Is there anything known about the following product? Is it a known function or related to a known function? $$\prod_{n\geqslant1}(1-q^{n^2})$$
$\begingroup$
$\endgroup$
3
-
3$\begingroup$ Well, it has an obvious combinatorial interpretation in terms of "partitions into squares" $\endgroup$– Sam HopkinsCommented Dec 14, 2021 at 16:56
-
$\begingroup$ @SamHopkins Yes indeed, this was in fact my original question. I do not know what is known about it. $\endgroup$– cocoCommented Dec 14, 2021 at 17:01
-
1$\begingroup$ Problem 3.28 of arxiv.org/abs/1808.08449 asks whether the coefficients of this power series are unbounded. I take this as evidence little is known about it. $\endgroup$– Sam HopkinsCommented Dec 16, 2021 at 4:27
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
Write $\theta(q)=\sum_{n \in \mathbb Z} q^{n^2}$. Taking logarithms, we get that the logarithm of your function is
$\sum_n \ln(1-q^{n^2}) = -\sum_n \sum_m \frac 1 m q^{n^2m} = -\sum_m \frac 1 m \frac {\theta(q^m)-1} 2 = \frac 1 2 \sum_{m=1}^\infty \frac {1 - \theta(q^m)} m$.
$\begingroup$
$\endgroup$
5
Following up on Sam's comment, A001156 in the OEIS gives the number of partitions of $n$ into squares; that generating function is the reciprocal of your product. There's an equinumerous type of partitions given: partitions of $n$ where each part $k$ occurs a multiple of $k$ times (which contributes a multiple of $k^2$ to the weight). You might look through the links and formula entries for the sequence to see if they lead to anything that interests you.
-
$\begingroup$ I tried Googling "pentagonal number theorem squares" and similar phrases and didn't get much that seemed relevant... $\endgroup$ Commented Dec 16, 2021 at 4:06
-
$\begingroup$ By the way, the coefficients of the OP's power series are also in the OEIS: oeis.org/A276516. Note that they are not always $0,\pm 1$! $\endgroup$ Commented Dec 16, 2021 at 4:18
-
$\begingroup$ @SamHopkins Far from it: the longer list shows that the 9996th entry is 18,786! $\endgroup$ Commented Dec 16, 2021 at 4:27
-
$\begingroup$ Yes; though as I commented above it is apparently unknown if the coefficients are unbounded. $\endgroup$ Commented Dec 16, 2021 at 4:28
-
$\begingroup$ @SamHopkins and BrianHopkins Many thanks for your help and for the references! $\endgroup$– cocoCommented Dec 16, 2021 at 15:53
$\begingroup$
$\endgroup$
Let $m$ be a positive integer. Let $f_{m}(z)=\prod_{k=1}^{\infty}(1-z^{k^{m}})$. Then $f(z)$ is analytic in the disk $\{z:|z|<1\}$ and satisfies the following functional equation:
$$\prod_{j=1}^{\infty}f_{m}(z^{j^{m}})^{\mu(j)}=1.$$
Here $\mu$ denotes the Mobius Mu function which is defined by setting $\mu(M^{2}N)=0$ whenever $M,N$ are positive integers with $M>1$ and $\mu(p_{1}\dots p_{s})=(-1)^{s}$ whenever $p_{1},\dots,p_{s}$ are distinct primes.
