School of Mathematical & Statistical Sciences Faculty Publications

Document Type

Article

Publication Date

2-27-2026

Abstract

Let A(q)=:∑∞n=0anqn and B(q)=:∑∞n=0bnqn be two eta quotients. In some previous papers, the present authors considered the problem of when

an=0⟺bn=0.

In the present paper we consider the “mod m” version of this problem, i.e. for which eta quotients A(q) and B(q) and for which integers m>1 do we have (non-trivially) that

an≡0(modm)⟺bn≡0(modm)?

(We say “non-trivially” as there are trivial situations where an≡bn(modm) for all n≥0). The m for which we found non-trivial (in the sense just mentioned) results were m=p2, p=2,3 and 5. For m=4 and m=9, we found results which apply to infinite families of eta quotients. One such is the following: Let A(q) be any eta quotient of the form

A(q)=f3j1+11∏3∤if3jii∏3|ifjii=:∞∑n=0anqn,B(q)=f3f31A(q)=:∞∑n=0bnqn

with fk=∏∞n=1(1−qkn). Then

a3n−b3n≡0(mod9),2a3n+1+b3n+1≡0(mod9),a3n+2+2b3n+2≡0(mod9).

Some of these theorems also had some combinatorial implications, one example being the following: Let p(3)2(n) denote the number of bipartitions (π1,π2) of n where π1 is 3-regular. Then

p(3)2(n)≡0(mod9)⟺nisnotageneralizedpentagonalnumber.

In the case of m=25, we do not have any general theorems that apply to an infinite family of eta quotients, such as the modulo 9 result stated above. Instead we give two tables of results that appear to hold experimentally. Proofs of results stated in these tables appear to need the theory of modular forms and are more complicated. We do prove some individual results, such as the following: Let the sequences {cn} and {dn} be defined by

f101=:∞∑n=0cnqn,f51f5=:∞∑n=0dnqn.

Then

cn≡0(mod25)⟺dn≡0(mod25).

Comments

https://rdcu.be/e9Oez

Publication Title

Research in Number Theory

DOI

10.1007/s40993-026-00707-4

Included in

Mathematics Commons

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