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






$\ \ \ \vdots$

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.

  • $\begingroup$ There are many ways proposed to "extend"/"generalize" Stirling numbers: yours is an approach that I did not see before. I need to digest it before being able to tell anything. $\endgroup$ – G Cab Aug 21 '16 at 23:21
  • $\begingroup$ The parameter $s$ looks superfluous, by defining $b_k=a_k^s$ all will reduce from $A^s$ to $B$: did I get it right ? $\endgroup$ – G Cab Aug 23 '16 at 12:45
  • $\begingroup$ Right! We can say that. $\endgroup$ – tyobrien Aug 23 '16 at 13:27
  • $\begingroup$ The normal Stirling numbers $ \left [^{n +1}_{n-k+1} \right]$ are given by the elementary symmetric polynomials of the first n natural numbers, $ e_k (1,2,\dotsc, n) $ , so the sum relation that you noted are just these polynomials in those parameters. $\endgroup$ – Paul LeVan Aug 23 '16 at 17:31

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.

  • $\begingroup$ Very well exposed, +1. $\endgroup$ – G Cab Aug 24 '16 at 11:21

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.


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