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Families of congruences for fractional partition functions modulo powers of primes

Abstract

Recently, Chan and Wang studied the fractional partition function and found several infinite classes of congruences satisfied by the corresponding coefficients. In this paper, we find new families of congruences modulo powers of primes using the Rogers-Ramanujan continued fraction and some dissection formulae of certain q-products. We also find analogous congruences in the coefficients of the fractional powers of the generating function for the 2-color partition function.

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Notes

  1. It is to be noted that the 2-color partition function \(p_{[1,\beta ;-1]}(n)\) and the 2-colored partition function \(p_{-2}(n)\) defined above are different.

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Acknowledgements

The authors would like to thank the anonymous referees for their helpful suggestions which significantly improved the presentation of the paper. The first author was partially supported by Grant no. MTR/2018/000157 of Science & Engineering Research Board (SERB), DST, Government of India under the MATRICS scheme. The second author was partially supported by Council of Scientific & Industrial Research (CSIR), Government of India under CSIR-JRF scheme. The authors thank both the funding agencies.

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Correspondence to Nayandeep Deka Baruah.

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Appendix

Appendix

Appendix A

The following congruences can be proved by employing the methods given in the proofs of Theorem 4 and Corollary 1. Let k and s be non-zero positive integers such that \(s\le \lfloor k/2\rfloor \) and b’s in (4.1)–(4.10) are co-prime to the moduli. Then, for all \(n\ge 0\), we have

$$\begin{aligned}&p_{-(5^{k}-2b)/b}\left( 5^{2s}\cdot n+5^{2s-1}\cdot r+\dfrac{5^{2s}-1}{12}\right) \nonumber \\&\equiv 0 ~\left( mod ~5^{k-2s+1}\right) ,\quad \text {where~} r\in \{1,2,3,4\}, \end{aligned}$$
(4.1)
$$\begin{aligned}&p_{-(7^{k}-2b)/b}\left( 7^{2s}\cdot n+7^{2s-1}\cdot r+\dfrac{7^{2s}-1}{12}\right) \nonumber \\&\equiv 0 ~\left( mod ~7^{k-2s+1}\right) ,\quad \text {where~} r\in \{1,2,\dots ,6\}, \end{aligned}$$
(4.2)
$$\begin{aligned}&p_{-(11^{k}-2b)/b}\left( 11^{2s}\cdot n+11^{2s-1}\cdot r+\dfrac{11^{2s}-1}{12}\right) \nonumber \\&\equiv 0 ~\left( mod ~11^{k-2s+1}\right) ,\quad \text {where~} r\in \{1,2,\ldots ,10\}, \end{aligned}$$
(4.3)
$$\begin{aligned}&p_{-(5^{k}-4b)/b}\left( 5^{2s}\cdot n+5^{2s-1}\cdot r+\dfrac{5^{2s}-1}{6}\right) \nonumber \\&\equiv 0 ~\left( mod ~5^{k-s+1}\right) ,\quad \text {where~} r\in \{1,2,3,4\}, \end{aligned}$$
(4.4)
$$\begin{aligned}&p_{-(5^{k}-8b)/b}\left( 5^{2s}\cdot n+5^{2s-1}\cdot r+\dfrac{2\cdot 5^{2s-1}-1}{3}\right) \nonumber \\&\equiv 0 ~\left( mod ~5^k\right) ,\quad \text {where~} r\in \{0,2,3,4\}, \end{aligned}$$
(4.5)
$$\begin{aligned}&p_{-(5^{k}-14b)/b}\left( 5^{2s}\cdot n+5^{2s-1}\cdot r+\dfrac{11\cdot 5^{2s-1}-7}{12}\right) \nonumber \\&\equiv 0 ~\left( mod ~5^k\right) ,\quad \text {where~} r\in \{0,1,3,4\}, \end{aligned}$$
(4.6)
$$\begin{aligned}&p_{-(7^{k}-6b)/b}\left( 7^{2s}\cdot n+7^{2s-1}\cdot r+\dfrac{3\cdot 7^{2s-1}-1}{4}\right) \nonumber \\&\equiv 0 ~\left( mod ~7^k\right) ,\quad \text {where~} r\in \{0,2,3,4,5,6\}, \end{aligned}$$
(4.7)

and when k is odd,

$$\begin{aligned} p_{-(5^{k}-8b)/b}\left( 5^{k}\cdot n+\dfrac{2\cdot 5^{k}-1}{3}\right)&\equiv 0 ~\left( mod ~5^k\right) , \end{aligned}$$
(4.8)
$$\begin{aligned} p_{-(5^{k}-14b)/b}\left( 5^{k}\cdot n+\dfrac{11\cdot 5^{k}-7}{12}\right)&\equiv 0 ~\left( mod ~5^k\right) , \end{aligned}$$
(4.9)
$$\begin{aligned} p_{-(7^{k}-6b)/b}\left( 7^{k}\cdot n+\dfrac{3\cdot 7^{k}-1}{4}\right)&\equiv 0 ~\left( mod ~7^k\right) . \end{aligned}$$
(4.10)

The identities, lemmas, and corollaries used to prove (4.1)–(4.10) are given in the following chart.

Appendix B

The following congruences can also be proved by using the methods explained in the proof of Theorems 4 and 6. Let k and s be positive integers such that \(s\le \lfloor k/2\rfloor \) and b’s in (4.11)–(4.19) are co-prime to the moduli. Then, for all \(n\ge 0\), we have

$$\begin{aligned}&p_{[1,3;-(5^{k}-b)/b]}\left( 5^{2s}\cdot n+5^{2s-1}\cdot r+\dfrac{5^{2s}-1}{6}\right) \nonumber \\&\equiv 0 ~\left( mod ~5^{k-2s+1}\right) ,\quad \text {where~} r\in \{1,2,3,4\}, \end{aligned}$$
(4.11)
$$\begin{aligned}&p_{[1,3;-(11^{k}-b)/b]}\left( 11^{2s}\cdot n+11^{2s-1}\cdot r+\dfrac{5\cdot 11^{2s-1}-1}{6}\right) \nonumber \\&\equiv 0 ~\left( mod ~11^{k-2s+1}\right) ,\quad \text {where~} r\in \{0,2,3,\ldots ,10\}, \end{aligned}$$
(4.12)
$$\begin{aligned}&p_{[1,4;-(7^{k}-b)/b]}\left( 7^{2s}\cdot n+7^{2s-1}\cdot r+\dfrac{11\cdot 7^{2s-1}-5}{24}\right) \nonumber \\&\equiv 0 ~\left( mod ~7^{k-2s+1}\right) , \quad \text {where~} r\in \{0,2,3,\dots ,6\}, \end{aligned}$$
(4.13)
$$\begin{aligned}&p_{[1,4;-(11^{k}-b)/b]}\left( 11^{2s}\cdot n+11^{2s-1}\cdot r+\dfrac{7\cdot 11^{2s-1}-5}{24}\right) \nonumber \\&\equiv 0 ~\left( mod ~11^{k-2s+1}\right) ,\quad \text {where~} r\in \{0,1,3,4,\dots ,10\}, \end{aligned}$$
(4.14)
$$\begin{aligned}&p_{[1,2;-(3^{k}-2b)/b]}\left( 3^{2s}\cdot n+3^{2s-1}\cdot r+\dfrac{3^{2s}-1}{4}\right) \nonumber \\&\equiv 0 ~\left( mod ~3^{k-s+1}\right) ,\quad \text {where } r\in \{1,2\}, \end{aligned}$$
(4.15)
$$\begin{aligned}&p_{[1,2;-(3^{k}-5b)/b]}\left( 3^{2s}\cdot n+3^{2s-1}\cdot r+\dfrac{7\cdot 3^{2s-1}-5}{8}\right) \nonumber \\&\equiv 0 ~\left( mod ~3^k\right) ,\quad \text {where } r\in \{0,2\}, \end{aligned}$$
(4.16)
$$\begin{aligned}&p_{[1,2;-(5^{k}-3b)/b]}\left( 5^{2s}\cdot n+5^{2s-1}\cdot r+\dfrac{7\cdot 5^{2s-1}-3}{8}\right) \nonumber \\&\equiv 0 ~\left( mod ~5^k\right) ,\quad \text {where } r\in \{0,2,3,4\}, \end{aligned}$$
(4.17)

and when k is odd,

$$\begin{aligned} p_{[1,2;-(3^{k}-5b)/b]}\left( 3^{k}\cdot n+\dfrac{7\cdot 3^{k}-5}{8}\right)&\equiv 0 ~\left( mod ~3^k\right) , \end{aligned}$$
(4.18)
$$\begin{aligned} p_{[1,2;-(5^{k}-3b)/b]}\left( 5^{k}\cdot n+\dfrac{7\cdot 5^{k}-3}{8}\right)&\equiv 0 ~\left( mod ~5^k\right) . \end{aligned}$$
(4.19)

The identities, lemmas, and corollaries used to prove (4.11)–(4.19) are given in the following chart.

Appendix C

The following steps are required to establish the conditions (ii)–(v) of Theorem 5.

Condition (ii). Use (2.9) of Lemma 6 to obtain the expression \(N=((4n+1)^2+2(4m+1)^2-3)/8,\) equivalent to \(8N+3=(4n+1)^2+2(4m+1)^2\). If \(\ell \equiv 5~or ~7~(mod ~8)\), then . It follows that \(8N+3\equiv 0~(mod ~\ell )\) if and only if \(4n+1\equiv 0~(mod ~\ell )\) and \(4m+1\equiv 0~(mod ~\ell ).\)

Condition (iii). Use (2.9) of Lemma 6 to obtain the expression \(N=((4n+1)^2+3(4m+1)^2-4)/8,\) equivalent to \(8N+4=(4n+1)^2+3(4m+1)^2\). If \(\ell \equiv 2~(mod ~3)\), then . It follows that \(8N+4\equiv 0~(mod ~\ell ),\) equivalently, \(2N+1\equiv 0~(mod ~\ell )\) if and only if \(4n+1\equiv 0~(mod ~\ell )\) and \(4m+1\equiv 0~(mod ~\ell ).\)

Condition (iv). Use (2.8) and (2.11) of Lemma 6 to obtain the expression \(N=((6n+1)^2+4(3m+1)^2-5)/12,\) equivalent to \(12N+5=(6n+1)^2+4(3m+1)^2\). If \(\ell \equiv 3~(mod ~4)\), then . It follows that \(12N+5\equiv 0~(mod ~\ell )\) if and only if \(\quad 6n+1\equiv 0~(mod ~\ell )\) and \(3m+1\equiv 0~(mod ~\ell ).\)

Condition (v). Use (2.9) of Lemma 6 to obtain the expression \(N=((4n+1)^2+4(4m+1)^2-5)/8,\) equivalent to \(8N+5=(4n+1)^2+4(4m+1)^2\). If \(\ell \equiv 3~(mod ~4)\), then . It follows that \(8N+5\equiv 0~(mod ~\ell )\) if and only if \(4n+1\equiv 0~(mod ~\ell )\) and \(4m+1\equiv 0~(mod ~\ell ).\)

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Baruah, N.D., Das, H. Families of congruences for fractional partition functions modulo powers of primes. Res. number theory 7, 57 (2021). https://doi.org/10.1007/s40993-021-00287-5

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Keywords

  • Partition
  • n-color partition
  • n-colored partition
  • Fractional partition function
  • Congruence
  • Rogers-Ramanujan continued fraction

Mathematics Subject Classification

  • 11P83
  • 05A15
  • 05A17