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On the number of partitions with distinct even parts
Discrete Mathematics 311 (2011) 940–943
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Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Note
On the number of partitions with distinct even parts Shi-Chao Chen Institute of Applied Mathematics, School of Mathematics and Information Science, Henan University, Kaifeng, 475001, PR China
article
abstract
info
Let ped(n) be the number of partitions of n wherein even parts are distinct (and odd parts are unrestricted). We obtain many congruences for ped(n) mod 2 and mod 4 by the theory of Hecke eigenforms. © 2011 Elsevier B.V. All rights reserved.
Article history: Received 11 August 2010 Received in revised form 19 February 2011 Accepted 21 February 2011 Available online 23 March 2011 Keywords: Partition congruences Integer partitions Hecke eigenforms
1. Introduction Let ped(n) be the number of partitions of n wherein even parts are distinct (and odd parts are unrestricted). The generating function for ped(n) is ∞ −
ped(n)qn =
n =0
∞ ∏ (1 − q4n ) n =1
(1 − qn )
.
(1)
Recently, Andrews et al. [1] obtained some arithmetic properties of the function ped(n) by Ramanujan’s 1 ψ1 summation formula. For example, for all integers α ≥ 1, n ≥ 0, ped(3n + 2) ≡ 0 (mod 2), ped(9n + 4) ≡ 0 (mod 4), ped(9n + 7) ≡ 0 (mod 4),
ped 32α+1 n +
17 × 32α − 1
≡ 0 (mod 2), 11 × 32α+1 − 1 ped 32α+2 n + ≡ 0 (mod 2), 8 19 × 32α+1 − 1 ped 32α+2 n + ≡ 0 (mod 2). 8
8
In this short note, we will prove more congruences for ped(n) mod 2 and mod 4. Before we state the main results, we recall some notation. The Dedekind η-function is given by the infinite product 1
η(z ) := q 24
∞ ∏
(1 − qn ),
n =1
E-mail address: [email protected]. 0012-365X/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2011.02.025
S.-C. Chen / Discrete Mathematics 311 (2011) 940–943
941
where q = exp(2π iz ) and z lies in the complex upper half plane H. The well-known ∆-function is denoted by
∆(z ) := η(z )24 =
∞ −
τ (n)qn .
n=1
We have the following general theorem. Theorem. Let ℓ and p be odd primes. Then the following statements hold. (i) Suppose that s is an integer satisfying 1 ≤ s < 8ℓ, s ≡ 1(mod 8) and ℓs = −1, where ℓs is the Legendre symbol. Then
s−1
ped ℓn +
8
≡ 0 (mod 2).
(2)
If τ (ℓ) ≡ 0(mod 2), then, for all n ≥ 0, α ≥ 1,
ped ℓ2α+1 n +
sℓ2α − 1
≡ 0 (mod 2).
8
(3)
(ii) Suppose that r is an integer such that 1 ≤ r < 8p, rp ≡ 1(mod 8), and (r , p) = 1. If τ (p) ≡ 0(mod 2), then, for all n ≥ 0, α ≥ 1,
ped p2α+2 n +
rp2α+1 − 1
8
≡ 0 (mod 2).
(4)
If c (p) ≡ 0(mod 4), then, for all n ≥ 0, α ≥ 1,
ped p
2α+2
n+
rp2α+1 − 1
8
≡ 0 (mod 4), η4 (16z )
where c (p) is the p-th coefficient of η(8z )η(32z ) :=
∑∞
n =1
(5) c (n)qn .
Remark. If p = ℓ = 3, then τ (3) = 252 and c (3) = 0. The theorem yields the results of Andrews et al. above. Examples. Since τ (5) = 4830 and c (5) = 0, our theorem gives the following new congruences. For all n ≥ 0, α ≥ 1, ped (5n + 2) ≡ 0 (mod 2), ped (5n + 4) ≡ 0 (mod 2),
ped 52α+1 n +
ped 52α+1 n +
17 × 52α − 1
8 33 × 52α − 1
8
≡ 0 (mod 2), ≡ 0 (mod 2).
For r = 13, 21, 29, 37,
ped 52α+2 n +
r × 52α+1 − 1
8
≡ 0 (mod 4).
2. Proof of theorem As usual, Mk (SL2 (Z)) (resp., Sk∑ (SL2 (Z))) is the complex vector space of weight k holomorphic modular (resp., cusp) forms ∞ n with respect to SL2 (Z). If f (z ) = n=0 a(n)q ∈ Mk (SL2 (Z)), then the action of the Hecke operator Tp,k on f (z ) is defined by f (z )|Tp,k :=
∞ − (a(pn) + pk−1 a(n/p))qn .
(6)
n =0
It is known that ∆(z ) ∈ S12 (SL2 (Z)) and ∑ is an eigenform for all Hecke operators; in other words, ∆(z )|Tp,12 = τ (p)∆(z ) ∞ n for any prime p. The coefficients of ∆(z ) = n=1 τ (n)q have the following properties (see page 164 of [2]):
τ (mn) = τ (m)τ (n) if (m, n) = 1,
(7)
τ (p ) = τ (p)τ (p
(8)
l
l −1
) − p τ (p 11
l −2
).
Now we begin to prove the theorem. Using the generating function (1) for ped(n), we find that ∞ − n =0
ped(n)q8n+1 = q
∞ ∏ (1 − q32n ) n=1
(1 − q8n )
≡ ∆(z ) =
∞ − n =1
τ (n)qn (mod 2).
(9)
942
S.-C. Chen / Discrete Mathematics 311 (2011) 940–943
The well-known Jacobi identity ∞ ∏
(1 − qn )3 =
∞ ∞ − − n(n+1) n(n+1) (−1)n (2n + 1)q 2 ≡ q 2
n=1
n =0
(mod 2)
n =0
shows that
∆(z ) = q
∞ ∏
(1 − qn )24 ≡
∞ −
n =1
q(2n+1)
2
(mod 2).
n =0
Therefore, we have ∞ −
∞ −
ped(n)q8n+1 ≡
n =0
q(2n+1)
2
(mod 2).
(10)
n=0
If s ≡ 1(mod 8) and ℓs = −1, then, for any n ≥ 0, 8ℓn + s cannot be an odd square. This implies that the coefficients of q8ℓn+s in the left-hand side of (10) must be even. It follows that
s−1
ped ℓn +
8
≡ 0 (mod 2).
(11)
Since τ (ℓ) ≡ 0(mod 2) and ∆(z ) is a Hecke eigenform, we have
∆(z )|Tℓ,12 = τ (ℓ)∆(z ) ≡ 0 (mod 2). By (6) and (9), we get ∞ −
ped(n)q
8n+1
∞ −
|Tℓ,12 ≡
ped
ℓn − 1
n=n0
n =0
8
n
ℓ −1 8
n
+ ped
ℓ
−1
8
∈ N, then ℓn8−1 = ℓ2 m + ℓ 8−1 . So, we have ℓ2 − 1 ped ℓ2 m + + ped(m) ≡ 0 (mod 2).
If we write m =
2
8
That is,
ped ℓ2 m +
ℓ2 − 1
8
≡ ped(m) (mod 2).
By induction, we find that
2α
ped ℓ m +
ℓ2α − 1
8
≡ ped(m) (mod 2).
Considering (11), we get
ped ℓ
2α+1
n+
sℓ2α − 1
8
≡ ped ℓn +
s−1 8
≡ 0 (mod 2).
This proves (3). Since τ (p) ≡ 0(mod 2), formula (8) gives
τ (p2α+1 ) ≡ 0 (mod 2). Applying (7), we have
τ (p2α+1 (8pn + r )) = τ (p2α+1 )τ (8pn + r ) ≡ 0 (mod 2). It follows from (9) that
ped p2α+2 n +
rp2α+1 − 1 8
≡ 0 (mod 2).
This proves (4). We denote f (z ) by the following eta-quotient: f (z ) =:
∞ − η4 (16z ) = c (n)qn . η(8z )η(32z ) n=1
qn ≡ 0
(mod 2).
S.-C. Chen / Discrete Mathematics 311 (2011) 940–943
Then f (z ) is a weight-1 Hecke eigenform with respect to Γ0 (256) with character χ (d) = of f (z ) satisfy the following similar properties as τ (n):
943
−256 d
(see [3]). The coefficients
c (mn) = c (m)c (n) if (m, n) = 1, c (pl ) = c (p)c (pl−1 ) − χ (p)c (pl−2 ). Using the fact that
(1 − q16n )4 ≡ 1 (mod 4), (1 − q32n )2 we find that f (z ) = q
∞ ∏ n=1
=q
∞ ∏ (1 − q32n ) (1 − q16n )4 n=1
≡q
(1 − q8n ) (1 − q32n )2
∞ ∏ (1 − q32n ) n=1
=
(1 − q16n )4 (1 − q8n )(1 − q32n )
∞ −
(1 − q8n )
ped(n)q8n+1
(mod 4) (mod 4).
n=0
Congruence (5) follows by similar arguments as the proof of (4). This completes the proof. Acknowledgements The author thanks the referee and the editor for numerous helpful suggestions. This research was supported by the National Natural Science Foundation of China (Grant No. 11026080) and the Natural Science Foundation of Education Department of Henan Province (Grant No. 2009A110001). References [1] G.E. Andrews, M.D. Hirschhorn, J.A. Sellers, Arithmetic properties of partitions with even parts distinct, Ramanujan J. 22 (2010) 273–284. [2] N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Springer-Verlag, New York, 1984. [3] Y. Martin, Multiplicative η-quotients, Trans. Amer. Math. Soc. 348 (1996) 4825–4856.
Contents lists available at ScienceDirect
Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Note
On the number of partitions with distinct even parts Shi-Chao Chen Institute of Applied Mathematics, School of Mathematics and Information Science, Henan University, Kaifeng, 475001, PR China
article
abstract
info
Let ped(n) be the number of partitions of n wherein even parts are distinct (and odd parts are unrestricted). We obtain many congruences for ped(n) mod 2 and mod 4 by the theory of Hecke eigenforms. © 2011 Elsevier B.V. All rights reserved.
Article history: Received 11 August 2010 Received in revised form 19 February 2011 Accepted 21 February 2011 Available online 23 March 2011 Keywords: Partition congruences Integer partitions Hecke eigenforms
1. Introduction Let ped(n) be the number of partitions of n wherein even parts are distinct (and odd parts are unrestricted). The generating function for ped(n) is ∞ −
ped(n)qn =
n =0
∞ ∏ (1 − q4n ) n =1
(1 − qn )
.
(1)
Recently, Andrews et al. [1] obtained some arithmetic properties of the function ped(n) by Ramanujan’s 1 ψ1 summation formula. For example, for all integers α ≥ 1, n ≥ 0, ped(3n + 2) ≡ 0 (mod 2), ped(9n + 4) ≡ 0 (mod 4), ped(9n + 7) ≡ 0 (mod 4),
ped 32α+1 n +
17 × 32α − 1
≡ 0 (mod 2), 11 × 32α+1 − 1 ped 32α+2 n + ≡ 0 (mod 2), 8 19 × 32α+1 − 1 ped 32α+2 n + ≡ 0 (mod 2). 8
8
In this short note, we will prove more congruences for ped(n) mod 2 and mod 4. Before we state the main results, we recall some notation. The Dedekind η-function is given by the infinite product 1
η(z ) := q 24
∞ ∏
(1 − qn ),
n =1
E-mail address: [email protected]. 0012-365X/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2011.02.025
S.-C. Chen / Discrete Mathematics 311 (2011) 940–943
941
where q = exp(2π iz ) and z lies in the complex upper half plane H. The well-known ∆-function is denoted by
∆(z ) := η(z )24 =
∞ −
τ (n)qn .
n=1
We have the following general theorem. Theorem. Let ℓ and p be odd primes. Then the following statements hold. (i) Suppose that s is an integer satisfying 1 ≤ s < 8ℓ, s ≡ 1(mod 8) and ℓs = −1, where ℓs is the Legendre symbol. Then
s−1
ped ℓn +
8
≡ 0 (mod 2).
(2)
If τ (ℓ) ≡ 0(mod 2), then, for all n ≥ 0, α ≥ 1,
ped ℓ2α+1 n +
sℓ2α − 1
≡ 0 (mod 2).
8
(3)
(ii) Suppose that r is an integer such that 1 ≤ r < 8p, rp ≡ 1(mod 8), and (r , p) = 1. If τ (p) ≡ 0(mod 2), then, for all n ≥ 0, α ≥ 1,
ped p2α+2 n +
rp2α+1 − 1
8
≡ 0 (mod 2).
(4)
If c (p) ≡ 0(mod 4), then, for all n ≥ 0, α ≥ 1,
ped p
2α+2
n+
rp2α+1 − 1
8
≡ 0 (mod 4), η4 (16z )
where c (p) is the p-th coefficient of η(8z )η(32z ) :=
∑∞
n =1
(5) c (n)qn .
Remark. If p = ℓ = 3, then τ (3) = 252 and c (3) = 0. The theorem yields the results of Andrews et al. above. Examples. Since τ (5) = 4830 and c (5) = 0, our theorem gives the following new congruences. For all n ≥ 0, α ≥ 1, ped (5n + 2) ≡ 0 (mod 2), ped (5n + 4) ≡ 0 (mod 2),
ped 52α+1 n +
ped 52α+1 n +
17 × 52α − 1
8 33 × 52α − 1
8
≡ 0 (mod 2), ≡ 0 (mod 2).
For r = 13, 21, 29, 37,
ped 52α+2 n +
r × 52α+1 − 1
8
≡ 0 (mod 4).
2. Proof of theorem As usual, Mk (SL2 (Z)) (resp., Sk∑ (SL2 (Z))) is the complex vector space of weight k holomorphic modular (resp., cusp) forms ∞ n with respect to SL2 (Z). If f (z ) = n=0 a(n)q ∈ Mk (SL2 (Z)), then the action of the Hecke operator Tp,k on f (z ) is defined by f (z )|Tp,k :=
∞ − (a(pn) + pk−1 a(n/p))qn .
(6)
n =0
It is known that ∆(z ) ∈ S12 (SL2 (Z)) and ∑ is an eigenform for all Hecke operators; in other words, ∆(z )|Tp,12 = τ (p)∆(z ) ∞ n for any prime p. The coefficients of ∆(z ) = n=1 τ (n)q have the following properties (see page 164 of [2]):
τ (mn) = τ (m)τ (n) if (m, n) = 1,
(7)
τ (p ) = τ (p)τ (p
(8)
l
l −1
) − p τ (p 11
l −2
).
Now we begin to prove the theorem. Using the generating function (1) for ped(n), we find that ∞ − n =0
ped(n)q8n+1 = q
∞ ∏ (1 − q32n ) n=1
(1 − q8n )
≡ ∆(z ) =
∞ − n =1
τ (n)qn (mod 2).
(9)
942
S.-C. Chen / Discrete Mathematics 311 (2011) 940–943
The well-known Jacobi identity ∞ ∏
(1 − qn )3 =
∞ ∞ − − n(n+1) n(n+1) (−1)n (2n + 1)q 2 ≡ q 2
n=1
n =0
(mod 2)
n =0
shows that
∆(z ) = q
∞ ∏
(1 − qn )24 ≡
∞ −
n =1
q(2n+1)
2
(mod 2).
n =0
Therefore, we have ∞ −
∞ −
ped(n)q8n+1 ≡
n =0
q(2n+1)
2
(mod 2).
(10)
n=0
If s ≡ 1(mod 8) and ℓs = −1, then, for any n ≥ 0, 8ℓn + s cannot be an odd square. This implies that the coefficients of q8ℓn+s in the left-hand side of (10) must be even. It follows that
s−1
ped ℓn +
8
≡ 0 (mod 2).
(11)
Since τ (ℓ) ≡ 0(mod 2) and ∆(z ) is a Hecke eigenform, we have
∆(z )|Tℓ,12 = τ (ℓ)∆(z ) ≡ 0 (mod 2). By (6) and (9), we get ∞ −
ped(n)q
8n+1
∞ −
|Tℓ,12 ≡
ped
ℓn − 1
n=n0
n =0
8
n
ℓ −1 8
n
+ ped
ℓ
−1
8
∈ N, then ℓn8−1 = ℓ2 m + ℓ 8−1 . So, we have ℓ2 − 1 ped ℓ2 m + + ped(m) ≡ 0 (mod 2).
If we write m =
2
8
That is,
ped ℓ2 m +
ℓ2 − 1
8
≡ ped(m) (mod 2).
By induction, we find that
2α
ped ℓ m +
ℓ2α − 1
8
≡ ped(m) (mod 2).
Considering (11), we get
ped ℓ
2α+1
n+
sℓ2α − 1
8
≡ ped ℓn +
s−1 8
≡ 0 (mod 2).
This proves (3). Since τ (p) ≡ 0(mod 2), formula (8) gives
τ (p2α+1 ) ≡ 0 (mod 2). Applying (7), we have
τ (p2α+1 (8pn + r )) = τ (p2α+1 )τ (8pn + r ) ≡ 0 (mod 2). It follows from (9) that
ped p2α+2 n +
rp2α+1 − 1 8
≡ 0 (mod 2).
This proves (4). We denote f (z ) by the following eta-quotient: f (z ) =:
∞ − η4 (16z ) = c (n)qn . η(8z )η(32z ) n=1
qn ≡ 0
(mod 2).
S.-C. Chen / Discrete Mathematics 311 (2011) 940–943
Then f (z ) is a weight-1 Hecke eigenform with respect to Γ0 (256) with character χ (d) = of f (z ) satisfy the following similar properties as τ (n):
943
−256 d
(see [3]). The coefficients
c (mn) = c (m)c (n) if (m, n) = 1, c (pl ) = c (p)c (pl−1 ) − χ (p)c (pl−2 ). Using the fact that
(1 − q16n )4 ≡ 1 (mod 4), (1 − q32n )2 we find that f (z ) = q
∞ ∏ n=1
=q
∞ ∏ (1 − q32n ) (1 − q16n )4 n=1
≡q
(1 − q8n ) (1 − q32n )2
∞ ∏ (1 − q32n ) n=1
=
(1 − q16n )4 (1 − q8n )(1 − q32n )
∞ −
(1 − q8n )
ped(n)q8n+1
(mod 4) (mod 4).
n=0
Congruence (5) follows by similar arguments as the proof of (4). This completes the proof. Acknowledgements The author thanks the referee and the editor for numerous helpful suggestions. This research was supported by the National Natural Science Foundation of China (Grant No. 11026080) and the Natural Science Foundation of Education Department of Henan Province (Grant No. 2009A110001). References [1] G.E. Andrews, M.D. Hirschhorn, J.A. Sellers, Arithmetic properties of partitions with even parts distinct, Ramanujan J. 22 (2010) 273–284. [2] N. Koblitz, Introduction to Elliptic Curves and Modular Forms, Springer-Verlag, New York, 1984. [3] Y. Martin, Multiplicative η-quotients, Trans. Amer. Math. Soc. 348 (1996) 4825–4856.