Because the factor in the exponential has the power series : \frac{\sqrt{1-t^2}-1}{t} = -\sum_{k\ge 0} C_k \left(\frac{t}{2}\right)^{2k+1} in terms of
Catalan numbers C_k, the coefficient in front of x^k of the polynomial can be written as :[x^k] s_n(x) =(-1)^k\frac{n!}{k!2^n}\sum_{n=l_1+l_2+\cdots +l_k}C_{(l_1-1)/2}C_{(l_2-1)/2}\cdots C_{(l_k-1)/2}, according to the general formula for
generalized Appell polynomials, where the sum is over all
compositions n=l_1+l_2+\cdots+l_k of n into k positive odd integers. The
empty product appearing for k=n=0 equals 1. Special values, where all contributing Catalan numbers equal 1, are : [x^n]s_n(x) = \frac{(-1)^n}{2^n}. : [x^{n-2}]s_n(x) = \frac{(-1)^n n(n-1)(n-2)}{2^n}. n and k=1 : [x]s_n(x) = -\frac{n!}{2^n}C_{(n-1)/2} --> By differentiation the recurrence for the first derivative becomes s'(x) =- \sum_{m=0}^{n-1} \frac{n!}{m!2^{n-m}} C_{(n-1-m)/2}s_m(x), where the sum is over the m such that (n-1-m)/2 is integer. --> : s'(x) =- \sum_{k=0}^{\lfloor (n-1)/2\rfloor} \frac{n!}{(n-1-2k)!2^{2k+1}} C_k s_{n-1-2k}(x). The first few of them are :s_0(x)=1; :s_1(x)=-\frac{1}{2}x; :s_2(x)=\frac{1}{4}x^2; :s_3(x)=-\frac{3}{4}x-\frac{1}{8}x^3; :s_4(x)=\frac{3}{2}x^2+\frac{1}{16}x^4; :s_5(x)=-\frac{15}{2}x-\frac{15}{8}x^3-\frac{1}{32}x^5; :s_6(x)=\frac{225}{8}x^2+\frac{15}{8}x^4+\frac{1}{64}x^6; ==Sheffer sequence==