Bernoulli Number analog using Cosine

I know that Bernoulli Numbers can be found with the generating function $$\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}x^n$$

I was wondering if any work has been done using a similar equation $$\frac{x^2}{\cos{x}-1}=\sum_{n=0}^{\infty}\frac{C_n}{n!}x^{2n}$$

I'm particularly interested in the $C_n$. Can anyone help me with a reference to work with this particular generating function or to the coefficients $C_n$?

The general formula you're looking for is

$$\dfrac{x^2}{\cos x-1}~=~\sum_{n=0}^\infty(-1)^n\frac{B_{2n}}{(2n-2)!~n}~x^{2n}$$

where $B_k$ is the $k^{th}$ Bernoulli number. Of course, the $0^{th}$ term has to be computed by evaluating the limit $\displaystyle\lim_{n\to0}\Big[(2n-2)!~n\Big]=-\frac12.$

Your sequence (negated) starts $$ 2, \frac{1}{6}, \frac{1}{60}, \frac{1}{504}, \frac{1}{3600}, \frac{1}{22176}, \frac{691}{82555200}, \frac{1}{570240}, \frac{3617}{8821612800}, \frac{43867}{414096883200}, \frac{174611}{5822264448000}, \ldots $$

We know that $$ \frac{1-\cos x}{x^2} = \frac{1}{2!} - \frac{1}{4!}x^2 + \frac{1}{6!}x^4 - \frac{1}{8!}x^6 + \cdots. $$ Let $c_n = C_n/n!$. The formula for the inverse of a power series gives $$ c_0 = 2, \quad c_n = 2\frac{c_{n-1}}{4!} - 2\frac{c_{n-2}}{6!} + 2\frac{c_{n-3}}{8!} - 2\frac{c_{n-4}}{10!} + \cdots. $$