Congruences modulo p2 and p3 for k dots bracelet partitions with k=mps

Accepted Manuscript Congruences modulo p2 and p3 for k dots bracelet partitions with k = mps

Nayandeep Deka Baruah, Zakir Ahmed

PII: DOI: Reference:

S0022-314X(15)00018-9 10.1016/j.jnt.2014.12.005 YJNTH 5049

To appear in:

Journal of Number Theory

Received date: 31 March 2014 Revised date: 12 December 2014 Accepted date: 12 December 2014

Please cite this article in press as: N.D. Baruah, Z. Ahmed, Congruences modulo p2 and p3 for k dots bracelet partitions with k = mps , J. Number Theory (2015), http://dx.doi.org/10.1016/j.jnt.2014.12.005

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Congruences modulo p2 and p3 for k dots bracelet partitions with k = mps NAYANDEEP DEKA BARUAH and ZAKIR AHMED Abstract. We find several new congruences modulo 2 for 11 dots bracelet partitions and congruences modulo p2 and p3 for k dots bracelet partitions for any prime p > 3 and k = mps with s ≥ 2 and s ≥ 3, respectively. We also find a relation between 5-core partitions and Ramanujan’s famous tau function τ (n) and several new congruences modulo 9 and 25 for 3- and 5-core partitions, respectively. Key Words: k dots bracelet partitions, t-core partitions, partition congruences. 2010 Mathematical Reviews Classification Numbers: Primary 11P83; Secondary 05A15, 05A17. 1. Introduction In 2007, Andrews and Paule [1] introduced Broken k-diamond partitions. Broken k-diamond partitions consist of two separated k-elongated partition diamonds of length n where in one of them, the source is deleted, as shown in Fig. 1. b(2k+1)n

b2k+1

b(2k+1)n+1

b7

b5

b3

a1

b2k+2

b(2k+1)n−1

a3

a2k+1

a2k+2

a2 a2k b6 b4 b2 Fig. 1. A broken k-diamond of length 2n.

b2k

a(2k+1)n

a(2k+1)n+1

a(2k+1)n−1

Definition 1.1. [1, Definition 4] For n, k ≥ 1, define ♦ := {(b2 , . . . , b(2k+1)n+1 , a1 , a2 , . . . , a(2k+1)n ) ∈ N(4k+1)n , ai and bi satisf y order relations in F ig. 1}, Hn,k ♦ h♦ n,k := hn,k (x2 , . . . , x(2k+1)n+1 , y1 , y2 , . . . , y(2k+1)n+1 )  b(2k+1)n+1 a1 a2 a(2k+1)n+1 := xb22 · · · x(2k+1)n+1 y1 y2 · · · y(2k+1)n+1 , ♦ (b2 ,...,b(2k+1)n+1 ,a1 ,a2 ,...,a(2k+1)n )∈Hn,k

and ♦ h♦ n,k (q) := hn,k (q, q, . . . , q).

Andrews and Paul found the generating function for the partition function h♦ ∞,k (q). Theorem 1.2. [1, Theorem 3] Let for n ≥ 0 and k ≥ 1, Δk (n) denote the total number of broken k-diamond partitions of n. Then h♦ ∞,k (q) = where (x; q)∞ :=

∞

n=0 (1

∞ 

Δk (n)q n =

n=0

− xq n ), |q| < 1. 1

(q 2 ; q 2 )∞ , (q; q)3∞ (−q 2k+1 ; q 2k+1 )∞

(1.1)

2

Baruah and Ahmed

For k = 1, they also proved the congruence Δ1 (2n + 1) ≡ 0 (mod 3).

(1.2)

In 2011, Fu [13] gave a combinatorial proof of (1.2) and also apply the combinatorial approach to generalise the broken k-diamond partitions which he called k dots bracelet partitions. Before defining k dots bracelet partitions, Fu defined infinite bracelet partitions which consists of repeating diamonds and dots with k − 2 dots between two consecutive diamonds as shown in the Fig. 2 and we see that an infinite bracelet partitions can be cut into k − 1 different ways with k dots in half. For any k ≥ 3, a k dots bracelet partitions consists of k − 1 different half bracelet as shown in Fig. 3. am−1

am−3

am+k−1

am+k−3

am

am−2

Fig. 2. Infinite bracelet with k dots.

am+k

am+k−2

Fu [13] denoted the number of k dots bracelet partitions for a positive integer n by Bk (n). The generating function for Bk (n) is given by ∞  0

Bk (n)q n =

(q 2 ; q 2 )∞ . (q; q)k∞ (−q k ; q k )∞

(1.3)

He also proved the following congruences for k dots bracelet partitions: (i) for n ≥ 0, k ≥ 3 if k = pr is a prime power, Bk (2n + 1) ≡ 0 (mod p), (ii) for any k ≥ 3, s an integer between 1 and p − 1 such that 12s + 1 is a quadratic nonresidue  modulo p and any n ≥ 0, if pk for some prime p ≥ 5, say k = pm, then Bk (pn + s) ≡ 0 (mod p), (iii) for any n ≥ 0, k ≥ 3 even, say k = 2m , where  is odd, Bk (2n + 1) ≡ 0 (mod 2m ). Radu and Sellers [17] extended the set of congruences given by Fu. They proved that for all n ≥ 0 B5 (10n + 7) ≡ 0 (mod 52 ), B7 (14n + 11) ≡ 0 (mod 72 ), and B11 (22n + 21) ≡ 0 (mod 112 ). More recently, Cui and Gu [12] found several congruences modulo 2 for 5 dots bracelet partitions and congruences modulo p for any prime p ≥ 5 for k dots bracelet partitions. Xia and Yao [18] also found several congruences modulo powers of 2 for 5 dots bracelet partitions.

Congruences modulo p2 and p3 for k dots bracelet partitions with k = mps

3

am+k−1

am+k−3

am

am+k

am+k−2 am+k−1

am+1

am+k−3

am+k

am+k−2

am+k−1

am+2k−1

am+k−3

am+k

am+2k−3

am+k−2

am+2k−2

am+k−1

am+2k−1

am+k

am+k−2

am+2k

am+2k−3

am+2k

am+2k−2 Fig. 3. k − 1 different half bracelet.

In this paper, we find several new congruences modulo 2 for 11 dots bracelet partitions and also find congruences modulo p2 and p3 for k dots bracelet partitions for any prime p > 3 by employing n Ramanujan’s theta functions and by finding the binomial expansion of (q; q)p∞ congruent modulo pn for n = 2 and n = 3 respectively.

4

Baruah and Ahmed

A t-core partition of a positive integer n is a partition in which none of its hook number is divisible by t. The generating function for at (n), the number of t-core partitions of a positive integer n, defined by Garvan, Kim and Stanton [14],is ∞  (q t ; q t )t∞ at (n)q n = . (1.4) (q; q)∞ n=0 They also gave arithmetic properties for 5-core and 7-core partitions by using combinatorial method. There are several arithmetic properties for 3- and 5-core partitions given by Baruah and Berndt [4], Hirschhon and Sellers [16] and Baruah and Nath [6, 5]. In the last section of this paper we find several congruences modulo 9 for 3-core partitions and also find that a5 (n − 1) ≡ τ (n) (mod 25), where τ (n) is Ramanujan’s famous tau function defined by ∞  q(q; q)24 = τ (n)q n . (1.5) n=1

In the remainder of this section, we introduce Ramanujan’s theta functions and some of their elementary properties, which will be used in our subsequent sections. Ramanujan’s general theta function f (a, b) is defined by ∞  f (a, b) := ak(k+1)/2 bk(k−1)/2 , |ab| < 1. (1.6) k=−∞

In this notation, Jacobi’s famous triple product identity takes the form f (a, b) = (−a; ab)∞ (−b; ab)∞ (ab; ab)∞ .

(1.7)

Three special cases of f (a, b) are [8, p. 36, Entry 22 ] ∞  2 ϕ(q) := f (q, q) = q k = (−q; q 2 )2∞ (q 2 ; q 2 )∞ , k=−∞ ∞ 

ψ(q) := f (q, q 3 ) =

q k(k+1)/2 =

k=0

(q 2 ; q 2 )∞ , (q; q 2 )∞

(1.8)

and f (−q) := f (−q, −q 2 ) =

∞ 

(−1)k q k(3k−1)/2 = (q; q)∞ ,

(1.9)

k=−∞

where the product representations in the above arise from (1.7) and the last equality is Euler’s famous pentagonal theorem. After Ramanujan, we also define χ(q) := (−q; q 2 )∞ .

(1.10)

By manipulating the q-products, one can easily arrive at the following representations: (q 2 ; q 2 )5∞ (q; q)2 (q 2 ; q 2 )2∞ (q; q)∞ (q 4 ; q 4 )∞ , ϕ(−q) = 2 2∞ , ψ(q) = , ψ(−q) = , 2 4 4 2 (q; q)∞ (q ; q )∞ (q ; q )∞ (q; q)∞ (q 2 ; q 2 )∞ (q; q)∞ (q 2 ; q 2 )2∞ , χ(−q) = 2 2 . (1.11) χ(q) = (q; q)∞ (q; q)∞ (q ; q )∞

ϕ(q) =

By (1.7), we also have [8, p. 51, Example (v); p. 350, Eq. (2.3)] f (q, q 5 ) = ψ(−q 3 )χ(q)

(1.12)

and f (q, q 2 ) = We also need the following result.

ϕ(−q 3 ) . χ(−q)

(1.13)

Congruences modulo p2 and p3 for k dots bracelet partitions with k = mps

5

Lemma 1.3. [8, p. 69, Eq. (36.8)] For an even integer μ and an integer ν with μ > ν ≥ 0, we have ψ(q μ+ν )ψ(q μ−ν ) = ϕ(q μ(μ

2



−ν 2 )

)ψ(q 2μ )

μ/2−1

+

q μm

2

−νm

f (q (μ+2m)(μ

2

−ν 2 )

, q (μ−2m)(μ

2

−ν 2 )

)f (q 2νm , q 2μ−2νm )

m=1

+ qμ

3

/4−μν/2

ψ(q 2μ(μ

2

−ν 2 )

)f (q μν , q 2μ−μν ).

(1.14)

The following two lemmas will be used in the subsequent sections. Lemma 1.4. For any prime p, 2

(q; q)p∞ ≡ (q p ; q p )p∞ (mod p2 ).

(1.15)

Proof. It is sufficient to prove that 2

(1 − q)p ≡ (1 − q p )p (mod p2 ). In fact, by the binomial theorem, it is enough to show that  2   p p ≡ (mod p2 ). np n

(1.16)

But from Bailey’s paper [3], we recall that, for any positive integer k, r and any prime p,     kp k ≡ (mod p2 ), rp r which immediately implies (1.16) by setting k = p and r = n.

2

By taking (npr−2 )-th power of the congruence (1.15), we also note that r

p p np (q; q)np ∞ ≡ (q ; q )∞

r−1

(mod p2 ),

(1.17)

for any n ∈ N and r ≥ 2. Lemma 1.5. For any prime p > 3, 3

2

(q; q)p∞ ≡ (q p ; q p )p∞ (mod p3 ).

(1.18)

Proof. The proof is similar to the above lemma. Here we use the following congruence from Bailey’s paper [2]: For any positive integer k, r and any prime p > 3     kp k ≡ (mod p3 ). rp r Setting k = p2 and r = n, we can easily arrive at (1.18).

2

By taking (nps−3 )-th power of the congruence (1.15), we also note that s

p p np (q; q)np ∞ ≡ (q ; q )∞

for any n ∈ N and s ≥ 3.

s−1

(mod p3 ),

(1.19)

6

Baruah and Ahmed

2. Congruence modulo 2 for 11 dots bracelet partitions Theorem 2.1. For any prime p ≥ 5, α ≥ 0 and n ≥ 0, we have   p2α + 5 q n ≡ (q; q)∞ (mod 2). B11 4 · p2α · n + 6 n=0 ∞ 

(2.1)

Proof. We note from (1.3) that ∞ 

B11 (n)q n =

n=0

=

(q 2 ; q 2 )∞ 11 (q; q)∞ (−q 11 ; q 11 )∞ (q 2 ; q 2 )∞ (q 11 ; q 11 )∞ . 22 22 (q; q)11 ∞ (q ; q )∞

(2.2)

Since (q; q)2∞ ≡ (q 2 ; q 2 )∞ (mod 2), we find that ∞ 

B11 (n)q n ≡

n=0

1 (mod 2). (q 8 ; q 8 )∞ (q; q)∞ (q 11 ; q 11 )∞

(2.3)

Now, setting μ = 6 and ν = 5 in (1.14), we have ψ(q)ψ(q 11 ) = ϕ(q 66 )ψ(q 12 ) + qf (q 88 , q 44 )f (q 2 , q 10 ) + q 14 f (q 20 , q −8 )f (q 110 , q 22 ) + q 15 ψ(q 132 )ϕ(q 6 ). Employing the trivial identity f (a, b) = af (a2 b, a−1 ), (1.12), and (1.13) in the above, we obtain ψ(q)ψ(q 11 ) = ϕ(q 66 ψ(q 12 ) + q 6

ψ(−q 6 )χ(q 2 )ϕ(−q 132 ) ψ(−q 66 )χ(q 22 )ϕ(−q 12 ) +q + q 15 ψ(q 132 )ϕ(q 6 ), 4 χ(−q ) χ(−q 44 )

which, by (1.11), is equivalent to  66 22 12 1 1 66 12 6 ψ(−q )χ(q )ϕ(−q ) = )ψ(q ) + q ϕ(q (q; q)∞ (q 11 ; q 11 )∞ (q 2 ; q 2 )2∞ (q 22 ; q 22 )2∞ χ(−q 4 )  6 2 132 ψ(−q )χ(q )ϕ(−q ) + q 15 ψ(q 132 )ϕ(q 6 ) . +q χ(−q 44 ) Since ϕ(q) ≡ 1 (mod 2), we arrive at  66 22 ψ(−q 6 )χ(q 2 ) 1 1 12 6 ψ(−q )χ(q ) ψ(q + q ≡ ) + q (q; q)∞ (q 11 ; q 11 )∞ (q 2 ; q 2 )2∞ (q 22 ; q 22 )2∞ χ(−q 4 ) χ(−q 44 )  + q 15 ψ(q 132 ) (mod 2). From (2.3) and (2.4), we have ∞  n=0

B11 (n)q n ≡

 66 22 1 1 12 6 ψ(−q )χ(q ) ) + q ψ(q (q 8 ; q 8 )∞ (q 2 ; q 2 )2∞ (q 22 ; q 22 )2∞ χ(−q 4 )  ψ(−q 6 )χ(q 2 ) +q + q 15 ψ(q 132 ) (mod 2). χ(−q 44 )

(2.4)

Congruences modulo p2 and p3 for k dots bracelet partitions with k = mps

7

Extracting the terms involving q 2n+1 from both sides of the above congruence, dividing both sides by q and then replacing q 2 by q, we find that

∞  1 ψ(−q 3 )χ(q) 1 7 66 + q B11 (2n + 1)q n ≡ 4 4 ψ(q ) (q ; q )∞ (q 2 ; q 2 )∞ (q 22 ; q 22 )∞ χ(−q 22 ) n=0 6 6 2 44 44

132 132 ; q )∞ 1 (q ; q )∞ (q ; q )∞ (q; q)∞ 7 (q ≡ 2 2 3 22 22 + q (q ; q )∞ (q ; q )∞ (q 2 ; q 2 )∞ (q 22 ; q 22 )∞ (q 3 ; q 3 )∞ (q 66 ; q 66 )∞ 6 6 2 22 22

1 (q ; q )∞ (q ; q )∞ (q; q)∞ 7 66 66 ≡ 2 2 3 22 22 (mod 2). + q (q ; q ) ∞ (q ; q )∞ (q ; q )∞ (q 2 ; q 2 )∞ (q 3 ; q 3 )∞ (2.5) Now, from Hirschhon and Roselin [15], we recall that 1 (q 8 ; q 8 )2 (q 12 ; q 12 )5 (q 4 ; q 4 )5 (q 24 ; q 24 )2∞ = 2 2 2 4 4 ∞ 6 6 4 ∞ 24 24 2 + 2 2 4 6 6 ∞ . 3 3 (q; q)∞ (q ; q )∞ (q ; q )∞ (q ; q )∞ (q ; q )∞ (q ; q )∞ (q ; q )∞ (q ; q )2∞ (q 8 ; q 8 )2∞ (q 12 ; q 12 )∞ (2.6) Multiplying both sides of the above by (q; q)2∞ and taking congruent modulo 2, we find that

(q 8 ; q 8 )2∞ (q 12 ; q 12 )5∞ (q 4 ; q 4 )5∞ (q 24 ; q 24 )2∞ (q; q)∞ 2 2 ≡ (q ; q ) + q ∞ (q 3 ; q 3 )∞ (q 2 ; q 2 )2∞ (q 4 ; q 4 )∞ (q 6 ; q 6 )4∞ (q 24 ; q 24 )2∞ (q 2 ; q 2 )4∞ (q 6 ; q 6 )2∞ (q 8 ; q 8 )2∞ (q 12 ; q 12 )∞ 2 2 5 12 12 2 (q ; q ) (q ; q ) ≡ 12 12 ∞ + q 2 2 ∞ (mod 2). (q ; q )∞ (q ; q )∞ Employing the above in (2.5), we obtain ∞ 

 (q 6 ; q 6 )2 (q 22 ; q 22 )  (q 2 ; q 2 )5 1 ∞ ∞ ∞ (q 2 ; q 2 )3∞ (q 22 ; q 22 )∞ (q 2 ; q 2 )∞ (q 12 ; q 12 )∞ (q 12 ; q 12 )2 (q 132 ; q 132 )∞  + q 2 2 ∞ + q 7 66 66 (q ; q )∞ (q ; q )∞ 12 12 3 ; q ) (q 66 ; q 66 )∞ (q (mod 2). ≡ (q 2 ; q 2 )∞ + q 2 2 5∞ + q 7 2 2 3 22 22 (q ; q )∞ (q ; q )∞ (q ; q )∞

B11 (2n + 1)q n ≡

n=0

Extracting the terms involving q 2n from both sides of the above congruence, and then replacing q 2 by q, we find that ∞  B11 (4n + 1)q n ≡ (q; q)∞ (mod 2). (2.7) n=0

Hence, (2.1) is proved for α = 0. Now, let (2.1) be true for some α ≥ 0, i.e.,   ∞  p2α + 5 2α q n ≡ (q; q)∞ (mod 2). B11 4 · p · n + 6 n=0

(2.8)

Recently, Cui and Gu [11] derived a p-dissection for (q; q)∞ for any prime p ≥ 5, namely, p−1

(q; q)∞ =

2 

(−1)k q

3k2 +k 2

f (−q

3p2 +(6k+1)p 2

, −q

±p−1 k=− p−1 2 ,k= 6

+ (−1) where

±p−1 6

q

p2 −1 24

2

2

(q p ; q p )∞ ,

⎧ ⎪ ⎨ p − 1, ±p − 1 6 := −p − 1 ⎪ 6 ⎩ , 6

if p ≡ 1 (mod 6); if p ≡ −1 (mod 6).

3p2 −(6k+1)p 2

)

(2.9)

8

Baruah and Ahmed

Furthermore, for

(p − 1) (±p − 1) −(p − 1) ≤k≤ and k = , 2 2 6 3k 2 + k p2 − 1 ≡ (mod p). 2 24

Using the above p-dissection in (2.8), then extracting the terms involving q pn+

p2 −1 24

from both sides

p2 −1

of the resulting congruence, dividing both sides by q 24 and then replacing q p by q, we find that   ∞  p2α + 5 p2 − 1 )+ q n ≡ (q p ; q p )∞ (mod 2), B11 4 · p2α (p · n + 24 6 n=0 i.e.,   p2 − 1 p2α + 5 + q n ≡ (q p ; q p )∞ (mod 2). B11 4 · p2α+1 · n + 4 · p2α 24 6 n=0 ∞ 

Again, extracting the terms involving q pn from both sides of the above and then replacing q p by q, we obtain   ∞  p2(α+1) + 5 q n ≡ (q; q)∞ (mod 2). B11 4 · p2(α+1) · n + (2.10) 6 n=0 Hence, (2.1) is true for α + 1 if it is true for α. So, by mathematical induction, we complete the proof of (2.1). 2 Corollary 2.2. For any prime p ≥ 5, α ≥ 0 and n ≥ 0, we have   p2α + 5 2α ≡ 0 (mod 2), B11 4 · p · n + 6 if n =

(2.11)

k(3k − 1) . 2 2

Proof. Readily follows from (2.1) and (1.9). 3. Congruence modulo p2 and p3 for k dots bracelet partitions

Theorem 3.1. Let k = mpr , where m ∈ N, p ≥ 5 and r ≥ 2. Then for any positive integer n, we have Bk (pn + ) ≡ 0 (mod p2 ), (3.1) where 1  ≤  ≤ p − 1 and 12 + 1 is quadratic nonresidue modulo p, i.e., in Legendre symbol  12 + 1 = −1. p Proof. We note that

∞ 

r

Bk (n)q n =

n=0

r

(q 2 ; q 2 )∞ (q np ; q np )∞ . r 2npr ; q 2npr ) (q; q)np ∞ (q ∞

(3.2)

Employing (1.17) in (3.2), we obtain ∞  n=0

r

Bk (n)q n ≡

r

(q 2 ; q 2 )∞ (q mp ; q mp )∞ mpr−1

(q p ; q p )∞

(q 2mpr ; q 2mpr )∞

(mod p2 ).

(3.3)

It is sufficient to find the coefficients of q pn+ , where 1 ≤  ≤ p − 1 and n ∈ N, in (q 2 ; q 2 )∞ only, because the coefficients of q pn+ in the other products in (3.3) are zero.

Congruences modulo p2 and p3 for k dots bracelet partitions with k = mps

9

From (1.9), we have (q 2 ; q 2 )∞ =

∞ 

(−1)k q k(3k−1) .

(3.4)

k=−∞

If the coefficients of q pn+ in (3.4) is nonzero, then for some k, we have pn +  = k(3k − 1), i.e.,  ≡ k(3k − 1) (mod p). Thus, 12 + 1 ≡ (6k − 1)2 (mod p),

(3.5)

which contradict the fact that 12 + 1 is a non quadratic residue modulo p. Thus, we complete the proof of (3.1). 2 Theorem 3.2. Let k = mps , where m ∈ N, p ≥ 5 and s ≥ 3. Then for any positive integer n, we have Bk (pn + j) ≡ 0 (mod p3 ), where 1 ≤  j ≤ p − 1 and 12j + 1 is quadratic nonresidue modulo p, i.e., in Legendre symbol  12j + 1 = −1. p Proof. The proof is similar to the above one. Here we use (1.19) in place of (1.17).

2

Theorem 3.3. Let k = mps , where m ∈ N, p ≥ 5 and s ≥ 3. Then for any positive integer n, we have   p2 − 1 ≡ 0 (mod p2 ), (3.6) Bk p(pn + j) + 12 for j = 1, 2, . . . , p − 1. Proof. We note that ∞ 

s

Bk (n)q n =

n=0

s

(q 2 ; q 2 )∞ (q np ; q np )∞ . s 2nps ; q 2nps ) (q; q)np ∞ (q ∞

Employing (1.19) in the above, we obtain ∞  n=0

s

Bk (n)q n ≡

s

(q 2 ; q 2 )∞ (q mp ; q mp )∞ (q p ; q p )mp ∞

s−1

(q 2mps ; q 2mps )∞

(mod p3 ).

(3.7)

Now, replacing q by q 2 in (2.9), employing the result in the above and then extracting the terms p2 −1

p2 −1

containing q pn+ 12 , dividing both sides by q 12 and then replacing q p by q in the above congruence, we arrive at   s−1 s−1 ∞ 2p 2p  ±p−1 (q p2 − 1 ; q )∞ (q mp ; q mp )∞ q n ≡ (−1) 6 Bk pn + (mod p3 ), s−1 mp 2mps−1 ; q 2mps−1 ) 12 (q; q) (q ∞ ∞ n=0 which is also true for modulo p2 , i.e.,   s−1 s−1 ∞ 2p 2p  ±p−1 (q p2 − 1 ; q )∞ (q mp ; q mp )∞ q n ≡ (−1) 6 Bk pn + (mod p2 ). mps−1 2mps−1 2mps−1 12 (q; q) (q ; q ) ∞ ∞ n=0

10

Baruah and Ahmed

Employing (1.17) in the above congruence, we find that   s−1 s−1 ∞  ±p−1 p2 − 1 (q 2p ; q 2p )∞ (q mp ; q mp )∞ q n ≡ (−1) 6 Bk pn + (mod p2 ) s−2 p ; q p )mp 2mps−1 ; q 2mps−1 ) 12 (q (q ∞ ∞ n=0

(3.8)

Comparing the coefficients of q pn+j , where j = 1, 2, . . . , p − 1, we complete the proof of (3.6).

2

Corollary 3.4. Let k = mps , where m ∈ N, p ≥ 5 and s ≥ 3. Then for any positive integer n, we have   ±p−1 p2 − 1 ≡ (−1) 6 Bk (n) (mod p2 ), (3.9) Bk p2 n + 12 

where k = mps−2 . Proof. Extracting the terms containing q pn from both sides of (3.8), and then replacing q p by q, we obtain   s−2 s−2 ∞  ±p−1 (q 2 ; q 2 )∞ (q mp ; q mp )∞ p2 − 1 q n ≡ (−1) 6 Bk p2 n + , s−2 12 (q; q)mp (q 2mps−2 ; q 2mps−2 )∞ ∞ n=0 i.e.,   ∞ ±p−1  p2 − 1 q n ≡ (−1) 6 Bk p2 n + Bk (n)q n (mod p2 ). 12 n=0 n=0 ∞ 

(3.10)

Comparing the coefficients of q n from both side of the above congruence, we readily arrive at (3.9). 2 4. Congruences for 3- and 5-core partitions In this section, we find some new congruences for 3-core partitions modulo 9 and a congruence modulo 25 between 5- core partitions and τ (n) by using (1.15). Theorem 4.1. For α ≥ 1 and n ≥ 0, we have   ∞  4α − 1 2α q n ≡ (q; q)8∞ (mod 9). a3 2 · n + 3 n=0

(4.1)

Proof. We note from (1.4) that ∞ 

a3 (n)q n =

n=0

(q 3 ; q 3 )3∞ , (q; q)∞

which, by (1.15), implies ∞ 

a3 (n)q n ≡

n=0

(q; q)9∞ (q; q)∞

≡ (q; q)8∞ (mod 9).

(4.2)

Now, from [8, p. 40, Entry 25], we have ϕ2 (−q) = ϕ2 (q 2 ) − 4qψ 2 (q 4 ), which is equivalent to (q; q)4∞ =

(q 4 ; q 4 )10 ∞ 2 (q ; q 2 )2∞ (q 8 ; q 8 )4∞

− 4q

(q 2 ; q 2 )2∞ (q 8 ; q 8 )4∞ . (q 4 ; q 4 )2∞

(4.3)

Congruences modulo p2 and p3 for k dots bracelet partitions with k = mps

From (4.2) and (4.3), we have ∞  a3 (n)q n ≡ n=0

(q 4 ; q 4 )10 (q 2 ; q 2 )2∞ (q 8 ; q 8 )4∞ ∞ − 4q 2 2 2 8 8 4 (q ; q )∞ (q ; q )∞ (q 4 ; q 4 )2∞

11

2 (mod 9).

Extracting the term containing q 4n+1 from both sides of the above congruence, dividing both sides by q and then replacing q 4 by q, we arrive at ∞ 

a3 (4n + 1)q n ≡ (q; q)8∞ (mod 9),

n=0

which is the case for α = 1 in (4.1). Now, let (4.1) be true for some α ≥ 1, i.e., ∞ 

a3 (22α · n +

n=0

4α − 1 n )q ≡ (q; q)8∞ (mod 9). 3

(4.4)

Now employing (4.3) in the above congruence and extracting the terms containing q 4n+1 from both sides of the resulting congruence, dividing both sides by q and then replacing q 4 by q, we find that ∞ 

a3 (22α+1 · n +

n=0

4α+1 − 1 n )q ≡ (q; q)8∞ (mod 9). 3

Hence, (4.1) is true for α + 1 if it is true for α. Thus, by mathematical induction, we complete the proof of (4.1). 2 Corollary 4.2. For any k ≥ 2 and n ≥ 0, we have   3 · 22k−1 + 4k − 1 a3 22k · n + ≡ 0 (mod 9). 3

(4.5)

Proof. From (4.1) we have   4α − 1 q n ≡ (q; q)8∞ (mod 9). a3 22α · n + 3 n=0 ∞ 

(4.6)

Employing (4.3) in the above congruence and extracting the term containing q 2n+1 from both sides of the resulting congruence, dividing both sides by q and then replacing q 2 by q, we obtain   ∞  4α − 1 q n ≡ (q 2 ; q 2 )8∞ (mod 9), a3 22α (2n + 1) + 3 n=0 i.e.,   4α − 1 q n ≡ (q 2 ; q 2 )8∞ (mod 9). a3 22α+1 · n + 22α + 3 n=0 ∞ 

Comparing the coefficients of q 2n+1 from both sides of the above congruence, we arrive at   4α − 1 q n ≡ 0 (mod 9), a3 22α+1 · (2n + 1) + 22α + 3 i.e.,   3 · 22α+1 + 4α+1 − 1 2(α+1) ≡ 0 (mod 9), ·n+ a3 2 3 which is equivalent to (4.5).

2

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Baruah and Ahmed

Theorem 4.3. For any positive integer k, we have   52k − 1 ≡ (−5)3k a3 (n) (mod 9), a3 52k · n + 3   22k − 1 ≡ (−8)k a3 (n) (mod 9). a3 22k · n + 3

(4.7) (4.8)

Proof. From (4.2), we have ∞ 

a3 (n)q n ≡ (q; q)8∞ =

n=0

∞ 

p8 (n)q n .

(4.9)

n=0

We note from Baruah and Sarmah’s paper [7] that   52k − 1 p8 52k · n + = (−5)3k p8 (n), 3   22k − 1 = (−8)k p8 (n). p8 22k · n + 3 Employing the above two identities in (4.9), we easily obtain (4.7) and (4.8). Theorem 4.4.

a5 (n − 1) ≡ τ (n) (mod 25).

Proof. We note from (1.4) that

∞ 

a5 (n)q n =

n=0

2 (4.10)

(q 5 ; q 5 )5∞ . (q; q)∞

Using (1.15) in the above expression, we find that ∞  a5 (n)q n ≡ (q; q)24 ∞ (mod 25). n=0

Multiplying both sides by q in the above congruence and using (1.5), we obtain ∞ ∞   a5 (n)q n+1 ≡ τ (n)q n (mod 25). n=0

n=1

By comparing the coefficients of q n in the above congruence, we complete the proof of (4.10).

2

There are several identities known for Ramamnujan’s tau function τ (n) for modulo 5, 25 and 125. for example: τ (n) ≡ nσ9 (n) (mod 5), τ (n) ≡ 5n2 σ7 (n) − 4nσ9 (n) (mod 125), if n is not a multiple of 5, where σs (n) denote the (sum of the) sth power of the divisor of n, i.e.,  ds . σs (n) = d|n

From the above identities for τ (n), we obtain the following congruences for 5-core partitions by using (4.10): a5 (5n − 1) ≡ nσ9 (n) (mod 5), a5 (n − 1) ≡ 5n2 σ7 (n) − 4nσ9 (n) (mod 25). For more identities of tau function, we refer to Ramanujan’s unpublished manuscript on the partition and tau functions with proofs and commentary by Berndt and Ono [9].

Congruences modulo p2 and p3 for k dots bracelet partitions with k = mps

Theorem 4.5.

  a5 2k+2 · n + 2k+2 − 1 ≡ rk a5 (2n + 1) + sk a5 (n) (mod 25),

13

(4.11)

where rk = −24rk−1 + sk−1 , sk = −2048rk−1 with r0 = −24 and b0 = −2048. Proof. We have ∞  n=0

a5 (n)q n =

(q 5 ; q 5 )5∞ . (q; q)∞

By using (1.17), taking modulo 25 on both sides of the above equation, we obtain ∞ 

a5 (n)q n ≡ (q; q)24 ∞ =

n=0

∞ 

p24 (n)q n .

n=0

Again, from Baruah and Sarmah’s paper [7], we have p24 (2k+2 · n + 2k+2 − 1) = rk p24 (2n + 1) + sk p24 (n), where rk = −24rk−1 + sk−1 , sk = −2048rk−1 with r0 = −24 and b0 = −2048. By using the above identities, we complete the proof of (4.11). 2 References [1] G.E. Andrews, P. Paule, MacMahon’s partition analysis XI. broken diamonds and modular forms, Acta Arith. 126 (2007) 281–294. [2] D.F. Bailey, Two p3 variations of Lucas’ Theorem, J. Number Theory 35 (1990) 208–215. [3] D.F. Bailey, Some binomial coefficient congruences, Appl. Math. Lett. 4 (1991) 1–5. [4] N.D. Baruah, B.C. Berndt, Partition identities and Ramanujan’s modular equations, J. of Combin. Theory Ser. A 114 (2007) 1024–1045. [5] N.D. Baruah, K. Nath, Two quotients of theta functions and arithmetic identities for 3-cores, in The Legacy of Srinivasa Ramanuajan, B.C. Berndt and D. Prasad (eds.), RMS Lecture Notes Series, Ramanujan Mathematical Society, 20 (2013) 99–110. [6] N.D. Baruah, K. Nath, Some results on 3-cores, Proc. Amer. Math. Soc. 142 (2014) 441–448. [7] N.D. Baruah, B.K. Sarmah, Identities and congruences for the general partition and Ramanujan’s tau function, Indian J. Pure Appl. Math. 44 (2013) 643–671. [8] B.C. Berndt, Ramanujan’s Notebooks, Part III, Springer, New York, 1991. [9] B.C. Berndt, K. Ono, Ramanujan’s unpublished manuscript on the partition and tau functions with proofs and commentary, Sem. Lotharingien de Combinatorie 42 (1999) 63 pp.; in The Andrews Festschrift, D. Foata and G.–N.Han, eds., Springer-Verlag, Berlin (2001) 39–110. [10] S.P. Cui, N.S.S. Gu, Congruences for broken 3-diamond and 7 dots bracelet partitions, Ramanujan J. 35 (2014) 165–178. [11] S.P. Cui, N.S.S. Gu, Arithmetic properties of -regular partitions, Adv. Appl. Math. 51 (2013) 507–523. [12] S.P. Cui, N.S.S. Gu, Congruences for k dots bracelet partition functions, Int. J. Number Theory 9 (2013) 1885–1894. [13] S. Fu, Combinatorial proof of one congruence for the broken 1-diamond partition and a generalization, Int. J. Number Theory 7 (2011) 133–144. [14] F. Garvan, D. Kim, D. Stanton, Cranks and t-cores, Invent. Math. 101 (1990) 1–17. [15] M.D. Hirschhorn, Roselin, On the 2-, 3-, 4- and 6- dissections of Ramanujan’s cubic continued fraction and its reciprocal, in Ramanujan Rediscovered, N. D. Baruah, B.C. Berndt, S. Cooper, T. Huber, M. Schlosser (eds.), RMS Lecture Notes Series, Ramanujan Mathematical Society, No. 14 (2010) 125–138. [16] M.D. Hirschhorn, J.A. Sellers, Elementary proofs of various facts about 3-cores, Bull. Aust. Math. Soc. 79 (2009) 507—512.

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[17] S. Radu, J.A. Sellers, Congruences modulo squares of primes for Fu’s k dots bracelet partitions, Int. J. Number Theory 9 (2013) 939–943. [18] E.X.W. Xia, O.X.M. Yao, Congruences modulo powers of 2 for Fu’s 5 dots bracele partitions, Bull. Aust. Math. Soc. 89 (2014) 360–372. Department of Mathematical Sciences, Tezpur University, Sonitpur-784028, Assam, INDIA E-mail address: [email protected] Department of Mathematical Sciences, Tezpur University, Sonitpur-784028, Assam, INDIA E-mail address: [email protected]