Suppose we have a sequence of distinct $a_i$'s: $\left\{ a_1, a_2, \ldots, a_n \right\}$. We also have a sequence of not necessarily distinct $b_i$'s: $\left\{ b_1, b_2, \ldots, b_n \right\}$. Each $b_i$ corresponds to how many $a_i$'s there are. For example, if $a_1 = 2$ and $b_1 = 3$, then there are $3$ $2$'s. The function I have in mind, denoted $\tau_j$, is a sum of all possible combinations of products of the $a_i$'s where we have all possible combinations such that the sum of the powers of each term equals $j$.

I provide the following example. Suppose we have $a_1 = a_2 = a_3$ and $a_4 = a_5$. Let's rewrite/rename this so we just have two distinct $a_i$'s: $a_1, a_2$. We can see that $b_1 = 3$ and $b_2 = 2$. Therefore:

\begin{equation} \tau_0 \left( a_1, a_2 \right) = 1 \end{equation}

\begin{equation} \tau_1 \left( a_1, a_2 \right) = a_1 + a_2 \end{equation}

\begin{equation} \tau_2 \left( a_1, a_2 \right) = a_1^2 + a_1 a_2 + a_2^2 \end{equation}

\begin{equation} \tau_3 \left( a_1, a_2 \right) = a_1^3 + a_1^2 a_2 + a_1 a_2^2 \end{equation}

\begin{equation} \tau_4 \left( a_1, a_2 \right) = a_1^3 a_2 + a_1^2 a_2^2 \end{equation}

\begin{equation} \tau_5 \left( a_1, a_2 \right) = a_1^3 a_2^2 \end{equation}

As we can see, the sum of the powers in each term in each $\tau_j$ is equal to $j$. Furthermore, any power of any $a_i$ cannot exceed its associated $b_i$. I am thinking for each term in the sum of each $\tau_j$, we have to cycle through powers of each $a_i$ from $0$ to $b_i$ under the condition that the sum of the powers must equal $j$.

But the question is: how do we write that? Can we have a sum symbol under a sum symbol, as a condition (by "under", I mean where we would typically have "$i=0$")? What is the proper notation for this? Any assistance is greatly appreciated.

  • $\begingroup$ More an approach than a notation, although it does use the generating function notation $[x^j]g(x)$ for the coefficient of $x^j$ of the power series of for $g(x)$: $$[x^j]\prod_{i=1}^n \frac{(a_ix)^{b_i+1}-1}{a_ix-1}$$ $\endgroup$ Dec 9, 2021 at 7:18
  • $\begingroup$ For your example, $b_1=3,b_2=2,$ $$(1+a_1x+a_1^2x^2+a_1^3x^3)(1+a_2x+a_2^2x^2)\\= 1 + (a_1+ a_2) x + (a_1^2 + a_1 a_2 + a_2 ^2) x^2 \\+ (a_1^3 + a_1^2 a_2 + a_1 a_2 ^2) x^3 + (a_1^3 a_2 + a_1^2 a_2 ^2) x^4 + a_1^3 a_2^2 x^5 ,$$ and $\tau_j$ is the coefficient of $x^j.$ $\endgroup$ Dec 9, 2021 at 7:33
  • $\begingroup$ @ThomasAndrews This is good. But let's expand on this concept. Suppose I want the sum of all the coefficients (i.e. $x = 1$). How would you write that? $\endgroup$ Dec 9, 2021 at 8:06
  • $\begingroup$ You can just substute $x=1$ into my formula:$$\sum_{j=0}^{b_1+b_2+\cdots+b_n} \tau_j=\prod\frac{a_i^{b_i+1}-1}{a_i-1}$$ $\endgroup$ Dec 9, 2021 at 16:35
  • $\begingroup$ For any $a_i=1,$ replace the undefined fraction in the product with $b_i+1=1+a_i+a_i^2+\cdots +a_i^{b_i}.$ $\endgroup$ Dec 9, 2021 at 16:44

1 Answer 1


You could use multi-index notation viz.$$\tau_j=\sum_{|\alpha|=j\land\alpha\le b}a^\alpha.$$Here $a^\alpha:=\prod_ia_i^{\alpha_i}$, $|\alpha|:=\sum_i\alpha_i$ (a multi-index $\alpha$ is required to satisfy $\alpha_i\ge0$), and $\alpha\le b$ abbreviates $\alpha_i\le b_i$ for all $i$.

  • $\begingroup$ I get it now! Thank you so much. One more question though... the definitions you gave... are those standard definitions (especially the product one) that are commonly known or would we have to explicitly state them? $\endgroup$ Dec 9, 2021 at 7:28
  • $\begingroup$ @RJOnyxMoonshadow They're all taken from the article, so yes, these are standard. $\endgroup$
    – J.G.
    Dec 9, 2021 at 7:33
  • $\begingroup$ I didn't see the product one, which is why I asked. Thanks again! $\endgroup$ Dec 9, 2021 at 7:36
  • 1
    $\begingroup$ It depends on your audience, but I suggest always defining notation. Just because it is on a Wikipedia page doesn’t mean all readers will know or remember the notation. @RJOnyxMoonshadow $\endgroup$ Dec 9, 2021 at 7:46

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