Let $c^*_{(1,r,a)}(n)$ be the generalization of the cubic partition function $c(n)$. In this paper, we prove some new congruences modulo odd prime $p$ by taking $r=3,4,5,7,11$ and $13$ using $q$-series identities. We study congruence properties of generalization of cubic partition function for different values of $a$ and give some particular cases as examples

Vidya, K. N.

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Note di Matematica ISSN 1123-2536, e-ISSN 1590-0932Note Mat. 42 (2022) no. 2, 7–17. doi:10.1285/i15900932v42n2p7New generalization of cubic partition of nK. N. VidyaDepartment Education in Science and Mathematics, Regional Institute ofEducation(NCERT), Mysuru 570 006, India.vidyaknagabhushan@gmail.comReceived: 18.3.2022; accepted: 29.10.2022.Abstract. Let c∗(1,r,a)(n) be the generalization of the cubic partition function c(n). In thispaper, we prove some new congruences modulo odd prime p by taking r = 3, 4, 5, 7, 11 and 13using q-series identities. We study congruence properties of generalization of cubic partitionfunction for different values of a and give some particular cases as examples.Keywords: Partitions; k-colors; Partition CongruencesMSC 2022 classification: Primary 11P83; Secondary 05A15, 05A17.1 IntroductionIn a paper [1], Chan started the study of cubic partitions by exhibiting a closerelation between a certain type of partition function and Ramanujan’s cubiccontinued fraction. For example, there are four cubic partitions of 3, namely3, 21 + 1, 22 + 1 and 1 + 1 + 1, where the subscripts 1 and 2 denote the colours.Cubic partition function c(n) is defined by∞∑n=0c(n)qn =1(q; q)∞(q2; q2)∞=1E(q)E(q2), (1.1)where E(q) is Euler’s product,E(q) = (q; q)∞ :=∞∏n=1(1− qn), | q |< 1.The function c(n) satisfies many Ramanujan type congruences, for examplec(3n + 2) ≡ 0 ( mod 3), ∀ n ≥ 0. Motivated by his works in [2, 3], manypartition congruences for analogous partition functions have been investigated.For example, Chen and Lin [4] found four new congruences modulo 7 by usingmodular forms, whereas Xiong [11] established sets of congruences modulo pow-ers of 5. In [6], Chern and Dastidar have presented two new congruences modulo11 for c(n). Furthermore, they have established a recursion for c(n), which ishttp://siba-ese.unisalento.it/ © 2022 Università del Salento8 K. N. Vidyaa special case of a broader class of recursions. Recently Hirschhorn [8] gave anelementary proof ofc(5αn+ δα) ≡ 0 (mod 5b(α/2)c),where α ≥ 2, n ≥ 0 and δα is the reciprocal of 8 modulo 5α.Zhao and Zhang [12] explored congruences for the following function:∞∑n=0cp(n)qn =1(q; q)2∞(q2; q2)2∞=1E2(q)E2(q2)(1.2)and proved that cp(5n+ 4) ≡ 0 ( mod 5), ∀ n ≥ 0. Since cp(n) counts a pairof cubic partitions, it is the number of cubic partition pairs. We can interpretcp(n) as the number of 4-colour partitions of n with colours r, y, o and b subjectto the restriction that the colours o and b appear only in even parts. RecentlyLin [9] studied the arithmetic properties of cp(n) modulo 27 and conjecturedthe following four congruences:cp(49n+ 37) ≡ 0 (mod 49),cp(81n+ 61) ≡ 0 (mod 243),∞∑n=0cp(81n+ 7)qn ≡ q(q2; q2)∞(q3; q3)2∞(q6; q6)∞(mod 81),∞∑n=0cp(81n+ 34)qn ≡ 36(q; q)∞(q6; q6)2∞(q3; q3)∞(mod 81)In two recent papers, Chern [5] and Lin, Wang and Xia [10] independentlyproved all the above four congruences.Let c∗(1,r,a)(n) be defined by∞∑n=0c∗(1,r,a)(n)qn =1[E(q)E(qr)]a(1.3)where a, r ≥ 1 are positive integers. c∗(1,r,a)(n) is the generalization of the cubicpartition function c(n).In this paper, we prove some quite interesting congruences modulo odd primep by taking r = 3, 4, 5, 7, 11 and 13 using q−series identities. We study congru-ence properties of generalization of cubic partition function for different valuesof a and give some particular cases as examples.In particular, some of theminvolve higher powers of the Euler function.New generalization of cubic partition of n 92 New congruences for c∗(1,r,a)(n)In this section, we prove six new congruence modulo an odd prime p. Toprove our congruences, we employ the following q-series identity from [7, equa-tion (0.46)]:E3(q) =∞∑n=−∞(4n+ 1)q[(4n+1)2−1]/8. (2.4)We also require the following congruence which follows from the binomial the-orem: For prime p and integer ` ≥ 1,Ep` ≡ Ep` (mod p). (2.5)Theorem 1. Suppose p is an odd prime divisor of a+ 3 and r is an integerwith 0 ≤ r < p. Suppose p and r satisfy the condition: 2r + 1 ≡ 0 ( mod p )and p ≡ 5 or 11 ( mod 12 ). Then, ∀n ≥ 0c∗(1,3,a)(pn+ r) ≡ 0 (mod p). (2.6)Proof. Since p divides a + 3, we can write a + 3 = pm, for some integer m.Setting r = 3 in (1.3), we find that∞∑n=0c∗(1,3,a)(n)qn =[E(q)E(q3)]3[E(q)E(q3)]pm. (2.7)Employing (2.5) in (2.7), we obtain∞∑n=0c∗(1,3,a)(n)qn =[E(q)E(q3)]3[E(qp)E(q3p)]m. (2.8)Using (2.4), we observe that[E(q)E(q3)]3 =∞∑m=−∞∞∑n=−∞(4n+ 1)(4m+ 1)q[(4n+1)2+3(4m+1)2−4]/8. (2.9)We note thatN = [(4n+ 1)2 + 3(4m+ 1)2 − 4]/8,which is equivalent to8N + 4 = (4n+ 1)2 + 3(4m+ 1)2.10 K. N. VidyaIf p ≡ 5 or 11 (mod 12), then the Legendre symbol(−3p)= −1. Therefore,it follows that8N + 4 ≡ 0 (mod p)or2N + 1 ≡ 0 (mod p)if and only if 4n + 1 ≡ 0 (mod p) and 4m + 1 ≡ 0 (mod p). Hence, thecongruences (2.6) now follows by employing (2.9) in (2.8) and then comparingthe coefficients of qpn+r. QEDCorollary 1. We havec∗(1,3,2)(5n+ 2) ≡ 0 (mod 5), (2.10)c∗(1,3,14)(17n+ 8) ≡ 0 (mod 17), (2.11)c∗(1,3,8)(11n+ 5) ≡ 0 (mod 11), (2.12)c∗(1,3,20)(23n+ 11) ≡ 0 (mod 23). (2.13)Proof. Take p = 5 and a = 2. Then, p is an odd prime, p ≡ 5 (mod 12) andp divides a+ 3.Therefore, using these in(2.6) we obtain (2.10). Similarly, takingp = 17 and a = 14 in (2.6) we obtain (2.11), taking p = 11 and a = 8 in (2.6) weobtain (2.12) and taking p = 23 and a = 20 in (2.6) we obtain (2.13). QEDTheorem 2. Suppose p is an odd prime divisor of a+ 3 and r is an integerwith 0 ≤ r < p. Suppose p and r satisfy the condition: 8r + 5 ≡ 0 (mod p)and p ≡ 3 (mod 4). Then, ∀n ≥ 0,c∗(1,4,a)(pn+ r) ≡ 0 (mod p). (2.14)Proof. Since p divides a + 3, we can write a + 3 = pm, for some integer m.Setting r = 4 in (1.3), we find that∞∑n=0c∗(1,4,a)(n)qn =[E(q)E(q4)]3[E(q)E(q4)]pm. (2.15)Employing (2.5) in (2.15), we obtain∞∑n=0c∗(1,4,a)(n)qn =[E(q)E(q4)]3[E(qp)E(q4p)]m. (2.16)New generalization of cubic partition of n 11Using (2.4), we observe that[E(q)E(q4)]3 =∞∑m=−∞∞∑n=−∞(4n+ 1)(4m+ 1)q[(4n+1)2+4(4m+1)2−5]/8. (2.17)We note thatN = [(4n+ 1)2 + 4(4m+ 1)2 − 5]/8,which is equivalent to8N + 5 = (4n+ 1)2 + 4(4m+ 1)2.If p ≡ 3 (mod 4), then the Legendre symbol(−4p)= −1. Therefore, it followsthat8N + 5 ≡ 0 (mod p)if and only if 4n+ 1 ≡ 0 (mod p) and 4m+ 1 ≡ 0 (mod p). Hence, the con-gruences (2.14) now follows by employing (2.17) in (2.16) and then comparingthe coefficients of qpn+r. QEDCorollary 2. We havec∗(1,4,4)(7n+ 2) ≡ 0 (mod 7), (2.18)c∗(1,4,8)(11n+ 9) ≡ 0 (mod 11). (2.19)Proof. Taking p = 7 and a = 4 in (2.14) we obtain (2.18) and taking p = 11and a = 8 in (2.14) we obtain (2.19). QEDTheorem 3. Suppose p is an odd prime divisor of a+ 3 and r is an integerwith 0 ≤ r < p. Suppose p and r satisfy any of the following two conditions:(1) 4r + 3 ≡ 0 (mod p), p ≡ 2 or 3 (mod 5) and p ≡ 1 (mod 4)(2) 4r + 3 ≡ 0 (mod p), p ≡ 1 or 4 (mod 5) and p ≡ 3 (mod 4)Then, ∀n ≥ 0,c∗(1,5,a)(pn+ r) ≡ 0 (mod p). (2.20)Proof. Since p divides a + 3, we can write a + 3 = pm, for some integer m.Setting r = 5 in (1.3), we find that∞∑n=0c∗(1,5,a)(n)qn =[E(q)E(q5)]3[E(q)E(q5)]pm. (2.21)12 K. N. VidyaEmploying (2.5) in (2.21), we obtain∞∑n=0c∗(1,5,a)(n)qn =[E(q)E(q5)]3[E(qp)E(q5p)]m. (2.22)Using (2.4), we observe that[E(q)E(q5)]3 =∞∑m=−∞∞∑n=−∞(4n+ 1)(4m+ 1)q[(4n+1)2+5(4m+1)2−6]/8. (2.23)We note thatN = [(4n+ 1)2 + 5(4m+ 1)2 − 6]/8,which is equivalent to8N + 6 = (4n+ 1)2 + 5(4m+ 1)2.Ifp ≡ 2 or 3 (mod 5) & p ≡ 1 (mod 4)orp ≡ 1 or 4 (mod 5) & p ≡ 3 (mod 4),then the Legendre symbol(−5p)= −1. Therefore, it follows that8N + 6 ≡ 0 (mod p)or4N + 3 ≡ 0 (mod p)if and only if 4n+ 1 ≡ 0 (mod p) and 4m+ 1 ≡ 0 (mod p). Hence, the con-gruences (2.20) now follows by employing (2.23) in (2.22) and then comparingthe coefficients of qpn+r. QEDCorollary 3. We havec∗(1,5,14)(17n+ 12) ≡ 0 (mod 17), (2.24)c∗(1,5,10)(13n+ 9) ≡ 0 (mod 13), (2.25)c∗(1,5,8)(11n+ 2) ≡ 0 (mod 11), (2.26)c∗(1,5,16)(19n+ 4) ≡ 0 (mod 19). (2.27)Proof. Setting p = 17 and a = 14 in (2.20) implies (2.24). For (2.25), we setp = 13 and a = 10 in (2.20). For (2.26), we put p = 11 and a = 8 in (2.20).Finally, by setting p = 19 and a = 16 in (2.20) we obtain (2.27). QEDNew generalization of cubic partition of n 13Theorem 4. Suppose p is an odd prime divisor of a+ 3 and r is an integerwith 0 ≤ r < p. Suppose p and r satisfy any of the following two conditions:(1) r + 1 ≡ 0 (mod p), p ≡ 3 or 5 or 6 (mod 7) and p ≡ 1 (mod 4)(2) r + 1 ≡ 0 (mod p), p ≡ 3 or 5 or 6 mod 7 and p ≡ 3 (mod 4)Then, ∀n ≥ 0,c∗(1,7,a)(pn+ r) ≡ 0 (mod p). (2.28)Proof. Since p divides a + 3, we can write a + 3 = pm, for some integer m.Setting r = 7 in (1.3), we find that∞∑n=0c∗(1,7,a)(n)qn =[E(q)E(q7)]3[E(q)E(q7)]pm. (2.29)Employing (2.5) in (2.29), we obtain∞∑n=0c∗(1,7,a)(n)qn =[E(q)E(q7)]3[E(qp)E(q7p)]m. (2.30)Using (2.4), we observe that[E(q)E(q7)]3 =∞∑m=−∞∞∑n=−∞(4n+ 1)(4m+ 1)q[(4n+1)2+7(4m+1)2−8]/8. (2.31)We note thatN = [(4n+ 1)2 + 7(4m+ 1)2 − 8]/8,which is equivalent to8N + 8 = (4n+ 1)2 + 7(4m+ 1)2.Ifp ≡ 3 or 5 or 6 (mod 7) & p ≡ 1 (mod 4)orp ≡ 3 or 5 or 6 (mod 7) & p ≡ 3 (mod 4)then the Legendre symbol(−7p)= −1. Therefore, it follows that8N + 8 ≡ 0 (mod p)orN + 1 ≡ 0 (mod p)if and only if 4n+ 1 ≡ 0 (mod p) and 4m+ 1 ≡ 0 (mod p). Hence, the con-gruences (2.28) now follows by employing (2.31) in (2.30) and then comparingthe coefficients of qpn+r. QED14 K. N. VidyaCorollary 4. We havec∗(1,7,14)(17n+ 16) ≡ 0 (mod 17), (2.32)c∗(1,7,2)(5n+ 4) ≡ 0 (mod 5), (2.33)c∗(1,7,10)(13n+ 12) ≡ 0 (mod 13), (2.34)c∗(1,7,28)(31n+ 30) ≡ 0 (mod 31), (2.35)c∗(1,7,16)(19n+ 18) ≡ 0 (mod 19). (2.36)Proof. Setting p = 17 and a = 14 in (2.28) we obtain (2.32). For (2.33), we setp = 5 and a = 2 in (2.28).For (2.34), we set p = 13 and a = 10 in (2.28). For(2.35), we set p = 31 and a = 28 in (2.28). Finally, by setting p = 19 and a = 16in (2.28) we obtain (2.36). QEDTheorem 5. Suppose p is an odd prime divisor of a+ 3 and r is an integerwith 0 ≤ r < p. Suppose p and r satisfy any of the following two conditions:(1) 2r + 3 ≡ 0 (mod p), p ≡ 2 or 6 or 7 or 8 or 10 (mod 11) and p ≡ 1(mod 4)(2) 2r + 3 ≡ 0 (mod p), p ≡ 2 or 6 or 7 or 8 or 10 (mod 11) and p ≡ 3(mod 4)Then, ∀n ≥ 0,c∗(1,11,a)(pn+ r) ≡ 0 (mod p). (2.37)Proof. Since p divides a + 3, we can write a + 3 = pm, for some integer m.Setting r = 11 in (1.3), we find that∞∑n=0c∗(1,11,a)(n)qn =[E(q)E(q11)]3[E(q)E(q11)]pm. (2.38)Employing (2.5) in (2.38), we obtain∞∑n=0c∗(1,11,a)(n)qn =[E(q)E(q11)]3[E(qp)E(q11p)]m. (2.39)Using (2.4), we observe that[E(q)E(q11)]3 =∞∑m=−∞∞∑n=−∞(4n+ 1)(4m+ 1)q[(4n+1)2+11(4m+1)2−12]/8. (2.40)New generalization of cubic partition of n 15We note thatN = [(4n+ 1)2 + 11(4m+ 1)2 − 12]/8,which is equivalent to8N + 12 = (4n+ 1)2 + 11(4m+ 1)2.Ifp ≡ 2 or 6 or 7 or 8 or 10 (mod 11) & p ≡ 1 (mod 4)orp ≡ 2 or 6 or 7 or 8 or 10 (mod 11) & p ≡ 3 (mod 4)then the Legendre symbol(−11p)= −1. Therefore, it follows that8N + 12 ≡ 0 (mod p)or2N + 3 ≡ 0 (mod p)if and only if 4n+ 1 ≡ 0 (mod p) and 4m+ 1 ≡ 0 (mod p). Hence, the con-gruences (2.37) now follows by employing (2.40) in (2.39) and then comparingthe coefficients of qpn+r. QEDCorollary 5. We havec∗(1,11,10)(13n+ 5) ≡ 0 (mod 13), (2.41)c∗(1,11,14)(17n+ 7) ≡ 0 (mod 17), (2.42)c∗(1,11,26)(29n+ 13) ≡ 0 (mod 29), (2.43)c∗(1,11,76)(79n+ 38) ≡ 0 (mod 79). (2.44)Proof. Setting p = 13 and a = 10 in (2.37) we obtain (2.41). For (2.42), we setp = 17 and a = 14 in (2.37). For (2.43), we set p = 29 and a = 26 in (2.37).Finally, by setting p = 79 and a = 76 in (2.37) we obtain (2.44). QEDTheorem 6. Suppose p is an odd prime divisor of a+ 3 and r is an integerwith 0 ≤ r < p. Suppose p and r satisfy any of the following two conditions:(1) 4r + 7 ≡ 0 (mod p), p ≡ 2 or 5 or 6 or 7 or 8 or 11 (mod 13) andp ≡ 7 (mod 4)(2) 4r + 7 ≡ 0 (mod p), p ≡ 1 or 3 or 4 or 9 or 10 or 12 (mod 13) andp ≡ 3 (mod 4)16 K. N. VidyaThen, ∀n ≥ 0,c∗(1,13,a)(pn+ r) ≡ 0 (mod p). (2.45)Proof. Since p divides a + 3, we can write a + 3 = pm, for some integer m.Setting r = 13 in (1.3), we find that∞∑n=0c∗(1,13,a)(n)qn =[E(q)E(q13)]3[E(q)E(q13)]pm. (2.46)Employing (2.5) in (2.46), we obtain∞∑n=0c∗(1,13,a)(n)qn =[E(q)E(q13)]3[E(qp)E(q13p)]m. (2.47)Using (2.4), we observe that[E(q)E(q13)]3 =∞∑m=−∞∞∑n=−∞(4n+ 1)(4m+ 1)q[(4n+1)2+13(4m+1)2−14]/8. (2.48)We note thatN = [(4n+ 1)2 + 13(4m+ 1)2 − 14]/8,which is equivalent to8N + 14 = (4n+ 1)2 + 13(4m+ 1)2.Ifp ≡ 2 or 5 or 6 or 7 or 8 or 11 (mod 13) & p ≡ 7 (mod 4)orp ≡ 1 or 3 or 4 or 9 or 10 or 12 (mod 13) & p ≡ 3 (mod 4)then the Legendre symbol(−13p)= −1. Therefore, it follows that8N + 14 ≡ 0 (mod p)or4N + 7 ≡ 0 (mod p)if and only if 4n+ 1 ≡ 0 (mod p) and 4m+ 1 ≡ 0 (mod p). Hence, the con-gruences (2.45) now follows by employing (2.48) in (2.47) and then comparingthe coefficients of qpn+r. QEDNew generalization of cubic partition of n 17Corollary 6. We havec∗(1,13,38)(41n+ 29) ≡ 0 (mod 41), (2.49)c∗(1,13,106)(109n+ 80) ≡ 0 (mod 109), (2.50)c∗(1,13,128)(131n+ 31) ≡ 0 (mod 131). (2.51)Proof. Setting p = 41 and a = 38 in (2.45) we obtain (2.49). For (2.50), we setp = 109 and a = 106 in (2.45). Finally, by setting p = 131 and a = 128 in (2.45)we obtain (2.51). QEDAcknowledgements. The author is thankful to the referee for his/hervaluable suggestions which has considerably improved the quality of the paper.References[1] H. C. Chan, Ramanujan’s cubic continued fraction and an analog of his “ most beautifulidentity ”, Int. J. Number Theory, 3 (2010), 673-680.[2] H. C. Chan, Ramanujan’s cubic partition function and Ramanujan type congruences forcertain partition functions, Int. J. Number Theory, 4 (2010), 819-834.[3] H. C. Chan, Distribution of a certain partition function modulo powers of primes, ActaMath. Sin(Engl. Ser), 27 (2011), 625-634.[4] W. Y. C. Chen, B. L. S. Lin, Congruences for the number of cubic partitions derivedfrom modular forms, Preprint, arXiv:0910.1263, 15 pp[5] S. Chern, Arithmetic properties of cubic partition pairs modulo powers of 3, Acta Math.Sin., 33(11) (2018), 1504-1512.[6] S. Chern, M. G. Dastidar, Congruences and recursions for the cubic partition, Ra-manujan J., 44 (2017), 559-66.[7] S. Cooper, Ramanujan’s theta functions. New York: Springer; 2017.[8] M. D. Hirschhorn, Cubic partitions modulo powers of 5, Ramanujan J. , 51(2020),67–84.[9] B.L.S. Lin, Congruences modulo 27 for cubic partition pairs, J. Number Theory, 171(2017), 31-42.[10] B.L.S. Lin, L. Wang, E. X. W. Xia, Congruences for cubic partition pairs modulopowers of 3 Ramanujan J., 46 (2018), 563-578.[11] X. H. Xiong, The number of cubic partitions modulo powers of 5 (Chinese), Sci. Sin.Math., 411 (2011), 1–15.[12] H. Zhao, Z. Zhang, Ramanujan type congruences for a partition function, The Elec-tronic Journal of Combinatorics, 18 (2011), p. 58.

New generalization of cubic partition of $n$

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New generalization of cubic partition of $n$

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New generalization of cubic partition of nnn

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New generalization of cubic partition of $n$