It would appear that sequences are not their own inverses. When we evaulate the Laurent series for the sequence \(2,6,48,594,10520\), we find \[...=\frac{2}{x}-\frac{12}{x^3}+\frac{648}{x^5}-\cdots\] with the next term of order \(100,000\). The only possible listed match is OEIS:A135397. The sequence \(1,6, 324, 157464\cdots\). The most important thing being, it is not the sequence of prime numbers, therefore the transform is not it’s own inverse!
We then describe the transform as a mapping from sequence \(S_k\) to sequence \(T_k\). By \[\underset{k=1}{\overset{\infty}{\large K \normalsize}} \frac{S_k}{x} = \sum_{i=k}^\infty \frac{(-1)^{k+1}T_k}{x^{2k-1}}\] where the \(T_k\) are defined by \[T_1 = \lim_{x\to\infty}\frac{\underset{k=1}{\overset{\infty}{\large K \normalsize}} \frac{S_k}{x}}{1/x}\\ T_2 = \lim_{x\to\infty}\frac{\left[\underset{k=1}{\overset{\infty}{\large K \normalsize}} \frac{S_k}{x}\right] -\frac{T_1}{x}}{1/x^3}\\ T_3 = \lim_{x\to\infty}\frac{\left[\underset{k=1}{\overset{\infty}{\large K \normalsize}} \frac{S_k}{x}\right] -\frac{T_1}{x}+\frac{T_2}{x^3}}{1/x^5}\\ T_n = \lim_{x\to\infty}\frac{\left[\underset{k=1}{\overset{\infty}{\large K \normalsize}} \frac{S_k}{x}\right] + \sum_{k=1}^{n-1} \frac{(-1)^kT_k}{x^{2k-1}}}{1/x^{2n-1}}\\\]
To test if an integer sequence is always produced when an integer sequence is input, a completely “Random” sequence, or at least random enough was used, the digits of \(\pi\). This results in a Laurent Series \[\pi(x) = \frac{3}{x} - \frac{3}{x^3} + \frac{15}{x^5} - \frac{87}{x^7} + \frac{565\pm1}{x^9} - \frac{\approx 4500}{x^11}\cdots\] if we divid the coefficients by \(3\) we extract the sequence \(1,1,5,29,\cdots\) which may be many on the OEIS, however, the term \(565\pm1\) does not fit any of them, and it is highly unexpected that the sequence of pi’s digits should transform into a meaningful sequence.