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This is a note in Sedgewick's Analytic Combinatorics: The number of lattice points with integer coordinates that belong to the closed ball of radius n in d-dimensional Euclidean space is $\displaystyle[z^{n^2}]\frac{1}{1-z}\Theta(z)^d$ where $\displaystyle\Theta(z) = 1 + 2\sum_{k=1}^{\infty} z^{k^2}$.

I've tried to figure out why this is true to no avail - perhaps $\Theta$ counts the number of ways to place points on a 1-dimensional ball of radius $k^2$, and raising it to the $d$ counts the cross product of all possibilities? I still don't know where the $\frac{1}{1-z}$ comes into play. Hints or explanations would be very much appreciated!

Batominovski
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Brian Burns
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1 Answers1

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The $z^m$ term in $\Theta(z)$ is the number of integers whose square is exactly $m$. (One for $m=0$, and two for each perfect square corresponding to its positive and negative roots.) So the $z^m$ term in $\Theta(z)^d$ is the number of $d$-tuples of integers the sum of whose squares is exactly $m$. Now, multiplying a power series by $1/(1-z)=1+z+z^2+z^3+\ldots$ does the following: $$ \frac{1}{1-z}\sum_m a_m z^m=\sum_m a_m \left(\sum_{k\ge 0}z^{k+m}\right)=\sum_m \left(\sum_{k\le m}a_{k}\right)z^m. $$ So the $z^m$ term in $\Theta^{d}/(1-z)$ is the number of $d$-tuples of integers the sum of whose squares is no greater than $m$. For the closed ball of radius $n$, replacing $m$ by $n^2$ gives the desired result.

mjqxxxx
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