# Ordinary Generating Function For the Unsigned Stirling numbers of the First Kind

On Wikipedia Here, the exponential generating function $$\sum_{n=k}^{\infty}{(-1)^{n-k}{n\brack k}\frac{z^n}{n!}}=\frac{1}{k!}(\log(1+z))^k$$ is given, where $${n\brack k}$$ is the unsigned Stirling numbers of the first kind. I have done a literature search to see if I could find a similar but ordinary generating function for the unsigned Stirling numbers of the first kind, but I haven't found any.

Could it be that I am not doing a proper search, or no ordinary generating functions for the unsigned Stirling numbers of the first kind are known? Can someone refer me to some examples they might have seen?

I should mention that I have seen this one: $$\sum_{k=0}^{n}{{n\brack k}x^k}=x(x+1)(x+2)(x+3)\cdots(x+n-1),$$ but I am talking about a generating function of a similar kind where the upper summation index is infinity, just as in the case of the exponential generating function I quoted earlier.

• Then you want to write down the OGF of what sequence? Aug 5, 2021 at 14:44
• There are generating functions at dlmf.nist.gov/26.8#ii and functions.wolfram.com/IntegerFunctions/StirlingS1/11 but I'm not sure if any of them fit your needs. Aug 5, 2021 at 15:44
• Thank you. I have taken a look at them...but I couldn't find what I was looking for. Aug 5, 2021 at 15:48
• In context, I think you will find what you want in Roman's "The Umbral Calculus" (it's cheap) Chapter 4, section 1, at the end of Misc. You have to do some flimflam to get the functional (1-e^(a*t)) into generating function form; but the formulas are at the beginning of section 1. If you want context/properties, you have to read the preceding chapters. If you want, I will write the formulas out. Aug 5, 2021 at 15:54
• I think I remember how to convert the EGF to OGF if needed. Aug 5, 2021 at 16:03

This my guess at what you want. The “answer” is at the end in Wolfram. If not please clarify.
As usual: Please read carefully, I am subject to a mathematical dyslexia :)
———————
$$x(x+1)\cdots(x+n-1)=\left(x\right)_{n}=\sum_{k=0}^{n}\left[\begin{array}{c} n\\ k \end{array}\right]x^{k}}$$
$$s(n,k)=\left(-1\right)^{n-k}\left[\begin{array}{c} n\\ k \end{array}\right]$$
Notice that $$\left(x\right)_{n}$$ is the rising product. Which is opposite of Wikipedia; this is a persistent annoyance, not isolated to Wikipedia.
Let:
$$G(n,k)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\left[\begin{array}{c} n\\ k \end{array}\right]x^{k}}\cdot\frac{t^{n}}{n!}=\sum_{n=0}^{\infty}}\left(x\right)_{n}\cdot\frac{t^{n}}{n!}}$$
Then we can use Wikipedia for
$$G\left(n,k\right)=_{1}F_{0}\left(\begin{array}{c} x\\ - \end{array};t\right)=\left(1-t\right)^{-x}$$ which rattles on (has an infinite number of terms) but does converge for t<1 (if your interested). We can eliminate the factorial by
$$G\left(n,k\right)=_{2}F_{0}\left(\begin{array}{c} 1,x\\ - \end{array};t\right)$$ which doesn't converge, but we can work term by term; if needed.
But
$$H\left(n,k\right)=\sum_{n=0}^{\infty}}\sum_{k=0}^{n}\left(-1\right)^{n-k}\cdot s(n,k)\cdot x^{k}}\cdot\frac{t^{n}}{n!}=\sum_{n=0}^{\infty}}\left(-x\right)_{n}\cdot\frac{\left(-t\right)^{n}}{n!}=_{1}F_{0}\left(\begin{array}{c} -x\\ - \end{array};-t\right)$$
$$=\left(1+t\right)^{x}=\sum_{n=0}^{x}\frac{x!}{\left(x-n\right)!\cdot n!}t^{n}}$$
Does if $$x\in N_{+}$$
Remark 1. The last term looks strange until we untangle it.
$$\sum_{n=0}^{x}\frac{x!}{\left(x-n\right)!\cdot n!}t^{n}}=\sum_{n=0}^{x}\frac{\Gamma\left(x+1\right)}{\Gamma\left(x-n+1\right)}\frac{t^{n}}{n!}}$$
and apply the Gamma duality $$\Gamma\left(z\right)=\frac{\pi}{\Gamma\left(1-z\right)\cdot sin(\pi\cdot z)}$$ which for integer z is $$\Gamma\left(z\right)=\frac{\left(-1\right)^{z}\cdot\pi}{\Gamma\left(1-z\right)}$$
which yields
$$\sum_{n=0}^{x}\frac{\Gamma\left(-x+n\right)}{\Gamma\left(-x\right)}\cdot\left(-1\right)^{n}\cdot\frac{t^{n}}{n!}=}\sum_{n=0}^{x}\left(-x\right)_{n}\cdot\frac{\left(-t\right)^{n}}{n!}}$$
Now when we remove the factorial we have
$$_{2}\,F_{0}\left(\begin{array}{c} 1,-x\\ - \end{array};-t\right)$$ does terminate and has representations at https://functions.wolfram.com/HypergeometricFunctions/HypergeometricPFQ/03/01/13/
Line 2 appears to work and is appealing; but that's due to a common programing bug. Don't "trust" computer algebra programs, find ways to double check them :)

• Thank you for your answer. It is not the same as what I was looking for, but it has given new insight. Aug 7, 2021 at 17:18
• There were several other insights I gathered along the way; to double-check my reasoning. But I didn't want to add a laundry list of things that nobody was interested in. If you are ever interested in actually calculating the Sheffer Sequences (Roman), Sheffer/Appell/Associated; let me know. I invented a mapping from Roman's Umbral Calculus to Matrices, which make some things a lot clearer; and the programming a lot less tedious and error-prone. Aug 7, 2021 at 20:01
• Okay. Thank you Aug 8, 2021 at 11:57