I found out about the Stirling numbers (first and second) when I studied a way to smooth the factorial function. This is the way I want to define them. Approach the factorial $(1+n)!$ as a product and analyse each individual sum. Each sum is a polynomial, these polynomials given the Stirling numbers. The Stirling numbers of the second kind are the diagonals of the negative integers.
I know the Stirling numbers of the second kind are according the following formula: $$S(n,k)=\sum_{m=1}^{k}\frac{(-1)^{k-m} \, m^{n}}{m! \, (k-m)!}$$ But I want to define my Stirling numbers as the vector that cancels out the Stirling numbers of the first kind, which are defined as the coefficients of $$\frac{(x+s-1)!}{(x-1)! \, s!}.$$ So, if I fill in for n=4 I'd get mirrored form the formal definition $S(4,k)={1,6,7,1}$. And if $n=7/2$ than I'd like to get $$S\left(\frac{7}{2},k\right) = \left\{ 1,\frac{35}{2^3},\frac{385}{2^7},\frac{105}{2^{10}},\frac{-21}{2^{15}},\frac{-483}{2^{18}},.... \right\}$$
Problem: The general formula shift the second Stirling numbers to the right (horizontally) and I want to shift it vertically. What is the "alternative" formula for the stirling numbers inorder to get these inbetween values?
I'm sorry for the vague formulation.