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As for the limit moments $\gamma_k$: if $\alpha=1$ then clearly ${\rm Tr}(A_\alpha^k) = n^k$ so $\gamma_k=\infty$ once $k>1$. So we assume $\alpha < 1$, and then we may as well take $\alpha \in {\bf C}$ with $|\alpha| < 1$. Then $\gamma_1$, $\gamma_2$, $\gamma_3$, $\gamma_4$, $\gamma_5$, etc. are $$ 1, \ \frac{1+\alpha}{1-\alpha},\ \frac{1+4\alpha+\alpha^2}{(1-\alpha)^2},\ \frac{1+9\alpha+9\alpha^2+\alpha^3}{(1-\alpha)^3},\ \frac{1+16\alpha+36\alpha^2+16\alpha^3+\alpha^4}{(1-\alpha)^4}, \ldots $$ and in general $\gamma_k = P_{k-1}(\alpha) / (1-\alpha)^{k-1}$ where $$ P_m(X) := \sum_{j=0}^m \left({m \atop j}\right)^2 X^j $$ is the polynomial obtained from the binomial expansion of $(1+X)^m$ by squaring each coefficient. These $P_m$ don't have an entirely elementary formula, but they can be written as hypergeometric polynomials, or (if memory serves) expressed in terms of Legendre polynomials, or manipulated using the generating function $$ \sum_{m=0}^\infty P_m(X) t^m = \left((\alpha-1)^2 t^2 - 2(\alpha+1)t + 1\right)^{-1/2} $$ if I did this right (I guessed the formula using the technique I described here a few weeks ago: Determining a generating function (of a restricted form)).

As for the limit moments $\gamma_k$: if $\alpha=1$ then clearly ${\rm Tr}(A_\alpha^k) = n^k$ so $\gamma_k=\infty$ once $k>1$. So we assume $\alpha < 1$, and then we may as well take $\alpha \in {\bf C}$ with $|\alpha| < 1$. Then $\gamma_1$, $\gamma_2$, $\gamma_3$, $\gamma_4$, $\gamma_5$, etc. are $$ 1, \ \frac{1+\alpha}{1-\alpha},\ \frac{1+4\alpha+\alpha^2}{(1-\alpha)^2},\ \frac{1+9\alpha+9\alpha^2+\alpha^3}{(1-\alpha)^3},\ \frac{1+16\alpha+36\alpha^2+16\alpha^3+\alpha^4}{(1-\alpha)^4}, \ldots $$ and in general $\gamma_k = P_{k-1}(\alpha) / (1-\alpha)^{k-1}$ where $$ P_m(X) := \sum_{j=0}^m \left({m \atop j}\right)^2 X^j $$ is the polynomial obtained from the binomial expansion of $(1+X)^m$ by squaring each coefficient. These $P_m$ don't have an entirely elementary formula, but they can be written as hypergeometric polynomials, or (if memory serves) expressed in terms of Legendre polynomials, or manipulated using the generating function $$ \sum_{m=0}^\infty P_m(X) t^m = \left((\alpha-1)^2 t^2 - 2(\alpha+1)t + 1\right)^{-1/2} $$ if I did this right (I guessed the formula using the technique I described here a few weeks ago: Determining a generating function (of a restricted form)).

In our case we can evalute the integral for $\gamma_k$ by writing it as the contour integral $$ \frac1{2\pi i}\oint_{|z|=1} f_\alpha(z)^k \frac{dz}{z}. $$ For each $k \geq 1$ the integrand has a pole of order $k$ at $z = \alpha$ and no other poles in $|z| \leq 1$; evaluating the residue at this pole yields the formula $\gamma_k = P_{k-1}(\alpha) / (1-\alpha)^{k-1}$ given above.

In our case we can evalute the integral for $\gamma_k$ by writing it as the contour integral

$$ \frac1{2\pi i}\oint_{|z|=1} f_\alpha(z)^k \frac{dz}{z}. $$

For each $k \geq 1$ the integrand has a pole of order $k$ at $z = \alpha$ and no other poles in $|z| \leq 1$; evaluating the residue at this pole yields the formula $\gamma_k = P_{k-1}(\alpha) / (1-\alpha)^{k-1}$ given above.

f(a,n) = factor(charpoly(matrix(n,n,i,j,a^abs(j-k))))
F = f(1/2,21)
vector(#F[,1], n, polgalois(F[n,1]))

f(a,n) = factor(charpoly(matrix(n,n,i,j,a^abs(i-j))))
F = f(1/2,21)
vector(#F[,1], n, polgalois(F[n,1]))