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I am partly repeating my self here.

The sequence $a_n$ generated by the Dirichlet series generating function: $\sum \limits_{n=1}^{\infty} \frac{a_n}{n^s}$

corresponding to $\zeta(s)^m$

seems to have the ordinary generating function:

$\sum \limits_{n=1}^{\infty} a_nx^n = x + {m \choose 1}\sum \limits_{a=2}^{\infty} x^{a} + {m \choose 2}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} + {m \choose 3}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} x^{abc} + {m \choose 4}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{d=2}^{\infty} x^{abcd} +...$

Is this a true result?

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1 Answer 1

up vote 1 down vote accepted

Yes it is true.

Since $$a_{n}=\sum_{i_{!}\cdot i_{2}\cdots i_{m}=n}1$$ we can just split into a sum based on the number of $i_j=1$. If $i_j=1$ for $m-1$ different $j$'s then we get the first term, where the $\binom{m}{1}$ comes from the fact that there are $m$ difference choices for the $i_j\neq 1$. Similarly if we have $i_j=1$ for $m-2$ indices, etc.

Alternatively, $a_n$ is the number of integer points which lie on the hyperbola-like curve $x_1 \cdot x_2 \cdots x_m=n.$ The sum you write down above counts the integer points by partitioning by the dimension they live it. Think of the point $(1,1,\dots,1)$ as the origin. Then the first term $x$ counts the 0-dimensional point $(1,1,\dots,1)$, which is only counted by $n=1$. The second term counts the number of integer points intersecting the axes and the curve, the third term counts the number of integer points which lie in both a two dimensional plane spanned by two axes, and the curve, etc. The sums start at 2 rather than 1 so that no point is double counted in this way.

Hope that helps,

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