<s>[INST] Looking for the logic of a sequence from convolution of probability distributions I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2
   \left(x^2+60\right)}{\left(x^2+12\right)^3},\frac{3 \left(x^4+162
   x^2+9477\right)}{\left(x^2+27\right)^4},\frac{4 \left(x^6+324 x^4+44928
   x^2+2847744\right)}{\left(x^2+48\right)^5},\frac{5 \left(x^8+564 x^6+141750
   x^4+19912500 x^2+1388390625\right)}{\left(x^2+75\right)^6},\frac{6 \left(x^{10}+900
   x^8+366120 x^6+87829920 x^4+13038019200
   x^2+998326798848\right)}{\left(x^2+108\right)^7}, \dots \right)$$
 [/INST] I played a little bit with Jack's answer. One can expand with the Newton formula
$$ (1 + \sqrt{3} t)^n = \sum_{k=0}^n \binom{n}{k} (\sqrt{3}t)^k $$
then the integrals can be performed using
$$ 
\int_0^\infty dt\, t^k e^{- t} \cos(s t) =  \frac{T_{1+k}\bigl(1/\sqrt{1 + s^2}\bigr) k!}{(1+s^2)^{(k+1)/2}} $$
where $T_a(x)$ is the T-Chebyshev polynomial. Then, the given sequence can be rewritten setting $s = x/\sqrt{3}$ as
$$ S_n = \frac{1}{18(s^2 + n^2)^{n+1}} \sum_{k=0}^n \frac{n!}{(n-k)!} (s^2 + n^2)^{n -\frac{k - 1}{2}} T_{1+k}\bigl(1/\sqrt{1 + s^2/n^2}\bigr)
$$
</s>