Arithmetic properties for (s,t)-regular bipartition functions

Abstract

Let B s,t (n) denote the number of (s, t)-regular bipartitions. Recently, Dou discovered an infinite family of congruences modulo 11 for B 3,11 (n). She also presented several conjectures on B s,t (n). In this paper, utilizing an theta function identity appeared in Berndt's book, we confirm three conjectures on B 3,7 (n) given by Dou. Moreover, we prove several infinite families of congruences modulo 3 and 5 for B 3,s (n) and B 5,s (n). In addition, we prove many infinite families of congruences modulo 7 for B 3,7 (n) by employing an identity of Newman.

WITHDRAWN: On New Congruences and Density Results for(k, 𝓁)-regular Bipartitions Modulo m

“…In [15], Wang studied the arithmetic properties of B 3,ℓ (n) and B 5,ℓ (n) and confirmed the conjectures proposed by Dou. Xia and Yao [24] also confirmed the conjectures of Dou and proved several infinite families of congruences for B s,t (n) modulo 3, 5 and 7. Adiga and Ranganatha [1] provided a simple proof for Ramanujan type congruence for the (3, 7)-regular bipartitions modulo 3 which was conjectured by Dou and also found some new infinite families of congruences for (3, 7)-regular bipartitions modulo 3.…”

Some new congruences for (s,t)-regular bipartition functions

“…Recently, Chandrashekar Adiga and Ranganatha [1] gave an elementary proof of (1.2) and they also proved several infinite families of congruences for B 3,7 (n) modulo 3. Most recently, Xia and Yao [15] confirmed the congruences (1.3) and (1.4). In the same paper, they presented a number of congruences for B 3,s (n), B 5,s (n) (s 1) and B 3,7 (n) modulo 3, 5 and 7, respectively.…”

Ramanujan type of congruences modulo m for (l, m)-regular bipartitions

“…Recently, Adiaga and Ranganatha [1] proved infinite families of congruences modulo 3 for B 3,7 (n). More recently, Xia and Yao [25] proved several infinite families of congruences modulo 3 for B 3,s (n), modulo 5 for B 5,s (n) and modulo 7 for B 3,7 (n). For example, let s be a positive integer and let p ≥ 5 be a prime, for n ≥ 0, B 3,s p 2α+1 n + (1 + s)(p 2α+2 − 1) 24 ≡ 0 (mod 3).…”