3-regular partitions and a modular \(K3\) surface (Q2782415)

A \(k\)-regular partition of \(n\) is a non-increasing sequence of positive integers whose sum is \(n\), with the condition that no summand is divisible by \(k\). Denote the number of such partitions by \(b_k(n)\); then we have the generating function NEWLINE\[NEWLINE\sum_{n=0}^\infty b_k(n)q^n=\prod_{n=1}^\infty \left(\frac{1-q^{kn}}{1-q^n}\right).NEWLINE\]NEWLINE The second author and \textit{K. Ono} have studied the function \(b_2(n)\) modulo powers of \(2\) [J. Comb. Theory, Ser. A 92, 138-157 (2000; Zbl 0972.05003)]. In the present paper the authors study the behavior of \(b_3(n)\) modulo \(3\) and \(9\). NEWLINENEWLINENEWLINEIf \(\eta^6(4z)=\sum a(n)q^n\), then the authors show that NEWLINE\[NEWLINEb_3(n)\equiv a(12n+1)\pmod 9.NEWLINE\]NEWLINE Using this fact, they prove several theorems. The first result states that if \(X\) is the \(K3\) surface defined by the equation \(s^2=x(x+1)y(y+1)(x+8y)\), and \(p\equiv 1\pmod {12}\) is prime, then NEWLINE\[NEWLINEb_3\left(\frac{p-1}{12}\right)\equiv \# X(\mathbb F_p)-(p+1)^2\pmod 9.NEWLINE\]NEWLINE Using the fact that \(\eta^6(4z)\) has complex multiplication, the authors go on to obtain an explicit formula modulo \(9\) for the number of \(3\)-regular partitions of \(n\) in terms of the prime factorization of \(12n+1\).NEWLINENEWLINEFor the entire collection see [Zbl 0980.00024].