Congruences modulo 3 for two interesting partitions arising from two theta function identities

Transcription

1 Note di Matematica ISSN , e-issn Note Mat no., 1 7. doi:10.185/i v3n1 Congruences modulo 3 for two interesting artitions arising from two theta function identities Kuwali Das Deartment of Mathamatical Sciences, Bodol University, Kokrajhar , Assam, INDIA kwldas90@gmail.com Received: ; acceted: Abstract. We find several interesting congruences modulo 3 for 5-core artitions two color artitions. Keywords: t-core artition, Theta function, Dissection, Congruence. MSC 000 classification: MSC 000 classification: rimary 11P83; secondary 05A17 Introduction A artition of n is a non-increasing sequence of ositive integers, called arts, whose sum is n. A artition of n is called a t-core of n if none of the hook numbers is a multile of t. If a t n denotes the number of t-cores of n, then the generating function for a t n is [5] a t nq n = qt ;q t t q;q, 1 where, here throughout the sequel, for any comlex number a q < 1, a;q := 1 q n. n=1 The following exact formula for n in terms of the rime factorization of n + 1 can be found in [5, Theorem ]. This theorem follows from an identity of Ramanujan recorded in his famous manuscrit on the artition function tau-function now ublished with his lost notebook [,. 139]. htt://siba-ese.unisalento.it/ c 01 Università del Salento

2 K. Das Theorem 1. Let n + 1 = 5 c a as s q b qat t be the rime factorization of n+1 into rimes i 1, mod 5, q j,3 mod 5. Then n = 5 c s i=1 a i+1 i 1 i 1 t j=1 q b j+1 j 1 q j 1. Many arithmetical identities congruences easily follow from the above theorem. For examle, for any ositive integer n non-negative integer r, we have 5 α n 1 = 5 α n 1 0 mod 5 α. Similarly, for a rime,3 mod 5 any non-negative integers n r, we can easily deduce that α n+1 1 = α α n 0 mod α α Next, let k n denote the number of -color artitions of n where one of the colors aears only in arts that are multiles of k. Then the generating function for k n is given by k nq n = 1 q;q q k ;q k. Recently, the following result was roved by Baruah, Ahmed Dastidar [1]. Theorem. If k {0, 1,, 3,, 5, 10, 15, 0}, then for any non-negative integer n, k 5n+l 0 mod 5, where k +l =. In this aer, we find some interesting congruences modulo 3 for 5-core artitions 5 n by emloying Ramanujan s theta functions their dissections. Since our roofs mainly rely on various roerties of Ramanujan s theta functions dissections of certain q-roducts, we end this section by defining a t-dissection Ramanujan s general theta function some of its secial cases. If Pq denotes a ower series in q, then a t-dissection of Pq is given by t 1 Pq = q k P k q t, k=0 where P k are ower series in q t. In the remainder of this section, we introduce

3 Congruences modulo 3 for two interesting artitions 3 Ramanujan s theta functions some of their elementary roerties, which will be used in our subsequent sections. For ab < 1, Ramanujan s general theta-function fa,b is defined by fa,b := n= a nn+1/ b nn 1/. In this notation, Jacobi s famous trile roduct identity [3,. 35, Entry 19] takes the form fa,b = a;ab b;ab ab;ab. Three imortant secial cases of fa,b are ϕq := fq,q = q ;q q;q q;q q ;q = q ;q 5 q;q q ;q, 3 ψq := fq,q 3 = q ;q q;q f q := f q, q = q;q, 5 where the roduct reresentations in 5 arise from the last equality in 5 is Euler s famous entagonal number theorem. After Ramanujan, we also define hence, χq := q;q = q ;q q;q q ;q χ q := q;q = q;q q ;q 7 Furthermore, the q-roduct reresentations of ϕ q ψ q can be written in the forms ϕ q = q;q q ;q, ψ q = q;q q ;q q ;q 8

4 K. Das Lemma 1. [, Theorem.] For any rime 5, q;q = where 1 k= 1 k ± 1 1 k q 3k +k f q 3 +k+1, q 3 k+1 ± 1 Furthermore, for 1 1, if 1 mod ; := 1, if 1 mod. k ± 1 q 1 q ;q, 9 k ± 1, 3k +k 1 mod. Some congruences modulo 3 for 5-core artitions We first introduce an imortant lemma which will be used later. Lemma. We have q ;q q 10 ;q 10 q q ;q q 30 ;q 30 +q ;q +q q 30 ;q 30 mod Proof. We note that [3,. 58], ϕ q ϕ q 5 = qχqf q 5 f q ϕ q 5 ϕ q = χqχ q 5 ψ q. 1 Multilying 11 1 ϕ qϕ q 5 ϕ q 5ϕ q 5 = 1qχ qχq 5 f q 5 f q 0 ψ q. 13 Emloying 5,, 7 8 in the last equation, we obtain ϕ qϕ q 5 ϕ q 5ϕ q 5 = 1qq ;q q 10 ;q 10. 1

5 Congruences modulo 3 for two interesting artitions 5 Taking congruences modulo 3, we obtain qq ;q q 10 ;q 10 ϕqϕq 3 +ϕq 5 ϕq 15 mod Recalling [3,.9, Corollary i], we have Relacing q by q 5 in the above equation, we have, ϕq = ϕq 9 +qfq 3,q 15 1 ϕq 5 = ϕq 5 +q 5 fq 15,q Emloying the above two equation in 15, we find that qq ;q q 10 ;q 10 ϕq 3 ϕq 9 +qϕq 3 fq 3,q 15 +ϕq 15 ϕq 5 Note that +q 5 ϕq 15 fq 15,q 75 mod ϕqfq,q 5 = q ;q 7 q 3 ;q 3 q 1 ;q 1 q;q 3 q ;q 3 q ;q 19 Since, q;q 3 q 3 ;q 3, so 19 can be written as Emloying 15 0 in 18, we obtain ϕqfq,q 5 q ;q mod 3. 0 qq ;q q 10 ;q 10 q 3 q ;q q 30 ;q 30 +qq ;q +q 5 q 30 ;q 30 mod 3. 1 From above congruence we can easily obtain 10. QED Theorem 3. We have, 3nq n q 3 ;q 3 q 5 ;q 5 mod 3, 3n+1q n q;q q 15 ;q 15 mod 3, 3 3n+q n nq n mod 3.

6 K. Das Proof. Putting t = 5 in 3.7 relacing q by q, we have Taking congruences modulo 3, nq n = q10 ;q 10 5 q ;q. 5 nq n q30 ;q 30 q ;q q 10 ;q 10 q ;q mod 3. With the hel of 10, can be rewritten as nq n q30 ;q 30 q ;q q q ;q q 30 ;q q ;q +q q 30 ;q 30 mod 3 q q ;q q 30 ;q q ;q 3 q 30 ;q q q30 ;q 30 5 q ;q mod 3 q q ;q q 90 ;q 90 +q 18 ;q 18 q 30 ;q q nq n mod 3. Comaring the terms involving q n, q n+ q n+ resectively from both sides of the above congruence, we can easily obtain. QED Alying the mathematical induction in, we can easily obtain the following Corollary 1. For any nonnegative integers k n, we have 3 k n+3 k 1 k n mod 3. Theorem. We have, 15n+ 0 mod 3, 30 15n+1 0 mod 3, 31 15n+9 3n+1 mod 3. 3

7 Congruences modulo 3 for two interesting artitions 7 Proof. From [3,. 70, Entry 1v], we recall that Aq q;q = q 5 ;q 5 5 Bq 5 q qbq5 Aq 5 33 where Aq = f q10, q 15 f q 5, q 0 Bq = f q5, q 0 f q 10, q 15. Relacing q by q 3 in 33 then emloying in, we obtain Aq 3nq n q 5 ;q 5 q 75 ;q Bq 15 q3 q Bq15 Aq 15 mod 3 3 Extracting the terms involving q 5n+3 from both sides of the congruence, we obtain, 35n+3q n q;q q 15 ;q 15 mod 3 35 Emloying 3 in 35, we can easily obtain 3. We have seen that in the right h side of the congruence 3, there is no terms involving q 5n+ q 5n+ hence we can easily obtain resectively. QED Theorem 5. We have, 15n+ 3n mod 3, 3 15n+10 0 mod 3, 37 Proof. Emloying 33 in 3, we obtain 15n+13 0 mod Aq 3n+1q n q 15 ;q 15 q 5 ;q 5 5 Bq 5 q qbq5 Aq 5 mod 3 39 Extracting the terms involving q 5n+1 from both sides of the congruence, we obtain,

8 8 K. Das 35n+1+1q n q 3 ;q 3 q 5 ;q 5 mod 3 0 Emloying in 0, we can easily obtain 3. It is clear that in the right h side of the congruence 39, there is no terms involving q 5n+3 q 5n+ hence we can easily obtain QED Theorem. For any rime 5 with 15 = 1 for any non- negative integers k n, 3 k n+ k 1 3n mod 3. 1 Proof. With the hel of 9, can be rewritten as [ 3nq n 1 k= 1 k ± 1 1 k q 3 3k +k f q 3 3 +k+1, q 3 3 k ± 1 q 3 1 f q 3 [ 1 k= 1 k ± 1 Now we consider the congruence ] 1 k q 5 3k +k f q 5 3 +k+1, q 5 3 k ± 1 q 5 1 f q 5 ] mod k +k +5 3m +m 8 1 mod, 3 15 where 1/ k, m 1/, with = 1. Since the above congruence is equivalent to 18k m+1 0 mod,

9 Congruences modulo 3 for two interesting artitions 9 15 = 1, there is only one solution k = m = ± 1/ for 3. That is, there are no other k m such that 3 3k +k +5 3m +m 8 1 are in the same residue class modulo. Therefore, equating the terms involving q n+8 1 Thus, 3 n+8 1 from both sides of, we deduce that q n = 3n+ 1 q n q 3 ;q 3 q 5 ;q 5 mod 3. 3 n+ 1 q n q 3 ;q 3 q 5 ;q 5 mod 3. From, we arrive at 3 n+ 1 3n mod 3. Now 1 can be established easily by mathematical induction. QED 15 Corollary. For any rime 5 with = 1 for any nonnegative integers k n, 3 k+ n+3i+ k mod 3, where i = 1,,..., 1. Proof. As in the roof of the revious theorem, it can also be shown that 3 k n+8 1 +k 1 q n q 3 ;q 3 q 5 ;q 5 mod 3, that is, 3 k+1 n+ k+ 1 q n q 3 ;q 3 q 5 ;q 5 mod 3. Since there are no terms on the right side of the above congruence in which the owers of q are congruent to 1,,..., 1 modulo, it follows, for i = 1,,..., 1, that 3 k+1 n+i+ k+ 1 0 mod 3, which is clearly equivalent to the roffered congruence. QED

10 70 K. Das 15 Corollary 3. For any rime 5 with = 1 for any nonnegative integers k n, 15 k n+3r +1 k 1 0 mod 3, where r =,. Proof. It can also be shown that 3 k n+ k 1 q n q 3 ;q 3 q 5 ;q 5 mod 3. In 33, relacing q by q 3, we can see that q 3 ;q 3 q 5 ;q 5 has no terms q 5n+r where r =,, it follows that 3 k 5n+r+ k 1 0 mod 3, which is clearly equivalent to the roffered congruence. QED 15 Theorem 7. For any rime 5 with = 1 for any nonnegative integers k n, 3 k n+ k 1 3n+1 mod 3. 5 Proof. With the hel of 9, 3 can be rewritten as [ 3n+1q n 1 k= 1 k ± 1 1 k q 3k +k f q 3 +k+1, q 3 k ± 1 q 1 f q [ Consider the congruence 1 k= 1 k ± 1 ] 1 k q 15 3k +k f q k+1, q 15 3 k ± 1 q 15 1 f q 15 ] mod 3. 3k +k +15 3m +m 1 1 mod, 7

11 Congruences modulo 3 for two interesting artitions 71 where 1/ k, m 1/, with congruence is equivalent to 15 = 1. Since the above k m+1 0 mod, 15 = 1, there is only one solution k = m = ± 1/ for 7. That is, there are no other k m such that 3k +k +15 3m +m 1 1 are in the same residue class modulo. Therefore, equating the terms involving q n+1 1 from both sides of, we deduce that Thus, 3 n q n = 3n+ 1 q n q ;q q 15 ;q 15 mod 3. 3 n+ 1 q n q;q q 15 ;q 15 mod 3. 8 From 8 3, we arrive at 3 n+ 1 3n+1 mod 3. Now 5 can be established easily by mathematical induction. QED The following results follow in a similar fashion. So we omit the roof. 15 Corollary. For any rime 5 with = 1 for any nonnegative integers k n, 3 k+ +3i+ k mod 3, where i = 1,,..., Corollary 5. For any rime 5 with = 1 for any nonnegative integers k n, 15 k +3r + k 1 0 mod 3, where r = 3,.

12 7 K. Das Some congruences modulo 3 for two color artitions 5 n Theorem 8. We have, 5 3nq n q3 ;q 3 q 5 ;q 5 mod 3, 9 5 3n+1q n q;q q 5 ;q 5 mod 3, n+q n q15 ;q 15 q;q mod where 5 nq n := 1 q;q q 5 ;q 5. Proof. We have 5 nq n = 1 q;q q 5 ;q 5 = q 5 ;q 5 5 q;q q 5 ;q 5. Taking congruences modulo 3 in the above equation, we obtain, 5 nq n q 5 ;q 5 5 q 15 ;q 15 q;q mod 3 1 q 15 ;q 15 nq n 5 Comaring the terms involving q 3n, q 3n+1 q 3n+ resectively from the both sides of the above congruence, we have 5 3nq n 1 q 5 ;q 5 3nq n mod 3, n+1q n 1 q 5 ;q 5 3n+1q n mod 3, 5

13 Congruences modulo 3 for two interesting artitions n+q n 1 q 5 ;q 5 3n+q n mod Emloying 3 in the above congruences, we can easily obtain the 8. Theorem 9. We have, QED 5 15n+9 5 3n+ mod 3, n+ 0 mod 3, n+1q n 0 mod Proof. Relacing q by q 3 in33 then emloying in9, we obtain 5 3nq n q75 ;q 75 q 5 ;q 5 Aq 15 Bq 15 q3 q Bq15 Aq 15 mod 3 59 Comaringthetermsinvolvingq 5n+3 frombothsidesoftheabovecongruence we can easily arrive at 5. There is no terms involving q 5n+, q 5n+ in the right h side of the congruence 59. So, we can easily obtain QED Theorem 10. For any k 1, we have 5 3 k 1 n+ 3k 1 +1 q n k+1 q;q q 5 ;q 5 mod 3. 0 Proof. We rove the result by mathematical induction. For k = 1, we obtain 5 3n+1q n q;q q 5 ;q 5 mod 3, 1 which is the 50. Let 0 be true for some ositive integer k. Therefore we can write 0 as 5 3 k 1 n+ 3k 1 +1 q n k+1 q 3 ;q 3 q 15 ;q 15 1 q;q q 5 ;q 5 mod 3 3 k+1 q 3 ;q 3 q 15 ;q 15 5 nq n mod 3.

14 7 K. Das Extracting the terms involving q 3n+ from both sides of the above congruence, we obtain 5 3 k 1 3n++ 3k 1 +1 q n k+1 q;q q 5 ;q 5 [ 5 3n+1q n + 5 3nq n 5 3n+q n] mod 3. Emloying 9, in, 5 3 k 1 3n++ 3k 1 +1 q n 5 [ k+1 q;q q 5 ;q 5 q;q q 5 ;q 5 + q3 ;q 3 q 15 ;q 15 ] q 5 ;q 5 mod 3 q;q k+ q;q q 5 ;q 5 mod 3. Hence the result is true for all k 1. QED Alying33wecanseeeasilythatq;q q 5 ;q 5 hasnotermscontaining q 5n+r for r = 3,. From the last result we can conclude the following Theorem 11. For any k 1, we have 5 3 k 1 n+ r +13k 1 +1 q n 0 mod 3, where, r = 3,. Theorem 1. For any rime 5 with 5 = 1 for any non- negative integers k n, 5 3 k n+ 3k n+1 mod 3. 7

15 Congruences modulo 3 for two interesting artitions 75 Proof. With the hel of 9, 50 can be rewritten as [ 5 3n+1q n 1 k= 1 k ± 1 1 k q 3k +k f q 3 +k+1, q 3 k ± 1 q 1 f q [ 1 k= 1 k ± 1 Consider the following congruence ] 1 k q 5 3k +k f q 5 3 +k+1, q 5 3 k ± 1 q 5 1 f q 5 ] mod k +k +5 3m +m 1 mod, 9 5 where 1/ k, m 1/, with = 1. Since the above congruence is equivalent to k m+1 0 mod, 5 = 1, there is only one solution k = m = ± 1/ for 7. That is, there are no other k m such that 3k +k +5 3m +m 1 are in the same residue class modulo. Therefore, equating the terms involving q n+ 1 from both sides of 8, we deduce that Thus, 5 3 n q n = 5 3n q n q ;q q 5 ;q 5 mod n q n q;q q 5 ;q 5 mod From 70 50, we arrive at 5 3 n n+1 mod 3.

16 7 K. Das Now 7 can be established easily by mathematical induction. QED The following results follow in a similar fashion. So we omit the roof. 5 Corollary. For any rime 5 with = 1 for any nonnegative integers k n, 5 3 k+ + 1i+3k mod 3, where i = 1,,..., 1. 5 Corollary 7. For any rime 5 with = 1 for any nonnegative integers k n, 5 15 k + 1r +3k +1 0 mod 3, where r = 3,. References [1] Z. Ahmed, N. D. Baruah M. G. Dastidar:New congruences modulo 5 for the number of -color artitions, J. Number Theory, , [] N. D. Baruah, J. Bora K. K. Ojah: Ramanujan s modular equations of degree 5, Proc. Indian Acad. Sci. Math. Sci., 1 01, [3] B. C. Berndt: Ramanujan s Notebooks, Part III, Sringer-verleg, New York, [] S. P. Cui N. S. S. Gu: Arithmetic roerties of l-regular artitions, Adv. Al. Math., , [5] F. Garvan, D. Kim D. Stanton: Cranks t-cores, Invent. Math., , [] S. Ramanujan: The Lost Notebook Other Unublished Paers, Narosa, New Delhi, 1988.