1 3 47 6 3 11 Journal of Integer Sequences, Vol. 0 (017), Article 17.4.3 Congruences Modulo Small Powers of 3 for Partitions into Odd Designated Summs B. Hemanthkumar Department of Mathematics M. S. Ramaiah University of Applied Sciences Bengaluru-560 058 India hemanthkumarb.30@gmail.com H. S. Sumanth Bharadwaj M. S. Mahadeva Naika 1 Department of Mathematics Central College Campus Bangalore University Bengaluru-560 001 India sumanthbharadwaj@gmail.com msmnaika@rediffmail.com Abstract Andrews, Lewis Lovejoy introduced a new class of partitions, partitions with designated summs. Let PD(n) denote the number of partitions of n with designated summs PDO(n) denote the number of partitions of n with designated summs in which all parts are odd. Andrews et al. established many congruences modulo 3 for PDO(n) by using the theory of modular forms. Baruah Ojah obtainednumerouscongruencesmodulo3, 4, 816forPDO(n)byusingthetafunction identities. In this paper, we prove several infinite families of congruences modulo 9, 16 3 for PDO(n). 1 Corresponding author. 1
1 Introduction Apartition ofapositiveintegernisanonincreasingsequenceofpositiveintegersλ 1,λ,...,λ m such that n = λ 1 + λ + + λ m, where the λ i s (i = 1,,...,m) are called parts of the partition. For example, the partitions of 4 are 4, 3+1, +, +1+1, 1+1+1+1. Let p(n) denote the number of partitions of n. Thus p(4) = 5. Andrews, Lewis Lovejoy [1] studied partitions with designated summs, which are constructed by taking ordinary partitions tagging exactly one of each part size. Thus the partitions of 4 with designated summs are given by 4, 3 +1, +, +, +1 +1, +1+1, 1 +1+1+1, 1+1 +1+1, 1+1+1 +1, 1+1+1+1. Let PD(n) denote the number of partitions of n with designated summs PDO(n) denote the number of partitions of n with designated summs in which all parts are odd. Thus PD(4) = 10 PDO(4) = 5. Recently, Chen et al. [5] obtained the generating functions for PD(3n), PD(3n + 1) PD(3n+) gave a combinatorial interpretation of the congruence PD(3n+) 0 (mod 3). Xia [11] proved infinite families of congruences modulo 9 7 for PD(n). For example, for all n 0 k 1 Throughout this paper, we use the notation PD( 18k 1 (1n+1)) 0 (mod 7). f k := (q k ;q k ) (k = 1,,3,...), where (a;q) := The generating function for PDO(n) satisfies (1 aq m ). m=0 PDO(n)q n = f 4f 6 f 1 f 3 f 1. (1) Using the theory of q-series modular forms Andrews et al. [1] derived PDO(3n)q n = f f6 4, () f1f 4 1 PDO(3n+1)q n = f4 f3f 3 1, (3) f1f 5 4 f6
They also established, for all n 0 PDO(3n+)q n = f3 f 6 f 1 f 4 1f 4. (4) PDO(9n+6) 0 (mod 3) PDO(1n+6) 0 (mod 3). Baruah Ojah [3] proved several congruences modulo 3, 4, 8 16 for PDO(n). For instance, PDO(8n+7) 0 (mod 8) PDO(1n+9) 0 (mod 16). The aim of this paper is to prove several new infinite families of congruences modulo 9, 16 3 for PDO(n). In particular, we prove the following Theorem 1. For all nonnegative integers α,β n, we have PDO( α+ 3 β (7n+66)) 0 (mod 144) (5) PDO( α+ 3 β (144n+138)) 0 (mod 88). (6) In Section, we list some preliminary results. We prove several infinite families of congruences modulo 9 for PDO(n) in Section 3, Theorem 1 many infinite families of congruences modulo 16 3 for PDO(n) in Section 4. Definitions preliminaries We will make use of the following definitions, notation results. Let f(a,b) be Ramanujan s general theta function [, p. 34] given by f(a,b) := n= a n(n+1) b n(n 1). Jacobi s triple product identity can be stated in Ramanujan s notation as follows: f(a,b) = ( a;ab) ( b;ab) (ab;ab). 3
In particular, ϕ(q) := f(q,q) = k= q k = f5, ϕ( q) := f( q, q) = f 1, (7) f1f 4 f ψ(q) := f(q,q 3 ) = f( q) := f( q, q ) = k=0 q k(k+1)/ = f f 1 (8) ( 1) k q k(3k 1)/ = f 1. (9) k= For any positive integer k, let k(k +1)/ be the k th triangular number k(3k ±1)/ be a generalized pentagonal number. Lemma. The following -dissections hold: f1 = f f8 5 f4f 16 1 f 1 q f f 16 f 8, (10) = f5 8 +q f 4f16, (11) ff 5 16 ff 5 8 f1 4 = f10 4 ff 8 4 1 f 4 1 = f14 4 f 14 f8 4 4q f f 4 8 f 4 +4q f 4f 4 8 f 10 (1). (13) Proof. Lemma is an immediate consequence of dissection formulas of Ramanujan, collected in Berndt s book [, Entry 5, p. 40]. Lemma 3. The following -dissections hold: f 3 1 = f3 4 3q f f1 3, (14) f 3 f 1 f 4 f6 f 3 3 = f3 4f6 +q f3 1, (15) f 1 ff 1 f 4 f 3 f1 3 = f6 4f6 3 +3q f 4f 6 f1 ff 9 1 f 7 (16) f 1 f3 3 = f f 4f 1 f 7 6 q f3 f1 6. (17) f4f 6 9 4
Proof. Hirschhorn et al. [6] established (14) (15). Replacing q by q in (14) (15), using the relation we obtain (16) (17). Lemma 4. The following -dissections hold: ( q; q) = f3 f 1 f 4, f 1 f 3 = f f8f 1 4 q f4 4f 6 f4, f4f 6 f4 f f8f 1 (18) 1 = f 8f1 5 f +q 4f 5 4 f 1 f 3 ff 4 f6f 4 4 ff 4 6f 8f 1 (19) f3 f1 = f4 4f 6 f 1 f 5 f 8 f 4 +q f 4f 6f 8 f 4 f 4 f 1. (0) Proof. Baruah Ojah [4] derived the above identities. Lemma 5. The following 3-dissections hold: ϕ( q) = ϕ( q 9 ) qf( q 3, q 15 ) (1) ψ(q) = f(q 3,q 6 )+qψ(q 9 ). () Proof. See Berndt s book [, p. 49] for a proof of (1) (). Lemma 6. The following 3-dissection holds: f 1 f = f 6f 4 9 f 3 f 18 Proof. Hirschhorn Sellers [8] have proved the above identity. qf 9 f 18 q f 3f18 4. (3) f 6 f9 Let t be a positive integer. A partition of n is called a t-core partition of n if none of the hook numbers of its associated Ferrers-Young diagram are multiples of t. Let a t (n) denote the number of t-core partitions of n. Then the generating function of a t (n) satisfies a t (n)q n = ft t f 1. (4) Many mathematicians have studied arithmetic properties of a 3 (n). See for example, Keith [9], Lin Wang [10]. Hirschhorn Sellers [7] obtained an explicit formula for a 3 (n) by using elementary methods proved 5
Lemma 7. Let 3n + 1 = k i=1 p α i i m j=1 q β j j, where p i 1 (mod 3) q j (mod 3) with α i,β j 0 be the prime factorization of 3n+1. Then k (α i +1), if all β j are even; a 3 (n) = i=1 0, otherwise. 3 Congruences modulo 9 In this section, we prove the following infinite families of congruences modulo 9 for PDO(n). Theorem 8. For all nonnegative integers α,β n, we have PDO(4 α (4n+16)) PDO(4n+16) (mod 9), (5) PDO( α 3 β (4n+4)) ( 1) α PDO(4n+4) (mod 9), (6) PDO(4 α (48n+40)) 0 (mod 9) (7) PDO( α 3 β (144n+10)) 0 (mod 9). (8) Theorem 9. For any nonnegative integer n, let 3n+1 = k q j (mod 3) are primes with α i,β j 0. Then, 6 k (α i +1) (mod 9), if all β j are even; PDO(48n+16) i=1 0 (mod 9), otherwise. i=1 p α i i m j=1 q β j j, where p i 1 (mod 3) (9) 3 k (α i +1) (mod 9), if all β j are even; PDO(7n+4) i=1 0 (mod 9), otherwise. (30) Corollary 10. Let p (mod 3) be a prime. Then for all nonnegative integers α n with p n, we have PDO(48p α+1 n+16p α+ ) 0 (mod 9) (31) PDO(7p α+1 n+4p α+ ) 0 (mod 9). (3) Theorem 11. If n cannot be represented as the sum of a triangular number three times a triangular number, then PDO(48n+4) 0 (mod 9). 6
Corollary 1. For any positive integer k, let p j 5, 1 j k be primes. If ( 3/p j ) = 1 for every j, then for all nonnegative integers n with p k n we have PDO(48p 1p p k 1p k n+4p 1p p k) 0 (mod 9). (33) By the binomial theorem, it is easy to see that for any positive integer m, f 3 m f 3m (mod 3) (34) f 9 m f 3 3m (mod 9). (35) Proof of Theorem 8. From (35), it follows that f3 3 f1 5 f 4 1 (mod 9). (36) In view of (36), we rewrite (3) as PDO(3n+1)q n f4 1f 4 f 1 f 4 f 6 (mod 9). (37) Substituting (1) in (37) extracting the terms containing odd powers of q, we get PDO(6n+4)q n 4 f6 1f 4 4f 6 f 3 f 3 (mod 9). (38) Employing (14) in (38) extracting the terms containing even powers of q, we derive PDO(1n+4)q n 4 f10 f 3 f 3 1f 6 (mod 9). (39) Substituting (16) in (39) extracting the terms containing odd powers of q, we get From (34), In view of the above two identities, PDO(4n+16)q n 1 f3 1f f 6 f 3 (mod 9). f 3 1f f 6 f 3 f3 6 f (mod 3). PDO(4n+16)q n 6 f3 6 f (mod 9), (40) 7
which implies that for all n 0 PDO(48n+40) 0 (mod 9) (41) PDO(48n+16)q n 6 f3 3 f 1 (mod 9). (4) Invoking (15) in (4) extracting the terms containing odd powers of q, By (40) (43), PDO(96n+64)q n 6 f3 6 f (mod 9). (43) PDO(96n+64) PDO(4n+16) (mod 9). (44) Congruence (5) follows from (44) mathematical induction. Congruence (7) follows from (41) (5). Employing (13) in (), which yields Applying (16) (45), which implies that PDO(3n)q n = f14 4 f6 4 +4q f 4f6f 4 8 4, f 1 f8f 4 1 ff 8 1 PDO(6n)q n = f14 f 4 3 f 1 1 f 4 4f 6 PDO(6n+3)q n = 4 f f 4 3f 4 4 f 8 1f 6 PDO(6n)q n = f14 f4f 4 6 From (35), f 0 f3 10 f1 f6 8 In view of the above two identities, ( f 6 4 f 3 6 f 9 f 1 PDO(1n)q n f0 f3 10 f1 f6 8 f f 4 3 f 4 1f 6 PDO(1n)q n f f 4 3 f 4 1f 6 8 (45). (46) ) +3q f 4f 6 f1 4, f 7 (mod 9). (mod 9). (mod 9). (47)
Substituting (15) (16) in (47), extracting the terms containing even powers of q, we have Using (34) (35) in (48), we get PDO(4n)q n f9 f 3 3 f 9 1f 3 6 +3q f f 5 6 f 5 1f 3 (mod 9). (48) PDO(4n+4)q n 3 f 1f f 5 6 f 3 3 (mod 9). (49) Substituting (17) in (49) using (34), we have PDO(4n+4)q n 3 f f 4f 1 f 6 3q f4 f 6 1 f 4f 4 6 3 f 4f 1 f f 6 3q f f 4 f 5 1 f 3 6 (mod 9), which yields By (49) (51), PDO(48n+4)q n 3ψ(q)ψ(q 3 ) (mod 9) (50) PDO(48n+48)q n 3 f 1f f 5 6 f 3 3 (mod 9). (51) Employing (3) in (49) using (34), PDO((4n+4)) PDO(4n+4) (mod 9). (5) PDO(4n+4)q n 3 f6 6f9 4 3q f5 6f 9 f 18 6q f4 6f18 4 f3f 4 18 f3 3 f3f 9 3 f3 9 f 3 +6q f3 18 f 6 +3q f 3f 6 f 5 18 f 3 9 (mod 9) which implies that PDO(7n+4)q n 3 f3 3 f 1 (mod 9), (53) PDO(7n+48)q n 6 f3 6 f (mod 9) (54) 9
From (54), By (49) (55), PDO(7n+7)q n 3 f 1f f 5 6 f 3 3 (mod 9). (55) PDO(144n + 10) 0 (mod 9). (56) PDO(3(4n+4)) PDO(4n+4) (mod 9). (57) Congruence (6) follows from (5), (57), mathematical induction. Congruence (8) follows from (56) (6). Proof of Theorem 9. From (4), (4) (53), it is clear that for all n 0 PDO(48n+16) 6a 3 (n) (mod 9), (58) PDO(7n+4) 3a 3 (n) (mod 9). (59) Congruence (9) follows from Lemma 7 (58). Congruence (30) follows from Lemma 7 (59). For any prime p any positive integer N, let υ p (N) denote the exponent of the highest power of p dividing N. Proof of Corollary 10. Suppose α 0, p (mod 3) p n, then it is clear that ( ) ) υ p (3 p α+1 n+ pα+ 1 ( +1 = υ p 3p α+1 n+p α+) = α+1. (60) 3 Congruences (31) (3) follow from (9), (30) (60). Proof of Theorem 11. From (8) (50), we have PDO(48n+4)q n 3 Theorem 11 follows from (61). Proof of Corollary 1. By (61), k=0 m=0 PDO(48n+4)q 48n+4 3 q k(k+1)/+3m(m+1)/ (mod 9). (61) k=0 m=0 q 6(k+1) +(6m+3) (mod 9), 10
whichimpliesthatif48n+4isnotoftheform6(k+1) +(6m+3), thenpdo(48n+4) 0 (mod 9). Let k 1 be an integer let p i 5, 1 i k be primes with ( 3 p i ) = 1. If N is of the form x +6y, then v pi (N) is even since ( 3 p i ) = 1. Let ( N = 48 p 1p p k 1p k n+ p 1p p k 1 p k 1 ) +4 = 48p 1p p k 1p k n+4p 1p p k 1p k. If p k n, then v pk (N) is an odd number hence N is not of the form x +6y. Therefore (33) holds. 4 Congruences modulo 4 5 In this section, we establish the following infinite families of congruences modulo 16 3 for PDO(n). Theorem 13. For all nonnegative integers α n, we have PDO(4 α (1n+8)) PDO(1n+8) (mod 4 ), (6) PDO(4 α (4n+3)) 0 (mod 4 ), (63) PDO(4 α (48n+14)) 0 (mod 4 ) (64) PDO(4n+17) 0 (mod 4 ). (65) Theorem 14. For all nonnegative integers α n, we have PDO( α (1n)) PDO(1n) (mod 4 ), (66) PDO( α (7n+4)) 0 (mod 4 ) (67) PDO( α (7n+66)) 0 (mod 4 ). (68) Theorem 15. For all nonnegative integers α n, we have PDO(4 α (4n) PDO(4n) (mod 5 ), (69) PDO(9 α (6n+3)) PDO(6n+3) (mod 5 ), (70) PDO(4n+9) 0 (mod 5 ), (71) PDO(9 α (16n+117)) 0 (mod 5 ), (7) PDO( α (7n+69)) 0 (mod 5 ), (73) PDO(3 α (7n+69)) 0 (mod 5 ) (74) PDO( α (144n+4)) 0 (mod 5 ). (75) 11
Theorem 16. If n cannot be represented as the sum of two triangular numbers, then for all nonnegative integers α r {1,6} we have PDO( α r(1n+3)) 0 (mod 4 ). Corollary 17. If p is a prime, p 3 (mod 4)), 1 j p 1 r {1,6}, then for all nonnegative integers α,β n, we have PDO( α p β+1 r(1pn+1j +3p)) 0 (mod 4 ). (76) For example, taking p = 3, we deduce that for all α,β,n 0, Combining (77) (67), Combining (78) (68), PDO( α 3 β (16n+16)) 0 (mod 4 ) (77) PDO( α 3 β (16n+198)) 0 (mod 4 ). (78) PDO( α 3 β (7n+4)) 0 (mod 4 ). (79) PDO( α 3 β (7n+66)) 0 (mod 4 ). (80) Theorem 18. If n cannot be represented as the sum of a triangular number four times a triangular number, then for all nonnegative integers α r {1,3} we have PDO( α r(48n+30)) 0 (mod 5 ). Corollary 19. If p is any prime with p 3 (mod 4)), 1 j p 1 r {1,3}, then for all nonnegative integers α,β n, we have PDO( α p β+1 r(48pn+48j +30p)) 0 (mod 5 ). (81) For example, taking p = 3 we find that for all α,n 0 β 1, Combining (8) (75), for all α,β,n 0, PDO( α 3 β (144n+4)) 0 (mod 5 ) (8) PDO( α 3 β (144n+138)) 0 (mod 5 ). (83) PDO( α 3 β (144n+4)) 0 (mod 5 ) (84) combining (83), (73) (74), for all α,β,n 0, PDO( α 3 β (7n+69)) 0 (mod 5 ). (85) 1
Theorem 0. If n cannot be represented as the sum of twice a pentagonal number three times a triangular number, then for any nonnegative integer α we have PDO(4 α (4n+11)) 0 (mod 4 ). Corollary 1. For any positive integer k, let p j 5, 1 j k be primes. If ( /p j ) = 1 for every j, then for all nonnegative integers α n with p k n we have PDO(6 4 α+1 p 1p p k 1p k n+11 4 α p 1p p k) 0 (mod 4 ). Theorem. If n cannot be represented as the sum of a pentagonal number six times a triangular number, then for any nonnegative integer α we have PDO(4 α (48n+38)) 0 (mod 4 ). Corollary 3. For any positive integer k, let p j 5, 1 j k be primes. If ( /p j ) = 1 for every j, then for all nonnegative integers n with p k n we have PDO(3 4 α+ p 1p p k 1p k n+38 4 α p 1p p k) 0 (mod 4 ). Theorem 4. If n cannot be represented as the sum of a pentagonal number four times a pentagonal number, then we have PDO(4n+5) 0 (mod 5 ). Corollary 5. For any positive integer k, let p j 5, 1 j k be primes. If ( 1/p j ) = 1 for every j, then for all nonnegative integers n with p k n we have PDO(4p 1p p k 1p k n+5p 1p p k) 0 (mod 5 ). Theorem 6. If n cannot be represented as the sum of a pentagonal number sixteen times a pentagonal number, then we have PDO(4n+17) 0 (mod 5 ). Corollary 7. For any positive integer k, let p j 5, 1 j k be primes. If ( 1/p j ) = 1 for every j, then for all nonnegative integers n with p k n we have PDO(4p 1p p k 1p k n+17p 1p p k) 0 (mod 5 ). By the binomial theorem, it is easy to see that for all positive integers k m, f k m f k 1 m (mod k ). (86) 13
Proof of Theorem 13. Using (13), we can rewrite (4) as which yields PDO(3n+)q n = f13 4 f 6 f 1 f 11 f 4 8 From (86) with k = 3 k =, we have PDO(6n+)q n = f13 f 3 f 6 f 11 1 f 4 4 +8q f 4f 6 f8f 4 1, f 7 PDO(6n+5)q n = 8 f f 3 f4f 4 6. f1 7 f 13 f 3 f 6 f 11 1 f 4 4 f f 3 f 6 f 3 1 (mod 3 ) Thus, f f 3 f 4 4f 6 f 7 1 f 3f 4 4f 6 f 3 1f (mod ). PDO(6n+)q n f f 3 f 6 f 3 1 (mod 4 ) PDO(6n+5)q n 8 f 3f 4 4f 6 f 3 1f (mod 5 ). Substituting (16) in the above two congruences extracting the terms containing even odd powers of q, we get PDO(1n+)q n f6 f 4 3 f 8 1f 6 PDO(1n+8)q n 6 f f 3f 6 f 6 1 PDO(1n+5)q n 8 f10 f 4 3 f 10 1 f 6 f f 4 3 f 6 8 f6 f 1 (mod 4 ), (87) (mod 4 ), (88) (mod 5 ) (89) PDO(1n+11)q n 4 f6 f 3f 6 f 8 1 4f 4 f 3 6 (mod 4 ). (90) 14
It follows from (90) that PDO(4n+11)q n 8f f 3 3 8f f 6 f 3 (mod 4 ) (91) PDO(4n+3) 0 (mod 4 ). (9) Employing (1) in (87) extracting the terms containing odd powers of q, which implies that PDO(4n+14)q n 8q f 1f 4 1 f 6 8qf f 3 1 (mod 4 ), PDO(48n+38)q n 8f 1 f 3 6 8f 1 f 1 f 6 (mod 4 ) (93) PDO(48n+14) 0 (mod 4 ). (94) Substituting (13) (0) in (88), extracting the terms containing even odd powers of q, we have PDO(4n+8)q n 6 f18 f3f 3 6 6 f f3f 3 6 (mod 4 ) (95) f1 17 f4f 5 1 f 1 f 4 f 1 PDO(4n+0)q n 1 f15 f3f 4 1 +4 f6 f3f 3 4f 3 6 f1 16 f4f 3 6 f1 13 f 1 1 f7 f 6 f 1 f 3 4 +4 f3 3f 3 4 f 1 (mod 4 ). (96) Substituting (15) in (95) (96), extracting even odd powers of q, we have PDO(48n+8)q n 6 f f 4 3 f 6 PDO(48n+3)q n 6 f 1f 3f 6 f (mod 4 ), (97) (mod 4 ) (98) PDO(48n+44)q n 8f f 3 6 8 f 4f 1 f 6 (mod 4 ). (99) 15
Again employing (1) in (97) extracting the terms containing odd powers of q, PDO(96n+56)q n 8q f 1f 4 1 f 6 8qf f 3 1 (mod 4 ). (100) By (99) (100), From (86), (88) (98), PDO(96n+9) 0 (mod 4 ) (101) PDO(19n+56) 0 (mod 4 ). (10) PDO(48n+3) PDO(1n+8) (mod 4 ). (103) Congruence (6) follows from (103) mathematical induction. Congruence (63) follows from (9), (101), (6). Similarly, congruence (64) follows from (94), (10), (6). By invoking (11) in (89) extracting the terms containing even odd powers of q, PDO(4n+5)q n 8 f 1f 5 4 f 8 8f 1 f 4 (mod 5 ) (104) Congruence (65) follows from (105). PDO(4n+17)q n 16 f 1f f 8 f 4 16f 1 f 16 (mod 5 ). (105) Proofs of Theorem 14 Theorem 15. From (86), f f 4 3f 4 4 f 8 1f 6 f4 3f 4 4 f f 6 (mod 3 ). (106) Using (106), we rewrite (46) as PDO(6n+3)q n 4 f4 3f 4 4 f f 6 (mod 5 ). (107) In view of (7) (8), PDO(6n+3)q n 4ψ (q )φ ( q 3 ) (mod 5 ). (108) Substituting () in (108) extracting the terms involving q 3n+1, we get PDO(18n+9)q n 4qψ (q 6 )φ ( q) (mod 5 ). (109) 16
Using (1) in (109) extracting the terms involving q 3n+1, PDO(54n+7)q n 4ψ (q )φ ( q 3 ) (mod 5 ). (110) Congruence (70) follows from (108), (110), mathematical induction. Applying (1) in (107) using (86), PDO(6n+3)q n 4 f4 4f1 10 16q 3f4 4f4 4 ff 6f 4 4 4 ff 1 4 f4 4f 1 f f 4 6 +16q 3f 8f 48 f 4 f 4 (mod 5 ) which implies that Congruence (71) follows from (11). From (11) (8), PDO(1n+3)q n 4 f4 f 6 f 1f 4 3 (mod 5 ) (111) PDO(1n+9)q n 16q f 4f 4 f f 1 (mod 5 ). (11) PDO(4n+1)q n 16ψ(q)ψ(q 6 ) (mod 5 ). (113) Invoking () in (113) extracting the terms containing q 3n+ q 3n+1, for all n 0 PDO(7n+69) 0 (mod 5 ) (114) PDO(7n+45)q n 16ψ(q )ψ(q 3 ) (mod 5 ). (115) Using () in (115) extracting the terms containing q 3n+1, Congruence (7) follows from (70) (116). PDO(16n+117) 0 (mod 5 ). (116) 17
Substituting (11) (13) in (111), extracting the terms containing odd powers of q, we get PDO(4n+15)q n 8 f f6 14 f8 +16q f5 4f6f 1 4 f 1 f3 1 f 4 f1 4 f 1 f3f 8 8 8ψ(q)ψ(q 4 )+16qψ(q)ψ(q 1 ) (mod 5 ). (117) Employing () in (117) extracting the terms involving q 3n+, we get PDO(7n+63)q n 16ψ(q 3 )ψ(q 4 )+8qψ(q 3 )ψ(q 1 ) (mod 5 ). (118) Again, extracting the terms involving q 3n+ in (118), we get PDO(16n+07) 0 (mod 5 ). (119) Congruence (74) follows from (114), (119), (70). Substituting (1) (13) in (45), extracting the terms containing even powers of q, we find PDO(1n)q n f38 f6 10 f1 8 f3f 4 1 4 f 4 1 f4 1f 6 f 4 1 f 4 3f 4 4f 6 6 (mod 4 ). (10) Using (1) (13) in (10), extracting the terms containing even powers of q, we have From (10) (11), PDO(4n)q n f4 1ff 6 6 18 f3 0 f4f 4 1 4 f4 1f 6 f 4 1 f 4 3f 4 4f 6 6 (mod 4 ). (11) PDO((1n)) PDO(1n) (mod 4 ). (1) Congruence (66) follows from (1) mathematical induction. Substituting (0) (13) in (45), using (86), we get PDO(6n)q n = f3 4 f 4 1 f 4 f8 10 f4 +16q f8 4f 6 8f 4 1 f 16 f 4 +64q 4f 4f 6f 10 8 f 4 f 14 f 1 +3q 3f14 4 f6f 8f 4 f 18 f1 +4q f6 4 f6f 4 f f8f 6 1 +3q f17 4 f 6 f 1 f 19 +4q f9 4 f 6 f 1 f 3 f 8 8 f8 f4 16 f1 4 +16q f f8 10 f 8f 4 16 +4q f 4f6f 8f 4 ff 6 1 +8qf4f 8 +4qf f 4 f 6 f 1 (mod 5 ), +8q f0 4 f 4 1 f 0 f 8f 4 +64q 3f5 4f 6 f 8 8f 1 f 15 18
which yields PDO(1n)q n f8 1f 16 f 4 6 f 10 4 f 1 +16qf 4f 8 +4q f f 3f 4f 1 f 6 1f 6 (mod 5 ) (13) PDO(1n+6)q n 8ff 4 +4f 1 f f 3 f 6 (mod 5 ). (14) Substituting (18) in (14) extracting the terms containing even odd powers of q, we get PDO(4n+6)q n 4 f 1f4f 6 4 +8f f 1f f 1 (mod 5 ) (15) By (86) (8), In view of above identities, PDO(4n+18)q n 4 f4 f 3f 1 f 4f 6 4 f 1f 4f 4 6 f f 1 f 4 f 3f 1 f 4f 6 ψ (q 3 ) (mod ) +8f 1f 4ψ (q) (mod 4 ). (mod 5 ). (16) PDO(4n+6)q n 4ψ (q) (mod 4 ) (17) PDO(4n+18)q n 4ψ (q 3 ) (mod 4 ). (18) It follows from (18) that PDO(7n+4) 0 (mod 4 ) (19) for all n 0 PDO(7n+66) 0 (mod 4 ) (130) PDO(7n+18)q n 4ψ (q) (mod 4 ). (131) 19
Employing (10) in (15) (16) extracting the terms containing odd powers of q, we get PDO(48n+30)q n 8 f f 4 3f 8 f 1 f 4 f 6 In view of (133), we have for all n 0 +16 f3 1f 8 f 4 8ψ(q)ψ(q 4 ) (mod 5 ) (13) PDO(48n+4)q n 8q f4 1f 6f 4 f f 3 f 1 8qψ(q 3 )ψ(q 1 ) (mod 5 ). (133) PDO(144n+4) 0 (mod 5 ) (134) PDO(144n+138) 0 (mod 5 ) (135) PDO(144n+90)q n 8ψ(q)ψ(q 4 ) (mod 5 ). (136) Substituting (1), (13), (0) in (13), extracting the terms containing even odd powers of q, we get PDO(4n)q n f1 1 f 10 f3 4 f4f 8 6 f4 3f 8 4 f 0 1 f 6 f 6 +16q f0 1 f 4 3f 8 4 f 14 f 6 +16q f8 f4f 3 6 4 +8q f17 f 6 f 1 f1 13 f 3 f 1 f1 16 f4 3 +16qf 6 4 +16q f3 1f 3 4f 1 f 3 +8qf f 4 f 6 f 1 (mod 5 ) (137) PDO(4n+1)q n 8 f16 1 f 4 3 f f 6 +4 f 0 f6 4 +16f f1 17 f 3 f f 4f 5 4 1 f 4 4 f6 4 +8f f 1 f 3 f 4 f f 4 (mod 5 ). (138) 1 Employing (19) in (138) extracting the terms involving even odd powers of q, we get PDO(48n+1)q n 4 f 1f4f 6 4 +8f f 1f f 1 (mod 5 ) (139) PDO(48n+36)q n 4 f4 f 3f 1 f 4f 6 (mod 5 ). (140) 0
In view of (15), (16), (139), (140), we have PDO((4n+6)) PDO(4n+6) (mod 5 ) (141) PDO((4n+18)) PDO(4n+18) (mod 5 ). (14) Congruence (67) follows from (19), (14), (66). Congruence (68) follows from (130), (14), (66). Substituting (13), (14), (0) in (137), extracting even odd powers of q, we get PDO(48n)q n f7 f 4 6 f1 7 f4 18 f1 +4q f66 f 3f 1 f1 70 f4 14 f6 48q f 1f f 4 6 f 3 f8 f6 4 +4q f f3f 4f 1 +16qf f1f 8 4f 1 f1f 6 6 f 4 f 6 f 1 (mod 5 ) (143) PDO(48n+4)q n 4 f69 f 3 f 6 f1 71 f4 16 +16 f60 f 4 6 f1 68 f4 10 f1 +8f 1 f f 3 f 6 1f 1 f f 3 f 6 +16f f 4 (mod 5 ). (144) Substituting (18) in (144) extracting the terms containing even odd powers of q, we get PDO(96n+4)q n 4 f 1f4f 6 4 +16f f 1f f 1 (mod 5 ) (145) From (16) (146), we have PDO(96n+7)q n 1 f4 f 3f 1 f 4f 4 6 (mod 5 ). (146) PDO( (4n+18)) 3PDO(4n+18) (mod 5 ). (147) Employing (10) in (145) extracting the terms containing odd powers of q, we get PDO(19n+10)q n 8 f f3f 4 8 8ψ(q)ψ(q 4 ) (mod 5 ). (148) f 1 f 4 f6 Substituting (13) (0) in (143), extracting the terms containing even odd power of q, we have PDO(96n)q n f6 f 4 3 f 0 1 f 8 4f 6 f4 3f 8 4 f 0 1 f 6 f 6 +16q f f 4 3f 8 4 f 1 1 f 6 +16q f8 f 3 4f 4 6 f 13 +8q f17 f 6 f 1 1 f 3 f 1 f1 16 f4 3 +16qf 6 4 +16q f3 1f 3 4f 1 f 3 +8qf f 4 f 6 f 1 (mod 5 ) (149) 1
PDO(96n+48)q n 8 f4 3f 14 f1 16 f6 +4 f 0 f6 4 +16f f1 17 f 3 f4f 5 1 f f 3 f 6 1 By (149) (137) 8ff 4 f 4 +4 f6 4 +16f 1 f f 3 f 6 (mod 5 ). (150) f 1 f 3 f 4 f 1 PDO(4(4n)) PDO(4n) (mod 5 ). (151) Congruence (69) follows from (151) mathematical induction. By substituting (19) (18) in (150), extracting the terms containing even odd powers of q, PDO(19n+48)q n 8f 1f +0 f 1f 4f 4 6 f f 1 From (15) (15), In view of (16) (153), PDO(19n+144)q n 1 f4 f 3f 1 f 4f 6 (mod 5 ) (15) (mod 5 ). (153) PDO( 3 (4n+6)) 3PDO(4n+6) (mod 5 ). (154) PDO( 3 (4n+18)) 3PDO(4n+18) (mod 5 ). (155) Congruence (73) follows from (114), (135), (14), (147), (155), (69). Congruence (75) follows from (134), (14), (147), (155), (69). Proof of Theorem 16. Using (86) (8) in (111), PDO(1n+3)q n 4ψ (q) (mod 4 ). (156) From (156), (17), (141) (66), PDO( α (4n+6))q n 4ψ (q) (mod 4 ) (157) From (131), (141) (66), PDO( α (7n+18))q n { 4ψ (q) (mod 4 ), if α = 0; 4ψ (q) (mod 4 ), if α 0. (158)
Combining (156), (157), (158), ( 1) r+1 4 q k(k+1)/+m(m+1)/ (mod 4 ), if α = 0; PDO( α r(1n+3))q n k,m=0 ( 1) r 4 q k(k+1)/+m(m+1)/ (mod 4 ), if α 0. k,m=0 Theorem 16 follows from (159). Proof of Theorem 18. From (13), (141), (148), (154) (69), we have (159) PDO( α (48n+30))q n 8ψ(q)ψ(q 4 ) (mod 5 ). (160) From (136), (14), (147), (155) (69), we have { PDO( α (144n+90))q n 8ψ(q)ψ(q 4 ) (mod 5 ), if α = 0; 8ψ(q)ψ(q 4 ) (mod 5 ), if α 0. (161) Theorem 18 follows from (160) (161). Proofs of Theorems 0,, 4 6. From (99), PDO(96n+44)q n 8f f 6 f 3 (mod 4 ). (16) Replacing n by 8n+3 in (6), we see that for all α,n 0, In view of (8), (91), (16) (163), PDO(4 α (96n+44)) PDO(96n+44) (mod 4 ). (163) PDO(4 α (4n+11)q n 8f ψ(q 3 ) (mod 4 ). (164) Theorem 0 follows from (164). From (100), PDO(19n+15)q n 8f 1 f 1 f 6 (mod 4 ). (165) Replacing n by 16n+1 in (6), we see that for all α,n 0, PDO(4 α (19n+15)) PDO(19n+15) (mod 4 ). (166) 3
In view of (8), (93), (165) (166), PDO(4 α (48n+38)q n 8f 1 ψ(q 6 ) (mod 4 ). (167) Theorem follows from (167). Theorem 4 follows from (9) (104). Theorem 6 follows from (9) (105). The proofs of Corollaries 17, 19, 1, 3, 5 7 are similar to the proof of Corollary 1; hence we omit the details. Proof of Theorem 1. Replacing n by n+1 in (8), PDO( α+ 3 β (7n+66)) 0 (mod 9). (168) Congruence (5) follows readily from (80) (168). Replacing n by 4n+3 in (8), PDO( α+ 3 β (144n+138)) 0 (mod 9). (169) Congruence (6) follows readily from (83) (169). 5 Acknowledgments The authors would like to thank the anonymous referee for his/her valuable comments suggestions. The second author was supported by CSIR Senior Research Fellowship (No. 09/039(0111)/014-EMR-I). References [1] G. E. Andrews, R. P. Lewis, J. Lovejoy, Partitions with designated summs, Acta Arith. 10 (00), 51 66. [] B. C. Berndt, Ramanujan s Notebooks, Part III, Springer-Verlag, 1991. [3] N. D. Baruah K. K. Ojah, Partitions with designated summs in which all parts are odd, Integers 15 (015), #A9. [4] N. D. Baruah K. K. Ojah, Analogues of Ramanujan s partition identities congruences arising from his theta functions modular equations, Ramanujan J. 8 (01), 385 407. [5] W. Y. C. Chen, K. Q. Ji, H. T. Jin, E. Y. Y. Shen, On the number of partitions with designated summs, J. Number Theory 133 (013), 99 938. 4
[6] M. D. Hirschhorn, F. Garvan, J. Borwein, Cubic analogue of the Jacobian cubic theta function θ(z, q), Canad. J. Math. 45 (1993), 673 694. [7] M. D. Hirschhorn J. A. Sellers, Elementary proofs of various facts about 3-cores, Bull. Aust. Math. Soc. 79 (009), 507 51. [8] M. D. Hirschhorn J. A. Sellers, A congruence modulo 3 for partitions into distinct non-multiples of four, J. Integer Sequences 17 (014), Article 14.9.6. [9] W. J. Keith, Congruences for 9-regular partitions modulo 3, Ramanujan J. 35 (014), 157 164. [10] B. L. S. Lin A. Y. Z. Wang, Generalisation of Keith s conjecture on 9-regular partitions 3-cores, Bull. Aust. Math. Soc. 90 (014), 04 1. [11] E. X. W. Xia, Arithmetic properties of partitions with designated summs, J. Number Theory 159 (016) 160 175. 010 Mathematics Subject Classification: Primary 11P83; Secondary 05A17. Keywords: partition with designated summ, congruence, theta function. (Concerned with sequences A07785 A10186.) Received October 5 016; revised version received January 13 017; January 3 017. Published in Journal of Integer Sequences, January 8 017. Return to Journal of Integer Sequences home page. 5