Variation on Stirling numbers of the first kind - What properties do they have?

Question

According to this paper by Daniel Grunberg and Wikipedia, the Stirling numbers of the first kind are related to the generalized Harmonic numbers.

For instance

$\begin{bmatrix}n+1\\0+1\end{bmatrix}=n!$

$\begin{bmatrix}n+1\\1+1\end{bmatrix}=n!H_n$

$\begin{bmatrix}n+1\\2+1\end{bmatrix}=n!\left[\frac{1}{2}(H_n)^2-\frac{1}{2}H_n^{(2)}\right]$

$\begin{bmatrix}n+1\\3+1\end{bmatrix}=n!\left[\frac{1}{6}(H_n)^3-\frac{1}{2}H_nH_n^{(2)}+\frac{1}{3}H_n^{(3)}\right]$

$\begin{bmatrix}n+1\\4+1\end{bmatrix}=n!\left[\frac{1}{24}(H_n)^4-\frac{1}{4}(H_n)^2H_n^{(2)}+\frac{1}{8}(H_n^{(2)})^2+\frac{1}{3}H_nH_n^{(3)}-\frac{1}{4}H_n^{(4)}\right]$

$\ \ \ \vdots$

For signed Stirling numbers we have the relation $\begin{bmatrix}n\\i\end{bmatrix}=(-1)^{n-i}s(n,i)$

These identities are defined via the partitions of $i$ in $\begin{bmatrix}k\\i+1\end{bmatrix}$ (see links above).

There are many fomulas that use the properties of the Stirling numbers of the first kind. I am interested in exploring the properties of the Stirling-type function that we get when we replace the generalized Harmonic number with a sum that runs over any set of numbers (not just the natural numbers), raised to the power $s$.

So for a set $A=\{a_1,a_2,a_3,...,a_n\}$ we will define ${\begin{bmatrix}n\\i\end{bmatrix}}_{A^s}$ to be the Stirling-type number that is related to $$f_A(s)=\sum_{k=1}^n\frac{1}{{a_k}^s}$$ in the same way that the regular Stirling number is related to the generalized Harmonic number. There is no reason (as far as I know) to still have the $+1$s in our new Stirling-type number relation so for simplicity we'll discard those taking ${\begin{bmatrix}n+1\\i+1\end{bmatrix}}\to{\begin{bmatrix}n\\i\end{bmatrix}}_{A^s}$ in our analogy so now we have these relations:

$\begin{bmatrix}n\\0\end{bmatrix}_{A^s}=\prod_{k=1}^n {a_k}^s$

$\begin{bmatrix}n\\1\end{bmatrix}_{A^s}=\left(\prod_{k=1}^n {a_k}^s\right)f_A(s)$

$\begin{bmatrix}n\\2\end{bmatrix}_{A^s}=\left(\prod_{k=1}^n {a_k}^s\right)\left[\frac{1}{2}(f_A(s))^2-\frac{1}{2}f_A(2s)\right]$

$\begin{bmatrix}n\\3\end{bmatrix}_{A^s}=\left(\prod_{k=1}^n {a_k}^s\right)\left[\frac{1}{6}(f_A(s))^3-\frac{1}{2}f_A(s)f_A(2s)+\frac{1}{3}f_A(3s)\right]$

$\begin{bmatrix}n\\4\end{bmatrix}_{A^s}=\left(\prod_{k=1}^n {a_k}^s\right)\left[\frac{1}{24}(f_A(s))^4-\frac{1}{4}(f_A(s))^2f_A(2s)+\frac{1}{8}(f_A(2s))^2+\frac{1}{3}f_A(s)f_A(3s)-\frac{1}{4}f_A(4s)\right]$

$\ \ \vdots$

So what properties can we prove about these Stirling-type numbers? Concerning their:

$\bullet$ recurrence relations (if any)?

$\bullet$ generating function?

$\bullet$ simple identities? I've discovered these

$$\begin{align} &\begin{bmatrix}n\\n\end{bmatrix}_{A^s}=1\\ &\begin{bmatrix}n\\n-1\end{bmatrix}_{A^s}=\sum_{k=1}^n {a_k}^s\\ &\begin{bmatrix}n\\n-2\end{bmatrix}_{A^s}=\sum_{k=1}^{n-1}\left({a_k}^s\sum_{j=k+1}^n{a_j}^s\right)\\ &\begin{bmatrix}n\\n-3\end{bmatrix}_{A^s}=\sum_{k=1}^{n-2}\left({a_k}^s\sum_{j=k+1}^{n-1}\left({a_j}^s\sum_{i=j+1}^n{a_i}^s\right)\right) \end{align}$$

And this pattern continues.

$\bullet$ sum or product identities? I've also discovered that $$\frac{1}{\prod_{k=1}^n{a_k}^s}\sum_{i=0}^n (-1)^{i}\begin{bmatrix}n\\i\end{bmatrix}_{A^s}=\prod_{k=1}^n\left(1-\frac{1}{{a_k}^s}\right)$$ which is interesting because it is the reciprocal of the zeta function when we input primes and take $n\to\infty$.

A further simple question I have would be, what uses for combinatorics (or other fields of math) does this have?

Any references to this topic are gladly welcomed.

Answer 1

In defining these varied Stirling numbers by their ordinary generating function, many of the identities can be shown to be equivalent. Let $ A = \{ a_0, a_1, \cdots, a_\infty\}$ and define: $$ \begin{bmatrix}n\\k\end{bmatrix}_A = \left [x^k \right] x^{\overline{n}}_A = \left [x^k \right] \prod_{p = 0}^{n-1} (x + a_p) $$ Where $ \left [x^k \right] $ denotes the coefficient of $ x^k $ in the power series expansion. The reason for the $ +1 $s in the original definition come from the fact that the normal Stirling numbers are defined over the set $ \{ 0, 1, \cdots, n - 1 \} $ giving the total of $ n $ numbers, but, to get to $ n $ in the formula, the $ +1 $s would be required.

Recurrence Formula: $$ \begin{bmatrix}n+1\\k\end{bmatrix}_A = a_n\begin{bmatrix}n\\k\end{bmatrix}_A + \begin{bmatrix}n\\k-1\end{bmatrix}_A $$ Proof - By the generating function: $$ x^{\overline{n+1}}_A = x^{\overline{n}}_A (x + a_n) = x \ x^{\overline{n}}_A + n x^{\overline{n}}_A $$ The coefficient of $ x^k $ on the far left is $ \begin{bmatrix}n+1\\k\end{bmatrix}_A $ , while on the right side it is $ \begin{bmatrix}n\\k-1\end{bmatrix}_A + a_n\begin{bmatrix}n\\k\end{bmatrix}_A $ , therefore the recurrence holds. $ \tag*{$\blacksquare$} $

Simple Identities:

The $ k $th elementary symmetric polynomial may be defined as: $$ e_k(a_0, \cdots , a_{n-1}) = \left [ x^{n-k} \right ] \prod_{p = 0}^{n-1} (x + a_p) = \begin{bmatrix}n\\n-k\end{bmatrix}_A $$ These given sums are then variations on the normal $ k $-tuple sum: $$ e_k(a_0, \cdots , a_{n-1}) = \sum_{0 \le j_1 \le \cdots \le j_k \le n-1} a_{j_1} \cdots a_{j_n} $$ $ \tag*{$\blacksquare$} $ Generalized Harmonic Numbers:

Let $ A_n! = \prod_{p=0}^{n-1} a_p $. Notice that for $ 0 \le k \le n-1 $ and nonzero $ a_k $: $$ e_{n-k} (a_0, \cdots , a_{n-1}) = A_n! \ e_k \left (\frac{1}{a_0}, \cdots , \frac{1}{a_{n-1}} \right) $$ By the Girard-Waring formulae, as in here: $$ e_k(\cdots) = \sum \prod_{i=1}^n \frac{(-1)^{m_i}}{m_i!} \left ( \frac{p_{i} (\cdots)}{i} \right )^{m_i} $$ Where the sum is over all non-negative $ m_i $ such that $ m_1 + 2 \ m_2 + \cdots + n \ m_n = n $. Now using the fact that: $$ p_k \left (\frac{1}{a_0}, \cdots , \frac{1}{a_{n-1}} \right) = \sum_{s = 0}^{n-1} \frac{1}{a_s^k} = H^{(k)}_{A_n} $$ Putting this all together gives exactly the forms from your links: $$ \begin{bmatrix}n+1\\k+1\end{bmatrix}_A = e_{n-k} (a_0, \cdots , a_n) = A_{n+1}! e_{k}\left (\frac{1}{a_0}, \cdots , \frac{1}{a_n} \right) = \sum \prod_{i=1}^n \frac{(-1)^{m_i}}{m_i!} \left ( \frac{H^{(i)}_{A_{n+1}}}{i} \right )^{m_i} $$ $ \tag*{$\blacksquare$} $ Lastly your sum-product identity is obtained from the generating function when $ x=-1 $ and then dividing through by $ A_n! $. It does give the form of a generic Euler Product, and, with a judicious choice of parameters, could give rise to many L-Functions. However, studying their properties may not be as easy to do with the Stirling approach as it is with their normal routes.

Answer 2

Just to complement Paul's answer, which already well put the picture in frame, let me add that the product $\prod_{p = 0}^{n-1} (x + a_p)$ can also be developed by Vieta's formulas, which give a straightforward expression for $\begin{bmatrix}n\\k\end{bmatrix}_A $. Also consider to define the recurrence $$ \left\{ \begin{gathered} n + 1 \\ m \\ \end{gathered} \right\}_{\;A} = a_{\,m} \left\{ \begin{gathered} n \\ m \\ \end{gathered} \right\}_{\;A} + \left\{ \begin{gathered} n \\ m - 1 \\ \end{gathered} \right\}_{\;A} \quad \left| {\;\left\{ \begin{gathered} 0 \\ m \\ \end{gathered} \right\}_{\;A} = \delta _{\,m,\,0} } \right. $$ and you will get $$ x^{\,n} = x_{\left\{ {0,0, \cdots } \right\}} ^{\,\overline {\,n\,} } = \sum\limits_k {\left( { - 1} \right)^{\,n - k} \left\{ \begin{gathered} n \\ k \\ \end{gathered} \right\}_A x_A ^{\,\overline {\,k\,} } } = \sum\limits_k {\left\{ \begin{gathered} n \\ k \\ \end{gathered} \right\}_A x_A ^{\,\underline {\,k\,} } } $$ So it is clear that a cascade of further developments is possible.