{1-3\alpha^2}7,$$ so that $n^\alpha(1-\beta/n)\lambda_n$ is a better estimate for $\gamma$ than just $n^\alpha\lambda_n$. $\endgroup$
    – metamorphy
    Commented Aug 4, 2022 at 8:55
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    $\begingroup$ Further, it seems that $\mu_n-\beta/n=O(n^{-\rho})$ with $\rho=(1-\alpha)(3-2\alpha)$. $\endgroup$
    – metamorphy
    Commented Aug 4, 2022 at 9:22


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