11-Dissection and Modulo 11 Congruences Properties for Partition Generating Function

Transcription

1 Int. J. Contemp. Math. Sciences, Vol. 9, 2014, no. 1, 1-10 HIKARI Ltd, Dissection and Modulo 11 Congruences Properties for Partition Generating Function Goksal Bilgici Kastamonu University, Education Faculty Department of the Computer Education and Instructional Technology 37100, Kastamonu, Turkey Ali Bulent Ekin Ankara University, Faculty of Science Department of Mathematics 06100, Tandogan, Ankara, Turkey Copyright c 2014 Goksal Bilgici and Ali Bulent Ekin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In a recent paper, we give 13-dissection and some congruences for modulo 13 for the partition generating function (1 q r ) 1 by using a method of Kolberg. In this paper, by following similar course, we develop an algoritmic approach and give 11-dissection for the partition generating function (1 q r ) 1. Then we re-obtain the congruences given by Atkin and Swinnerton-Dyer. Mathematics Subject Classification: 11P83 Keywords: Partition, Dissections, q - Equivalance

2 2 Goksal Bilgici and Ali Bulent Ekin 1 Introduction A partition of a positive integer n is a non-increasing sequence of positive integers whose sum is n. The number of partitions of n is denoted p(n) and p(0) is assumed as 1. Euler gave the following generating function for the series {p(n)} n=0 p(n)q n 1 = n=0 1 q. r Throughout this paper, m>1 always denotes a positive integer prime to 6, and the variables y and q are always related to y = q m ( q < 1). We define F:= p(n)q n. For k =1, 2,..., m 1, we define k-th component of F as follows: F (k,m) := q k n=0 n=0 p(mn + k)y n, then we have a dissection for partition generating function, namely m 1 F= p(n)q n = F (k,m). n=0 Kolberg gave 5-dissection and 7-dissection for F. A simpler form of 7- dissection for F was given by Ekin in his doctoral thesis [3]. In a recent paper [2], the authors obtained 13-dissection for F and some congruences for the components F (k,13) where 0 k 12. In this paper, by following the same course in [2], we obtain 11-dissection for F. The 11-dissection for F may appear for the first time. After obtaining the components, using a q-equivalence given in [2], we obtain the congruences for the components F (k,11) (mod 11) given by Atkin and Swinnerton-Dyer in [1]. We prefer the following notation: k=0 P (a) = (y a ; y m ) (y m a ; y m ) P (0) = (y m,y m ) where (z; q) = (1 zq r 1 ) and a is not a multiple of m. P (a) satisfies P (m a) =P (a), P( a) =P (m + a) = y a P (a). For m = 11, Atkin and Swinnerton-Dyer gave

3 components and congruences for partition function 3 Theorem 1.1 For m =11, we have F (0,11) P (0) P (1) F (1,11) P (0)P (5) q P (2)P (3) F (2,11) 2 P (0)P (3) 2q P (1)P (4) F (3,11) 3 P (0)P (2) 3q P (1)P (3) F (4,11) 5q 4 P (0) P (2) F (5,11) 5 P (0)P (4) 7q P (2)P (5) F (6,11) 0 F (7,11) 4q 7 P (0) P (3) F (8,11) 8 P (0)P (1) 6yq P (4)P (5) F (9,11) 8q 9 P (0) P (4) F (10,11) 10 P (0) 9q P (5) (mod 11). They use the following congruence to calculate Theorem 1.1: 10 k=0 F (k,11) q k =(q; q) 1 { (1 q r ) 3} 3 (1 q r ) / (1 y r ) (mod 11). (1) We calculate 11-dissection for the partition generating function at first. Then, we obtain the Theorem 1.1 by using the components with an algorithmic approach. 2 Preliminaries Kolberg defines, for s =0, 1,..., m 1 g s := ( 1) n q 1 2 n(3n+1) s (mod m) 1 2 n(3n+1)

4 4 Goksal Bilgici and Ali Bulent Ekin and h s := These definitions give 1 n(n +1) s (mod m), 2 n 0 ( 1) n (2n +1)q 1 2 n(n+1). m 1 (1 q r m 1 )= g s and (1 q r ) 3 = s=0 s=0 By these equations, we conclude the following relation (g 0 + g g m 1 ) 3 = h 0 + h h m 1. (2) We have the following lemma from [4]. Lemma 2.1 0, if 24s +1 is a quad. non-residue mod m g s = ( 1) [ 1 (m+1)] 6 q 1 24 (m2 1) P (0), if 24s +1 0 (mod m) (3) and 0, if 8s +1 is a quad. non-res. mod m h s = ( 1) [ 1 (m 1)] (4) 2 mq 1 8 (m2 1) P 3 (0), if 8s +1 0 (mod m). Using the following lemma which is given by the authors in [2], we can determine the g s in terms of P (a). Lemma 2.2 Let 24s +1 is a quadratic residue mod m and m =6λ + μ where λ is a positive integer and μ = ±1. Then we have g s =( 1) c+λ q 1 2 (3c2 mc+3λ 2 +μλ) P (2c) (5) P (c) where c is a solution of the congruence x 2 (4s μλ)/6 (mod m). Kolberg also gives Lemma 2.3 For s =0, 1,..., m 1 F (s,m) =( 1) (m 1)s P (0) D (y; y) m+1 s (6) where D s is the following determinant; g s g s+1 g s+m 2 g s 1 g s g s+m 3 (7) g s m+2 g s m+3 g s we put g r = g s when r s (mod m) in (7). h s.

5 components and congruences for partition function 5 We define where m is prime and 24k +1 0 For the denominator of (8), we have A s := gk 1 m D s (mod m). So we have F (s,m) = q 1 24 (m3 m 2 m+1) P m (0) A (y; y) m+1 s. (8) (y; y) = (1 y r )=P(0)P (1)P (2) P((m 1)/2). (9) We use the following lemma which is given by the authors in [2] to obtain the congruence properties of components. Lemma 2.4 If m Z + is a prime, then [ ] m (1 q r ) 3 P m+2 (0)P m+3 (1)P m+3 (2) P m+3 ((m 1)/2) (mod m). (10) For the left-hand side on Eq.(10), we need the following lemma which is Lemma 3 in [1]: Lemma 2.5 We have (m 3)/2 (1 q r ) 3 P (0) ( 1) c (2c +1)q 1 2 c(c+1) P c=0 ( m 1 2 ) c (mod m). (11) 3 Components and Congruences for m =11 In this section we give the components F (k,11) and find the congruences given by Atkin and Swinnerton-Dyer. For m = 11, from (3) and (4) we have g 3 = g 6 = g 8 = g 9 = g 10 =0, g 5 = q 5 P (0) and We set h 2 = h 5 = h 7 = h 8 = h 9 =0, h 4 = 11q 15 P 3 (0). α := g 0 g 1 5, β := g 1 g 1 5, γ := g 2 g 1 5, θ := g 4 g 1 5, δ := g 7 g 1 5. (12)

6 6 Goksal Bilgici and Ali Bulent Ekin From (2) we find 3(2g 2 g 4 g 7 +2g 1 g 5 g 7 + g 0 g1 2 + g2 4 g 5 + g0 2 g 2) = h 2 =0, 3(2g 0 g 1 g 4 +2g 4 g 5 g 7 + g 2 g7 2 + g 1 g2 2 + g0g 2 5 ) = h 5 =0, 3(2g 0 g 2 g 5 +2g 1 g 2 g 4 + g 4 g7 2 + g2 0 g 7 + g1 2 g 5) = h 7 =0, 3(2g 0 g 1 g 7 +2g 1 g 2 g 5 + g2 2 g 4 + g 5 g7 2 + g 0g4 2 ) = h 8 =0, 3(2g 0 g 4 g 5 +2g 0 g 2 g 7 + g2 2 g 5 + g1 2 g 7 + g 1 g4 2 ) = h 9 =0, 3(g 1 g7 2 + g0g g 0 g2 2 + g4g g1g 2 2 )+g5 3 = h 4 = 11g5. 3 By the help of (12), these equations become, respectively 2γθδ + αβ 2 + α 2 γ + θ 2 +2βδ =0, (13) 2αβθ + γδ 2 + βγ 2 + α 2 +2θδ =0, (14) 2βγθ + θδ 2 + α 2 δ + β 2 +2αγ =0, (15) 2αβδ + αθ 2 + γ 2 θ + δ 2 +2βγ =0, (16) 2αγδ + βθ 2 + β 2 δ + γ 2 +2αθ =0, (17) α 2 θ + β 2 γ + θ 2 δ + γ 2 α + δ 2 β = 4. (18) Now we put x 1 := α 2 θ, x 2 := β 2 γ, x 3 := θ 2 δ, x 4 := γ 2 α, x 5 := δ 2 β. (19) Thus (18) becomes x 1 + x 2 + x 3 + x 4 + x 5 = 4. (20) After multiplying A s (s =0, 1,.., 10) by α, α 1, β 1, γ 1, δ, θ 1,1,θ, δ 1, γ and β respectively, A s can be written in terms of x i. From (3) and (5), we get P (0)P (4) P (0)P (2) 2 P (0)P (5) g 0 =, g 1 = q, g 2 = q, P (2) P (1) P (3) g 4 = q 4 P (0)P (1) y, g 5 = q 5 7 P (0)P (3) P (0), g 7 = q. P (5) P (4) These equations give αβγθδ = 1 and x 1 x 2 x 3 x 4 x 5 = 1. (21) Lemma 3.1 We have x 1 x 2 = x 4 +1, (22) x 2 x 3 = x 5 +1, (23) x 3 x 4 = x 1 +1, (24) x 4 x 5 = x 2 +1, (25) x 5 x 1 = x (26)

7 components and congruences for partition function 7 Proof. We define A := x 3 x 4 x 5, B := x 1 x 2 x 5, C := x 1 x 4 x 5, D := x 1 x 2 x 3 and E := x 2 x 3 x 4. Multiplying equations (13), (14), (15), (16) and (17) by δ, θ, γ, β and α respectively give us 2A + B + C = x 3 +2x 5, 2D + A + E = x 1 +2x 3, 2E + A + C = x 2 +2x 4, 2B + D + E = x 5 +2x 2, 2C + D + B = x 4 +2x 1. The solution of this equations system is A = x 3 x 4 x 5 = 1 4 ( x 1 x 2 x 4 +3x 3 +3x 5 ), B = x 1 x 2 x 5 = 1 4 ( x 1 +3x 2 x 4 x 3 +3x 5 ), C = x 1 x 4 x 5 = 1 4 (3x 1 x 2 +3x 4 x 3 x 5 ), D = x 1 x 2 x 3 = 1 4 (3x 1 x 2 x 4 +3x 3 x 5 ), E = x 2 x 3 x 4 = 1 4 ( x 1 +3x 2 +3x 4 x 3 x 5 ). Using equation (20), we get x 3 x 4 x 5 = x 3 + x 5 +1, (27) x 1 x 2 x 5 = x 2 + x 5 +1, (28) x 1 x 4 x 5 = x 1 + x 4 +1, (29) x 1 x 2 x 3 = x 1 + x 3 +1, (30) x 2 x 3 x 4 = x 2 + x (31) Multiplying both sides of (27) by x 1 x 2 and using (21), we find 1 =x 1 x 2 x 3 + x 1 x 2 x 5 + x 1 x 2. (32) Substituting for (28), (30) and (20) into the equation (32), we obtain x 1 x 2 = x (33) and we find the others similarly. Algorithm 1. Let U be a linear combination of x i 1 1 x i 2 2 x i 3 3 x i 4 4 x i 5 5 where each i r is a non-negative integer.

8 8 Goksal Bilgici and Ali Bulent Ekin 1. Substitute for the equations (22)-(26) into U. This step turns all terms into the form x a i or x a i xb j where (i, j)=(1,4), (2,5), (4,2), (3,1), (5,3). 2. If b>1 then substitute for x 2 j,ifa>1 and b=0 substitute for x 2 i into U. This step writes U as sums of x a i x j and x i. To evaluate this step we use the following equations which can be easily found by (20): x 2 1 = x 1 x 4 x 3 x 1 4x 1 x 4 x 3 2, (34) x 2 2 = x 4 x 2 x 2 x 5 4x 2 x 4 x 5 2, (35) x 2 3 = x 3 x 1 x 5 x 3 4x 3 x 5 x 1 2, (36) x 2 4 = x 1 x 4 x 4 x 2 4x 4 x 1 x 2 2, (37) x 2 5 = x 5 x 3 x 2 x 5 4x 5 x 2 x 3 2. (38) By using Algorithm 1, we get the A s in simple forms: αa 0 = x 4 4 x 2 +31x 3 1 x 4 +42x 3 3 x 1 42x 3 4 x 2 5x 3 5 x x 2 1 x x 2 3 x 1 +10x 2 2 x 5 296x 2 4 x 2 64x 2 5 x x 1 x x 3 x 1 332x 4 x x 2 x 5 429x 5 x 3 362x x x 3 116x x , (39) α 1 A 1 = 7x 4 2 x 5 8x 3 4 x 2 +3x 3 1 x 4 22x 3 5 x x 3 2 x 5 +2x 3 3 x 1 +28x 2 4 x 2 +85x 2 1 x 4 180x 2 5 x x 2 2 x 5 +74x 2 3 x 1 113x 4 x x 4 x x 5 x 2 524x 5 x x 1 x x x x x 1 158x , A 6 = 11[ x 3 1 x 4 x 3 2 x 5 x 3 4 x 2 x 3 3 x 1 x 3 5 x 3 14x 2 1 x 4 14x 2 2 x 5 14x 2 4 x 2 14x 2 3 x 1 14x 2 5 x 3 29x 1 x 4 29x 2 x 5 29x 2 x 4 29x 1 x 3 29x 3 x ]. The last equation proves the famous congruence p(11n +6) 0 (mod 11) of Ramanujan. If we observe the indices of the terms in A s, having αa 0 and α 1 A 1 is enough to obtain other components via the permutation (12345). This permutation gives the following relations αa 0 βa 10 θa 7 γa 9 δa 4 αa 0 (40) α 1 A 1 β 1 A 2 θ 1 A 5 γ 1 A 3 δ 1 A 8 α 1 A 1 (41) A 6 A 6. (42) Using these relations, that is, changing indices in convenient order and making the same operations, we get the other components. For m = 11, with the help of Eq.(10) and Lemma 2.5, we have P 11 (5) + 8yP 11 (4) + 5y 3 P 11 (3) + 4y 6 P 11 (2) + 9y 10 P 11 (1) P 2 (0)P 14 (1)P 14 (2)P 14 (3)P 14 (4)P 14 (5) (mod 11). (43)

9 components and congruences for partition function 9 For abbreviation, we define (a 1,a 2,,a (m+1)/2 ):=y a 1 P a 2 (1)P a 3 (2)P a (m+1)/2 ((m 1)/2). We write (43) in terms of x i in 20 different ways by dividing (43) by (a, b, c, d, e, f) where 1 b, c, d, e, f 5 and b + c + d + e + f = 11. Five of them are useful for us and the remaining 15 of them are linearly independent on these five. These can be found dividing (43) by (3, 1, 2, 3, 2, 3), (4, 2, 2, 3, 3, 1), (4, 2, 3, 1, 3, 2), (4, 3, 1, 2, 2, 3) and (5, 3, 3, 2, 1, 2): 7x 6 2 x5 3 x 1x 4 +2x 6 3 x4 4 x 2 +3x 5 1 x4 2 x 3 +10x 5 4 x3 5 x 3 +6x 4 5 x3 1 (y; y) 2 ( 3, 13, 12, 11, 12, 11) 9x 6 3x 5 4x 2 x 5 + x 6 4x 4 5x 3 +7x 5 2x 4 3x 4 +5x 5 5x 3 1x 4 +3x 4 1x 3 2 (y; y) 2 ( 4, 12, 12, 11, 11, 13) x 6 4x 5 5x 1 x 3 +5x 6 5x 4 1x 4 +2x 5 3x 4 4x 5 +3x 5 1x 3 2x 5 +4x 4 2x 3 3 (y; y) 2 ( 4, 12, 11, 13, 11, 12) 8x 6 1 x5 2 x 3x 5 +7x 6 2 x4 3 x 1 +5x 5 5 x4 1 x 2 +10x 4 4 x3 5 +2x5 3 x3 4 x 2 (y; y) 2 ( 4, 11, 13, 12, 12, 11) 5x 6 5x 5 1x 2 x 4 +3x 6 1x 4 2x 5 +10x 5 4x 4 5x 1 +4x 5 2x 3 3x 1 +9x 4 3x 3 4 (y; y) 2 ( 5, 11, 11, 12, 13, 12) (mod 11). By using Algorithm 1 to the left hand side of these congruences, we obtain 3x 4 1x 4 +5x 3 1x 4 +6x 1 x x 3 2x 5 +6x 3 x x 2 1x 4 + x 1 x x 2 2x 5 +3x 2 x x 3x x 1x 4 +5x 1 x 3 +8x 2 x 4 +6x 3 x 5 +6x 1 +8x 2 +7x 4 +7x 3 +2x 5 +4 (y; y) 2 ( 3, 13, 12, 11, 12, 11) (44) 7x 4 2x 5 +3x 3 1x 4 +2x 1 x x 3 2x 5 +3x 2 x x 2 1x 4 +8x 1 x x 2 2x 5 +6x 2 x x 3 x x 1 x 4 +8x 2 x 4 +3x 2 x 5 +4x 3 x 5 + x 1 +3x 2 +9x 4 +4x 3 +9x 5 +2 (y; y) 2 ( 4, 12, 12, 11, 11, 13) (45) 2x 1 x x 1x x3 2 x 5 +10x 2 x x 3x x2 1 x 4 +9x 1 x x2 2 x 5 +7x 2 x x 3x x 1x 4 +4x 1 x 3 +4x 2 x 5 +7x 3 x 5 + x 1 +5x 2 +9x 4 +4x 3 + x (y; y) 2 ( 4, 12, 11, 13, 11, 12) (46) 5x 3 x x 3 1x 4 +10x 3 2x 5 +10x 2 x x 3 x x 2 1x 4 +5x 1 x x 2 2x 5 + x 2 x x 3 x x 1 x 3 +10x 2 x 4 + x 2 x 5 +10x 3 x 5 +6x 1 +8x 2 +7x 4 +8x 3 +10x 5 +3 (y; y) 2 ( 4, 11, 13, 12, 12, 11) (47) 10x 2 x x3 1 x 4 +9x 1 x x 2x x 3x x2 1 x 4 +2x 1 x x2 2 x 5 +x 2 x x 3x x 1x 4 +9x 1 x 3 +9x 2 x 4 + x 2 x 5 +5x 1 +5x 2 +9x 4 +3x 3 + x 5 +6 (y; y) 2 ( 5, 11, 11, 12, 13, 12) (mod 11). (48) Now we can obtain the congruences given by Atkin and Swinnerton-Dyer. From (39), we have αa 0 10x 2 x x 3 1x 4 +9x 1 x x 2 x x 3 x x 2 1x 4 +2x 1 x x 2 2x 5

10 10 Goksal Bilgici and Ali Bulent Ekin From (48), we find +x 2 x x 3x x 1x 4 +9x 1 x 3 +9x 2 x 4 + x 2 x 5 + x 1 + x 2 +5x 4 +10x 3 +8x (mod 11) A 0 α 1 (y; y) 2 ( 5, 11, 11, 12, 13, 12) (mod 11). (49) Then the equations (8) and (49) give us F (0,11) P (0) P (1) (mod 11). The other congruences in Theorem 1.1 can be obtained similarly. References [1] A.O.L. Atkin and H.P.F. Swinnerton-Dyer, Some Properties of Partitions, Proc. London Math. Soc., 3 (4) (1954), [2] G. Bilgici and A.B. Ekin, Some Congruences for Modulus 13 Related to Partition Generating Function, The Ramanujan Journal, (to be appear). [3] A.B. Ekin, The Rank and the Crank in the Theory of Partition, D.Phil Thesis, University of Sussex, [4] O. Kolberg, Some Identities Involving the Partition Function, Math. Scand., 5 (1957), Received: November 1, 2013