The $$M_{2}$$ M 2 -rank of partitions without repeated odd parts modulo $$6$$ 6 and $$10$$ 10

Mao, Renrong

doi:10.1007/s11139-014-9578-3

Cited by 15 publications

(15 citation statements)

References 13 publications

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“…Continuing on their work, Mao [6,7] extended the results for Dyson rank differences modulo 10 and M 2 rank differences modulo 6 and 10. He obtained several interesting inequalities based on his results such as N (1, 10, 5n + 1) > N (5, 10, 5n + 1), N 2 (0, 6, 3n + 1) + N 2 (1, 6, 3n + 1) > N 2 (2, 6, 3n + 1) + N 2 (3, 6, 3n + 1).…”

Section: Introduction and Resultsmentioning

confidence: 76%

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Proof of a limited version of Mao’s partition rank inequality using a theta function identity

Barman

Sachdeva

2016

Res. Number Theory

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Section: Introduction and Resultsmentioning

confidence: 76%

“…Mao also gave some conjectures in [6,7] based on computational evidence, both for the Dyson rank and M 2 rank for partitions with unique odd parts. …”

Section: Introduction and Resultsmentioning

confidence: 99%

Proof of a limited version of Mao’s partition rank inequality using a theta function identity

Barman

Sachdeva

2016

Res. Number Theory

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“…We give these identities for c = 7. Similar formulas were determined by Lovejoy and Osburn for c = 3 and c = 5 in [16] and for c = 6 and c = 10 by Mao in [18]. Rather than using harmonic Maass forms, those formulas used the q-series techniques developed by Atkin and Swinnerton-Dyer [2] to determine rank difference formulas for the rank of partitions.…”

Section: Statement Of Resultsmentioning

confidence: 96%

The generating function of the $M_2$-rank of partitions without repeated odd parts as a mock modular form

Jennings-Shaffer

2018

Trans. Amer. Math. Soc.

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Abstract. By work of Bringmann, Ono, and Rhoades it is known that the generating function of the M 2 -rank of partitions without repeated odd parts is the so-called holomorphic part of a certain harmonic Maass form. Here we improve the standing of this function as a harmonic Maass form and show more can be done with this function. In particular we show the related harmonic Maass form transforms like the generating function for partitions without repeated odd parts (which is a modular form). We then use these improvements to determine formulas for the rank differences modulo 7. Additionally we give identities and formulas that allow one to determine formulas for the rank differences modulo c, for any c > 2.

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“…Most recently, Mao [7,8] has derived generating function formulas for Dyson's rank on partitions modulo 10, and the M 2 rank on partitions without repeated odd parts modulo 6 and 10. In this work he proves a number of inequalities, including for example N (0, 10, 5n + 1) > N (4, 10, 5n + 1),…”

Section: Introduction and Statement Of Resultsmentioning

confidence: 99%

“…We will also make use of the following theorem for M 2 rank of partitions without repeated odd parts. Theorem 2.3 (Mao [8]). We have that ∞ n=0 N 2 (0, 10, n) + N 2 (1, 10, n) − N 2 (4, 10, n) − N 2 (5, 10, n) q n 45,100…”

Section: 50mentioning

confidence: 99%

A proof of some conjectures of Mao on partition rank inequalities

2016

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Based on work of Atkin and Swinnerton-Dyer on partition rank difference functions, and more recent work of Lovejoy and Osburn, Mao has proved several inequalities between partition ranks modulo 10, and additional results modulo 6 and 10 for the M2 rank of partitions without repeated odd parts. Mao conjectured some additional inequalities. We prove some of Mao's rank inequality conjectures for both the rank and the M2 rank modulo 10 using elementary methods.Furthermore, we will make use of the following notation of Mao. 1 For positive integers a < b, defineLemma 2.1 (Mao [7]). Given positive integers a < b, the q-series coefficients of L a,b are all nonnegative.Mao proved rank difference formulas that we will use in our proof of Theorem 1.3. First, for unrestricted partitions, Mao proved the following theorem. Theorem 2.2 (Mao [7]). We have that ∞ n=0 N (0, 10, n) + N (1, 10, n) − N (4, 10, n) − N (5, 10, n) q n = J 25 J 5 50 J 2 20,50 J 4 10,50 J 3 15,50 + 1 J 25 ∞ n=−∞ (−1) n q 75n(n+1)/2+5 1 + q 25n+5 + q J 25 J 5 50 J 5,50 J 2 10,50 J 2 15,50 + q 2 J 25 J 5 50 J 2 5,50 J 15,50 J 2 + q 2 J 25 J 5 50 J 20,50 J 3 10,50 J 3 15,50 + q 3 J 25 J 5 50 J 5,50 J 10,50 J 2 15,50 J 20,50 + q 4 J 25 J 5 50 J 2 20,50 J 25,50 2q 5 J 4 10,50 J 4 1 We note that our definition of L a,b differs from Mao's in that the roles of a and b are reversed.3 = 2q 5 J 15 100 J 10,100 J 50,100 J 3 5,100 J 2 15,100 J 2 20,100 J 3 25,100 J 30,100 J 2 35,100 J 3 45,100 + 1 J 25,100 ∞ n=−∞ (−1) n q 50n 2 +25n 1 + q 50n+10 + q J 15 100We see that S, T 1 , . . . , T 4 all have nonnegative coefficients. Thus to prove (3), it suffices to show that J 2 5 J 5 10 J 2 4,10 J 3 3,10 J 4 2,10has positive coefficients. Let T 1 + T 2 + T 3 + T 4 = ∞ n=1 a(n)q n , and let J 2 5 J 5 10 J 2 4,10 J 3 3,10 J 4 2,10 = 1 + ∞ n=1 b(n)q n .We will show that b(n) > a(n) for all n ≥ 1. 5 = 1 J 5,20

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The $$M_{2}$$ M 2 -rank of partitions without repeated odd parts modulo $$6$$ 6 and $$10$$ 10

Cited by 15 publications

References 13 publications

Proof of a limited version of Mao’s partition rank inequality using a theta function identity

Proof of a limited version of Mao’s partition rank inequality using a theta function identity

The generating function of the $M_2$-rank of partitions without repeated odd parts as a mock modular form

A proof of some conjectures of Mao on partition rank inequalities

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