Transcription

1 This article was originally published in a journal published by Elsevier, and the attached copy is provided by Elsevier for the author s benefit and for the benefit of the author s institution, for non-coercial research and educational use including without liitation use in instruction at your institution, sending it to specific colleagues that you know, and providing a copy to your institution s adinistrator. All other uses, reproduction and distribution, including without liitation coercial reprints, selling or licensing copies or access, or posting on open internet sites, your personal or institution s website or repository, are prohibited. For exceptions, perission ay be sought for such use through Elsevier s perissions site at:

2 Journal of Cobinatorial Theory, Series A 114 (2007) The divisibility odulo 24 of Kloosteran sus on GF(2 ), odd Pascale Charpin a, Tor Helleseth b, Victor Zinoviev c a INRIA, Doaine de Voluceau-Rocquencourt, BP , Le Chesnay, France b Departent of Inforatics, University of Bergen, N-5020 Bergen, Norway c Institute for Probles of Inforation Transission, Russian Acadey of Sciences, Bol shoi Karetnyi per. 19, GSP-4, Moscow , Russia Received 18 October 2005 Available online 12 July 2006 Counicated by Vera Pless Abstract In a previous paper, we studied the cosets of weight 4 of binary extended 3-error-correcting BCH codes of length 2 (where is odd). We expressed the nuber of codewords of weight 4 in such cosets in ters of exponential sus of three types, including the Kloosteran sus K(a), a F. In this paper, we derive soe congruences which link Kloosteran sus and cubic sus. This allows us to study the divisibility of Kloosteran sus odulo 24. More precisely, if we know the traces of a and of a 1/3, we are able to evaluate K(a) odulo 24 and to copute the nuber of those a giving the sae value of K(a) odulo Elsevier Inc. All rights reserved. Keywords: 3-Error-correcting BCH code; Coset; Melas code; Irreducible code; Divisibility; Kloosteran su; Cubic su; Inverse-cubic su; Quadratic apping; Jacobi sybol 1. Introduction This paper is a natural continuation of our previous papers [4,5,9]. Here we exploit the connection between binary priitive (in narrow sense) extended BCH codes of length 2 ( odd) with inial distance 8 and Kloosteran sus over finite fields of order 2. In [4], the coset This work was supported by INRIA-Rocquencourt, by the Norwegian Research Council and also by the Russian fund of fundaental researches (project ). E-ail address: pascale.charpin@inria.fr (P. Charpin) /$ see front atter 2006 Elsevier Inc. All rights reserved. doi: /j.jcta

3 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) weight distributions of such BCH codes was described. It was proved that all these distributions are known as soon as they are known for cosets of weight 4. We indicated, for = 7, the surprising distribution of the nuber of leaders in each coset of weight 4. In [9], the coset weight distributions of the Z 4 -linear Goethals codes are considered. For the cosets of weight 4, the nuber of codewords of weight 4 was expressed in ters of Kloosteran sus. Using this last approach, and considering again the BCH codes, we found in [5] the exact expression of the nuber of codewords of weight 4 (in cosets of weight 4) in ters of exponential sus of three different types including Kloosteran sus. In this paper, we exploit the deep connection between cubic and Kloosteran sus which appeared in [5]. The paper is organized as follows. After soe preliinaries, we present in Section 3 our ain expression established in [5] as a congruence (Theore 1 and (8)). We derive soe properties, notably on the divisibility of the inverse-cubic sus (Lea 5). By Theore 2 we state the key congruences which will allow us to copute the divisibility of Kloosteran sus odulo 24. All the results are given by Theore 3 providing six different cases which depend each on the for of. In Section 5, we recall the connection between Kloosteran sus and the [2 + 1, 2] irreducible code and explain the consequences of our results on the weight distribution of this code. Finally, in the last section, we copute for each case provided by Theore 3 the nuber of ties this case occurs (Table 2). We then contribute to the knowledge of the spectru of Kloosteran sus which is, equivalently, the knowledge of the weight distribution of Melas codes or of the [2 + 1, 2] irreducible codes. A part of our results in this paper was presented in [6]. 2. Preliinaries In this paper, the appings are generally defined on F 2, the finite field of order 2, with odd and 5. 1 For any such apping f, we will consider the Boolean function x Tr(f (x)) where Tr is the trace-function fro F 2 to F 2. For siplicity, we will use this notation: e ( f(x) ) = ( 1) Tr(f (x)). Also #T will denote the cardinality of any set T Exponential sus Now, we need to define several exponential sus on F 2. Definition 1. The Kloosteran sus are: K(a) = ( ) 1 e x + ax, a F 2. x F 2 The cubic sus are: C(a,b) = x F 2 e ( ax 3 + bx ), a F 2,b F 2. 1 Note that for = 3 the Kloosteran sus are cubic sus and the 3-error-correcting BCH codes do not exist.

4 324 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) The inverse-cubic sus are: G(a, b) = ( ) a e x 3 + bx, a F 2,b F 2. x F 2 For siplicity, G(a,a) will be denoted by G(a) and C(1,b)by C(b). The Kloosteran sus and the inverse-cubic sus are generally defined on F 2,theultiplicative group of F 2. We extend the to 0, assuing that e(x 1 ) = e(x 3 ) = 1forx = 0. Actually, we consider the Boolean functions on F 2 : ( ) 1 Tr = Tr ( x ) ( ) 1 and Tr = Tr ( x ). x x 3 Note that, since the appings x x 2 1 1, x x and x x 3 are perutations on F 2 when is odd, we have (for any a): K(0) = G(a, 0) = C(a,0) = 0. The next lea is directly obtained fro the result of Lachaud and Wolfann [11, Theore 3.4] which is suitable for any (even or odd). We only replace the su on F 2, denoted K(a) in [11], by K(a) = K(a)+ 1. Lea 1. The set K(a), a F 2, is the set of all the integers s 0 (od 4) with s 1 in the range [ 2 (+2)/2, 2 (+2)/2 ]. Moreover, the following result is due to Helleseth and Zinoviev [9]. Lea 2. For any (odd or even) 4 (od 8) if Tr(a) = 1, K(a) 0 (od 8) if Tr(a) = Quadratic sus Let i>0and a F 2. The spectru of quadratic sus on F 2 of the for S(i,a,b) = e ( ax 2i +1 + bx ), b F 2 (1) x F 2 is well-known. It is obtained by coputing the rank of the syplectic for of the Boolean function x Tr(ax 2i +1 ). The values S(i,a,b), when b runs through F 2, and the nuber of ties they occur, only depend on this rank, usually denoted by r = 2h. That is: 0 occurs 2 2 2h ties; 2 h occurs h 1 ties; 2 h occurs 2 2 h 1 ties. (2) These properties are explained in [12, Chapter 15] and [10,13]. In this paper, is odd. We ainly need the sus S(1, 1,b), b F 2 (the cubic sus), and the su S(3, 1, 1) (see Proposition 5). For clarity, the spectru of the cubic su is recalled in

5 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) Table 1 The spectru of C(b) (Definition 1) Value Nuber it occurs (+1)/ ( 3)/2 2 (+1)/ ( 3)/2 Table 1. It corresponds to the rank 1, i.e., h = ( 1)/2. When i = 3 in (1), and for odd, the corresponding rank is r = 3 if 3 divides and r = 1 otherwise (see [12, Fig. 15.2]). On the other hand, it is generally difficult to know the sign of S(i,a,b) for any fixed (i, a) and for a given b. Fori = 1 (and for odd ) this was specified by Carlitz in [3] by eans of the Jacobi sybol ( 2 ) (also called the quadratic character). Recall that the Jacobi sybol is a generalization of the Legendre sybol which was introduced for any odd prie p: ( ) 2 1, if 2 is a square odulo p, = (3) p 1, otherwise. We will need the exact expression of the Jacobi sybol. Proposition 1. Let be odd. Set = 1 s, where s 1 and each i is an odd prie. Then ( ) ( ) ( ) =. 1 s Moreover, ( ) 2 1 for ±1 (od 8), = ( 1) (2 1)/8 = (4) 1 for ±3 (od 8). In the next lea, we give a slightly different version of the result of Carlitz [3, Theore 2]. Lea 3. Let be odd. Recall that C(b) = x F 2 e(x3 + bx), b F 2 (Definition 1). Then we have: (i) C(1) = ( 2 )2(+1)/2, where ( 2 ) is the Jacobi sybol; (ii) if Tr(b) = 1(with b 1), then C(b) = e ( β 3)( ) 2 2 (+1)/2, where β is unique in F 2 satisfying b = β 4 + β + 1 with Tr(β) = 0; (iii) if Tr(b) = 0, then C(b) = 0. Proof. Coparing to the theore of Carlitz, only (ii) is slightly odified. In [3] the expression of C(b) is C(b) = e ( β 3 + β )( ) 2 2 (+1)/2, where b = β 4 + β + 1forsoeβ.

6 326 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) The apping g : x x 4 + x + 1 is two-to-one on F 2 and g(x) = g(x + 1). More precisely, for any b such that Tr(b) = 1 there are exactly two eleents β and β +1 such that b = β 4 +β +1. But Tr ( (β + 1) 3 + β + 1 ) = Tr ( β 3 + β 2) = Tr ( β 3 + β ). Thus we can define β as the unique eleent in the pair (β, β + 1) which satisfies Tr(β) = 0. Moreover, e(β 3 + β) = e(β 3 ) copleting the proof of (ii). In Section 6, we will need specific results on quadratic sus. It is first the next forula: C(1) = 2 e ( x 3) = 2 e ( x 3). (5) x F 2, Tr(x)=0 This forula holds because e ( x 3 + x ) = x F 2 x F 2, Tr(x)=0 x F 2, Tr(x)=1 e ( x 3) x F 2, Tr(x)=1 e ( x 3). But C(0) = 0 eaning that Tr(x)=0 e(x3 ) = Tr(x)=1 e(x3 ). We will need a basic property of quadratic functions. A proof of the next lea can be found in [2, Proposition II-5]. First recall that any Boolean function f on F 2 is said to be balanced if the nuber of those x such that f(x)= 1, i.e., the weight of f is equal to 2 1. In this case, we have obviously ( 1) f(x) = 0. x F 2 Lea 4. Let f be any quadratic Boolean function f on F 2. Then f is balanced if and only if there is u F 2 such that f(x)+ f(x+ u) = 1, x. In the next exaple, we point out the balanceness of a special class of quadratic Boolean functions. Exaple 1. Set f(x)= i I Tr(x2i +1 ) where I 0, 1,..., 1}, odd. Then we have for any x F : f(x)+ f(x+ 1) = Tr ( x 2i + x + 1 ) = Tr(1) i I i I #I (od 2). Thus, when is odd, such f is balanced if the cardinality of I is odd. 3. Cosets of the 3-error-correcting extended BCH code Let us denote by B the binary BCH code of length n = 2, odd and 5, with inial distance 8. Finding the nuber of words in a coset of weight 4 of B needs to solve the following syste of equations over F 2 : x + y + z + u = a, x 3 + y 3 + z 3 + u 3 = b, (6) x 5 + y 5 + z 5 + u 5 = c.

7 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) Here x,y,z and u are pairwise distinct eleents of F 2 and a,b,c F 2 are fixed. Let μ(a,b,c) be the nuber of different such 4-tuples (x,y,z,u)which are solutions to the syste (6). Assue that a 0. Then there are ɛ 0, 1} and η F 2 such that μ(a, b, c) = μ(1,ɛ,η).tobeore precise ( ) b ɛ = Tr a 3 and η = c a 5 + b2 a 6 + b a 3 are uniquely defined, as explained in [5]. The condition a 0 eans that the corresponding coset is contained in the Reed Muller code of order 1 and not contained in the Reed Muller code of order 2, i.e., (x,y,z,u) is not a flat. The reader can refer to [4] for a coplete explanation of the types of cosets of 3-error-correcting extended BCH codes. In [5] we established that μ(1,ɛ,η)can be expressed in ters of exponential sus. Theore 1. Let μ(1,ɛ,η), ɛ 0, 1} and η F 2, as above defined. Then μ(1,ɛ,η)is even and can be expressed as follows: for η 1 24 μ(1,ɛ,η)= 2 8 ( 1 + ( 1) ɛ+1) + 3 G(η + 1) + ( 1) ɛ+1 2 (K(η + 1) + C ( (η + 1) 1/3)). (7) Furtherore, when η = 1 then μ(1,ɛ,1) = 0. Reark 1. The expression of μ(1,ɛ,η)in [5] is of the for G + ( 1) ɛ+1 2 (K + C ( (η + 1) 1/3) 3 ), where G = G(η + 1) 1 and K = K(η + 1) 1 (these sus were coputed on F 2, not on F 2 ). Fro now on μ(ɛ, η) = μ(1,ɛ,η)and G(a,a) is denoted G(a), for siplicity. There are several consequences of the previous theore. We deduce easily that for 9 there are solutions of (6) except for the (a,b,c) corresponding to (1,ɛ,1). This leads to the deterination of the nuber of cosets of weight 4, as we will explain in a forthcoing paper [7]. Corollary 1. When 9 then μ(ɛ, η) > 0 for any ɛ 0, 1} and for any η 1. Consequently, for any such (ɛ, η) and for 9: 2 8 ( 1 + ( 1) ɛ+1) + 3 G(η + 1) + ( 1) ɛ+1 2 (K(η + 1) + C ( (η + 1) 1/3)) 0 (od 48). (8) Proof. Fro Leas 1 and 3, we have: C ( (η + 1) 1/3) 0, ±2 (+1)/2} and K(η + 1) < 2 (/2)+1 + 1, where λ denotes the absolute value of λ. Moreover, we proved in [5] that G(η + 1) < 2 (/2) We consider the equality (7) and we have to prove that the ter on the right is strictly positive. We consider A = 3 G(η + 1) + ( 1) ɛ+1 2 (K(η + 1) + C ( (η + 1) 1/3)). Fro the previous bounds we get: A < 3 2 (/2) ( 2 (/2) (+1)/2) + 5

8 328 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) which iplies A < 4 2 (/2) (+3)/ We want to prove that 2 8(1 + ( 1) ɛ+1 ) + A > 0. Clearly this is true as soon as (/2) + 4 <, that is + 8 < 2 which is 8 <. Thus, we know that, for 9, μ(ɛ, η) is a nonzero positive even nuber. In accordance with (7) we directly obtain (8). Reark 2. Let B be the 3-error correcting extended BCH code. When = 5, B is in fact the self-dual Reed Muller code. The weight distribution of cosets of B is known [1]. When = 7, the weight distribution of cosets of B was coputed in [4]. Note that for <9 the values K(a) are easily coputed (see Tables 4 and 5). Moreover, we are able to deduce a property on the divisibility of inverse-cubic sus G(a). Note that it is suitable for odd only. Recall that G(a) = x F 2 e(a/x3 + ax). Lea 5. For any odd, 5, 8 (od 16), if Tr(a) = 1, G(a) 0 (od 16), if Tr(a) = 0. Proof. We use (7) with a = η + 1, a 0. Note that G(0) = 2 is a trivial case. The cubic su C(a 1/3 ) is congruent to 0 odulo 2 (+1)/2. Thus 2C(a 1/3 ) is congruent to 0 odulo 16 as soon as ( + 3)/2 4 which is 5. Since 2 8(1 + ( 1) ɛ+1 ) is 0 odulo 16 for any ɛ 0, 1}, we then deduce fro (7): ±2 K(a) + 3 G(a) 0 (od 16). Now we apply Lea 2. If Tr(a) = 1 then 2K(a) 8 odulo 16 iplying that G(a) 8 odulo 16 too. When Tr(a) = 0, 2K(a) 0 odulo 16 and then G(a) 0 odulo Divisibility of Kloosteran sus The following relations between cubic and Kloosteran sus are deduced fro (7) (with a = η + 1, a 0), Leas 2 and 3. Recall that 5. Theore 2 (The key congruences). For any a, a F 2, we have: If Tr(a) = 0 then else C ( a 1/3) + K(a) 16 (od 24) (9) C ( a 1/3) + K(a) 4 (od 24). (10) Proof. First note that for any integer r, r 1, we have: 2 r ( 1) r (od 3). Thus, since is odd, we deduce fro (7): 2 + ( 1 + ( 1) ɛ+1) + ( 1) ɛ+1 2 (K(a) + C ( a 1/3)) 0 (od 3).

9 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) This gives the sae congruence for ɛ = 0 and ɛ = 1: K(a) + C ( a 1/3) 1 (od 3). (11) Now C(a 1/3 ) 0 odulo 8 as soon as 5 (Lea 3). Hence K(a) + C ( a 1/3) K(a) (od 8). We apply Lea 2. Set L(a) = K(a) + C(a 1/3 ).IfTr(a) = 0 then L(a) = 8R, forsoeinteger R, which leads to L(a) 2R odulo 3. According to (11) we get R 2 odulo 3. Consequently L(a) 16 odulo 24. Siilarly, if Tr(a) = 1 then L(a) = 8R + 4 leads to L(a) 2R + 1 odulo 3. Then we get R 0 odulo 3 which iplies L(a) 4 odulo 24, copleting the proof. In accordance with the previous theore, it appears that the divisibility of the sus K(a) odulo 24 is strongly related with those of the cubic sus. This fact will be specified by proving the next three corollaries. Corollary 2. Assue that Tr(a 1/3 ) = 0, a F 2. Then 16 (od 24) if Tr(a) = 0, K(a) 4 (od 24) if Tr(a) = 1. Proof. Recall that C(a 1/3 ) = 0 if and only if Tr(a 1/3 ) = 0. Thus, one siply rewrites (9) and (10) with C(a 1/3 ) = 0. When C(b) 0, for soe b F 2, then C(b) =±2 ρ with ρ = ( + 1)/2. Since 2 ρ = 8(3 1) ρ 3, we get 16 (od 24) if ρ is even, 2 ρ (12) 8 (od 24) if ρ is odd. Set = 4h + τ with τ 1, 3}. Since 2ρ = 4h + τ + 1, we have clearly ρ is odd = 4h + 1 and ρ is even = 4h + 3. (13) The next proposition coes directly fro (12) and (13). Proposition 2. Let b F 2 such that C(b) 0. Thus C(b) =±2 (+1)/2 with: 16 (od 24) if = 4h + 3, 2 (+1)/2 8 (od 24) if = 4h + 1 and 8 (od 24) if = 4h + 3, 2 (+1)/2 16 (od 24) if = 4h + 1. Now we are able to treat (9) and (10) when C(a 1/3 ) 0. Aong the next corollaries, we will prove the first only. The proof of the other is syetrically obtained. Corollary 3. Assue that a F 2 is such that Tr(a 1/3 ) 0 with C(a 1/3 ) = 2 (+1)/2. Then 0 (od 24) if = 4h + 3, Tr(a) = 0 K(a) 8 (od 24) if = 4h + 1

10 330 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) and 12 (od 24) if = 4h + 3, Tr(a) = 1 K(a) 20 (od 24) if = 4h + 1. Proof. If Tr(a) = 0 then (9) is satisfied. According to Proposition 2, we get 16 + K(a) 16 when = 4h + 3 and 8 + K(a) 16 when = 4h + 1. If Tr(a) = 1 then (10) is satisfied. Thus we get respectively 16 + K(a) 4 and 8 + K(a) 4, copleting the proof. Corollary 4. Assue that a F 2 is such that Tr(a 1/3 ) 0 with C(a 1/3 ) = 2 (+1)/2. Then 8 (od 24) if = 4h + 3, Tr(a) = 0 K(a) 0 (od 24) if = 4h + 1 and 20 (od 24) if = 4h + 3, Tr(a) = 1 K(a) 12 (od 24) if = 4h + 1. Using the terinology of Lea 3, we suarize our results as follows. Theore 3. Let a be any nonzero eleent of F 2, where is odd and 5. Then we have (1) If Tr(a 1/3 ) = 0 then (a) if Tr(a) = 0 then K(a) 16 (od 24); (b) if Tr(a) = 1 then K(a) 4 (od 24). (2) When Tr(a 1/3 ) = 1 then there is a unique β such that Tr(β) = 0 and a 1/3 = β 4 + β + 1. Hence (a) if Tr(a) = 0 then (i) if = 4h + 3 then 0 (od 24) if e(β 3 )( 2 K(a) ) = 1, 8 (od 24) if e(β 3 )( 2 ) = 1; (ii) if = 4h + 1 then 8 (od 24) if e(β 3 )( 2 K(a) ) = 1, 0 (od 24) if e(β 3 )( 2 ) = 1; (b) if Tr(a) = 1 then (i) if = 4h + 3 then 12 (od 24) if e(β 3 )( 2 K(a) ) = 1, 20 (od 24) if e(β 3 )( 2 ) = 1; (ii) if = 4h + 1 then 20 (od 24) if e(β 3 )( 2 K(a) ) = 1, 12 (od 24) if e(β 3 )( 2 ) = 1. There is a surprising consequence of the previous theore concerning the cases where K(a) is zero which can be expressed as follows.

11 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) Corollary 5. If a F 2 is such that K(a) = 0 then Tr(a) = 0 and Tr(a 1/3 ) = 1. Moreover, in this case, C(a 1/3 )>0 for = 4h + 3 and C(a 1/3 )<0 for = 4h On the [2 + 1, 2] irreducible binary code In this section, we will denote by Tl k the trace function fro F 2 k to F 2l, where l divides k. Also, we denote by γ a priitive (2 + 1)th root of unity in F 2 2 and by α a priitive root of F 2. For any Boolean function f, the weight of f is the Haing weight of the vector of its values. Let us denote by C the binary irreducible [2 +1, 2] code. The codewords of C are precisely the vectors: c u = T1 2 (u), T1 2 (uγ),...,t1 2 ( uγ 2 )}, u F2 2. (14) Since gcd(2 + 1, 2 1) = 1, each such u can be written u = α i γ j. Thus c u is a shift of c α i. Therefore, for studying the weights of C we only have to consider the weights of those c u with u F 2. It was first proved by Dillon in [8] that the weights of the code C are exactly those of the dual of the Melas code, say M. The code M is the cyclic binary [2 1, 2] code whose nonzeros are α and α 1 only. Since the words of M are described by the functions x T 1 (bx 1 + ax) on F 2, the connection between the weights of M and the Kloosteran sus is clear. The reader can see [11] for ore details and its generalization to any characteristic in [14]. We now present the result due to Dillon in our context. Note that the Haing weight of any vector v is denoted wt(v). Theore 4. [8] Let w u = wt(c u ), u F 2, where c u is defined by (14). Let us denote by λ u the weight of the function x T 1 (x 1 + ux) defined on F 2. Then w u = 2 λ u. Therefore the Kloosteran sus K(u) and the weights w u are related by: K(u) = ( 2 ) 2 2w u = ( 1) T 1 2 (uγ j ). (15) j=1 In other words, to copute the nonzero weights of C is exactly to copute the K(u), u F 2. Proof. Let us denote by G the cyclic subgroup of F 2 2 generated by γ. We have for any u F 2 and for any j [1, 2 ]: T1 2 ( uγ j ) = T1 T 2 ( uγ j ) = T1 ( ( u γ j + γ j )), since γ 2 +1 = 1. Set S =x F 2 T1 (x 1 ) = 1}. The function g fro G to F 2, defined by g(y) = y + y 1, is zero for y = 1 only and takes exactly twice each value in S. This is because, for any b, the equation y 2 + by + 1 = 0 has two solutions in the quadratic extension of F 2 if and only if T1 (1/b) = 1. Hence we have: w u = # j [ 1, 2 ] ( ( T 1 u γ j + γ j )) = 1 } = 2# x F 2 T 1 (ux) = 1 and T1 ( x 1 ) = 1 }.

12 332 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) Table 2 This table is explained at the end of Section 6 = 4h + 3 = 8s + 7 = 8s + 3 N 1 [16] ( 3)/ ( 3)/2 1 N 2 [4] ( 3)/ ( 3)/2 N 3 [0] 2 3 E/ E/8 + 2 ( 3)/2 N 4 [8] E/8 2 ( 3)/2 2 3 E/8 N 5 [12] E/8 + 2 ( 3)/2 2 3 E/8 N 6 [20] 2 3 E/ E/8 2 ( 3)/2 = 4h + 1 = 8s + 5 = 8s + 1 N 1 [16] ( 3)/ ( 3)/2 1 N 2 [4] ( 3)/ ( 3)/2 N 3 [0] 2 3 E/ E/8 2 ( 3)/2 N 4 [8] E/8 + 2 ( 3)/2 2 3 E/8 N 5 [12] E/8 2 ( 3)/2 2 3 E/8 N 6 [20] 2 3 E/ E/8 + 2 ( 3)/2 Thus the weight of x T1 (x 1 + ux) on F 2 equals 2 w u which iplies w u = 2 λ u. Now K(u) = ( 1) T 1 (x 1 +ux) = 2 2λ u. x F 2 Thus K(u) = 2 + 2w u. The proof is copleted since ( 2 ) 2 2w u = ( 1) T 1 2 (uγ j ). j=1 The weights w u of C can be coputed by eans of H(u)= ( 1) T 1 2(yu), u F 2, y G\1} and we know fro (15) that H(u) = K(u). Consequently, Theore 3, as well as Table 2, provide soe properties of the weights of the code C. Corollary 6. For each t 0 odulo 4 in the range [0, 20] we have H(u) t (od 24) K(u) 24 t (od 24). Consequently, those u such that H(u) t are described by Theore 3 and the nuber of ties they occur ( for a fixed t) is given by Table On the spectru of Kloosteran sus Now we are going to copute for each case of Theore 3 the nuber of those a which occur. Proposition 3. Let a F 2. For any set T, recall that #T is the cardinality of T. Set N 1 = # a K(a) 16 (od 24) }

13 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) and N 2 = # a K(a) 4 (od 24) }. Then ( ) 2 N 1 = ( 3)/2 1 and N 2 = N 1, where ( 2 ) is the Jacobi sybol given by (4). Proof. Fro Theore 3: N 1 = #a Tr(a) = 0 and Tr(a 1/3 ) = 0}. By setting a = b 3 (with b 0), N 1 is exactly: N 1 = # b Tr ( b 3) = 0 and Tr(b) = 0 }. (16) Siilarly N 2 is the nuber of those b such that Tr(b 3 ) = 1 and Tr(b) = 0. Since N 1 + N 2 = # b F 2 Tr(b) = 0 } = 2 1 1, then N 2 = N 1.Letf be the Boolean function b Tr(b 3 ) and denote by w the nuber of ties f(b)= 1 when Tr(b) = 0. According to (16), N 1 = 2 1 w 1 (note that we do not take b = 0). According to (5), we have C(1) = 2(2 1 2w), so that 4w = 2 C(1). Thus, since C(1) = ( 2 )2(+1)/2, we get ( ) 2 N 1 = C(1)/4 = ( 3)/2 1 which copletes the proof. Now we are going to treat the case where Tr(a 1/3 ) = 1 which is ore coplicated. We want to copute the N i as follows stated: for a F 2, we define N 3 = # a } K(a) 0 (od 24) ; (17) N 4 = # a } K(a) 8 (od 24) ; (18) N 5 = # a } K(a) 12 (od 24) ; (19) N 6 = # a } K(a) 20 (od 24). (20) Proposition 4. For any odd, we have: ( ) 2 N 3 + N 4 = ( 3)/2, ( ) 2 N 5 + N 6 = ( 3)/2, where ( 2 ) is given by (4). Moreover, ( 3)/2 if = 4h + 3, N 3 + N 5 = ( 3)/2 if = 4h + 1.

14 334 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) Proof. Notice that 6 i=3 N i = 2 1 since this su equals the cardinality of those a such that Tr(a 1/3 ) = 1. Recall that C(1) = ( 2 )2(+1)/2. Fro Theore 3, setting a = b 3 (with b 0), we have: N 3 + N 4 = # b Tr ( b 3) = 0 and Tr(b) = 1 }. Let f be the Boolean function x Tr(x 3 ) and denote by τ the nuber of x such that Tr(x) = 1 and f(x)= 1. Fro (5), we get: C(1) = 2 e ( x 3) = 2 ( 2 1 2τ ). Thus x F 2, Tr(x)=1 ( ) 2 N 3 + N 4 = 2 1 τ = C(1)/4 = ( 3)/2. Since the su of the N i is equal to 2 1, we directly deduce N 5 + N 6. Now, because C(b) 0 if and only if Tr(b) = 1, we have: } C(b) > 0if = 4h + 3 N 3 + N 5 = # b. C(b) < 0if = 4h + 1 Hence N 3 + N 5 is equal to the nuber of those b such that C(b) is positive (respectively negative). This nuber is given by Table 1, copleting the proof. Now, we are going to copute the N i defined by (17) (20). In accordance with Theore 3, each such N i depends on. We have to consider = 4h + 3 and = 4h + 1 for each i. These cases differ by the value of e(β 3 )( 2 ) which is the sign of C(a1/3 ). We will denote this sign by λ i. Thus we have, according to Theore 3(2), i 3, 5} if = 4h + 3 then λ i = 1elseλ i = 1, i 4, 6} if = 4h + 3 then λ i = 1elseλ i = 1. (21) Lea 6. For i 3, 4, 5, 6},letɛ i F 2 and λ i ±1}, where λ i is given by (21). Then N i = # β Tr(β) = 0, Tr( β 9 + β ) = ɛ i,e ( β 3)( ) } 2 = λ i, where if i 3, 4} then ɛ i = 0 else ɛ i = 1. Proof. With notation of Theore 3, we set b = a 1/3, a F 2. In accordance with the case (2) of this theore we have Tr(b) = 1 and for any i: N i = # b Tr(b) = 1, Tr( b 3) = ɛ i,e ( β 3)( ) } 2 = λ i, where β is the sole eleent such that Tr(β) = 0 and b = β 4 + β + 1; ɛ i equals 0 when i 3, 4} and 1 otherwise. Thus the value of N i, as stated above, is exactly the cardinality of the set of those β such that Tr(β) = 0, Tr((β 4 + β + 1) 3 ) = ɛ i and e(β 3 )( 2 ) = λ i. It reains to copute Tr(b 3 ):

15 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) Tr (( β 4 + β + 1 ) 3) (( = Tr β 4 + β + 1 )( β 8 + β )) = Tr ( β 3 + β 9 + β + β 3 + β 3 + β + 1 ) = Tr ( β 9 + β ). Fro Proposition 4, we know that to copute the N i given by (17) (20) is reduced to copute N 3. So, we are going to copute N 3. Proposition 5. Set E = x F 2 e(x9 + x); s is soe nonzero integer. Then 2 3 E/8 if 8s + 7, 8s + 5}, N 3 = E/8 ( 2 )2( 3)/2 if 8s + 3, 8s + 1}, where ( 2 ) is given by (4) and ±2 (+1)/2 if 0 (od 3), E = ±2 (+3)/2 otherwise. Proof. To copute N 3, defined by (17), is to copute those a corresponding to the case of Theore 3 where Tr(a 1/3 ) = 1, Tr(a) = 0, C(a 1/3 )>0if = 4h + 3 and C(a 1/3 )<0if = 4h + 1. As before, set b = a 1/3 and b = β 4 + β + 1. Thus, according to Lea 6, N 3 = # β Tr(β) = 0, Tr( β 9 + β ) = 0, e ( β 3)( ) } 2 = λ 3, where λ 3 = 1if = 4h+3 and λ 3 = 1if = 4h+1. This eans that e(β 3 ) = ( 2 ) and ( 2 ), respectively. Using (4) we ust count precisely those β such that 1 if = 4h + 3 = 8s + 7, e ( β 3) 1 if = 4h + 3 = 8s + 3, = (22) 1 if = 4h + 1 = 8s + 5, 1 if = 4h + 1 = 8s + 1, where h = 2s + 1 and h = 2s, respectively, for each for of. Then N 3 = 1 ( )( ( 1 + e(β) 1 e β 9 + β 3))( λe ( β 3) + 1 ), 8 β F 2 where λ depends on only; it is the sign of e(β 3 ) given in (22). So 8N 3 = 2 + β e(β) + λ β e ( β 3) β e ( β 9 + β 3) β e ( β 9 + β 3 + β ) λ β e ( β 9) + λ β e ( β 3 + β ) λ β e ( β 9 + β ). Using Lea 4 and Exaple 1 the sus of e(β), e(β 3 ), e(β 9 ) and e(β 9 + β 3 + β) equal zero. On the other hand, the su of e(β 9 + β 3 ) is obviously equal to those of e(β 3 + β). Thus, we get 8N 3 = 2 + (λ 1)C(1) λ β e ( β 9 + β ).

16 336 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) Table 3 Divisibility of K(a), a 0, coputed on F 2 9 K(a) (od 24) Nuber it occurs 0 N 3 = 48 4 N 2 = N 4 = N 5 = N 1 = N 6 = 64 Since C(1) = ( 2 )2(+1)/2 and replacing λ by its value given by (22), we obtain the expected expression of N 3. As recalled in Section 2.2, E = S(3, 1, 1). When b runs through F 2,wehave:S(3, 1,b) 0, ±2 (+3)/2 } if 3 divides and S(3, 1,b) 0, ±2 (+1)/2 } otherwise. Particularly S(3, 1, 1) cannot be zero. Indeed, let f be the Boolean function x Tr(x 9 + x) on F 2 and take any u F 2. Then f(x)+ f(x+ u) = Tr ( x 8 u + xu 8 + u 9 + u ) = Tr (( u u 8) x ) + Tr ( u 9 + u ). The function x f(x)+ f(x+ u) is constant if and only if u 26 + u = 0. This is possible if and only if gcd(, 6) = gcd(, 3) = 3. But in this case u 8 = u and then Tr(u 9 + u) = 0. We deduce that for any u the function x f(x)+ f(x+ u) cannot be equal to 1. Fro Lea 4, it cannot be balanced, i.e., E 0, copleting the proof. Explanation of Tables 2 and 3: The values of the N i are coputed in Table 2. This table includes the nuber of those a occurring in each case described by Theore 3. Recall that is odd and 5. The N i are precisely defined by Proposition 3, (17) (20). In Table 2, we use the next notation: N i [ν]=# a K(a) ν(od 24) }. Notation E is used for the su: E = x F 2 e(x9 + x). The knowledge of the sign of E for each kind of in Table 2 would give the precise values of the N i (see Proposition 5). As an exaple, we copute separately the N i for = 9. In this case E = x F 2 9 e(x9 +x) = 64. These results are given in Table 3 and can be checked by eans of Table 2. Note that = 8 + 1(s = 1inTable2). Acknowledgents The authors thank the anonyous reviewers who, by a nuber of coents and suggestions, contributed to iproving the anuscript. Appendix We copute here the Kloosteran sus for = 5 and = 7 but also the values of the exponential sus G and C, according to (7). For a F 2 with Tr(a 1/3 ) = 1, we denote by β one

17 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) Table 4 = 5; p(x) = x 5 + x a 1/3 Tr(a 1/3 ) a Tr(a) β g Tr(g) K C G α 0 α α 3 1 α 9 1 α 2 α α 5 1 α 15 0 α 14 α α 7 0 α α 11 1 α 2 0 α 3 α α 15 0 α Table 5 = 7; p(x) = x 7 + x a 1/3 Tr(a 1/3 ) a Tr(a) β g Tr(g) K C G α 0 α α 3 0 α α 5 0 α α 7 1 α 21 1 α 30 α α 9 0 α α 11 1 α 33 0 α 66 α α 13 0 α α 15 1 α 45 0 α 22 α α 19 1 α 57 1 α 51 α α 21 1 α 63 0 α 23 α α 23 1 α 69 1 α 38 α α 27 1 α 81 0 α 112 α α 29 0 α α 31 1 α 93 1 α 16 α α 43 0 α α 47 0 α α 55 1 α 38 1 α 52 α α 63 0 α of the two eleents in F 2 such that a 1/3 = β 4 + β + 1. In Tables 4 and 5, the results are given for a set of representatives of the cyclotoic cosets only (since it is the sae for all eleents fro such coset). The star * is given to separate the cases with Tr(a 1/3 ) = 0, i.e., when there is no β such that a 1/3 = β 4 + β + 1. For short, we use the following notation: g = β 3 + β, K = K(a), C = C(a 1/3 ) and G = G(a). The Jacobi sybol is equal to 1for = 5 and to 1 for = 7. We denote by p the priitive polynoial generating F 2. References [1] P. Caion, B. Courteau, A. Montpetit, The weight enuerators of the extreal self-dual binary codes of length 32, in: Proceedings of EUROCODE 92, in: CISM Courses and Lectures, vol. 339, Springer-Verlag, 1993, pp [2] A. Canteaut, C. Carlet, P. Charpin, C. Fontaine, On cryptographic properties of the cosets of R(1,), IEEE Trans. Infor. Theory 47 (4) (2001) [3] L. Carlitz, Explicit evaluation of certain exponential sus, Math. Scand. 44 (1979) 5 16.

18 338 P. Charpin et al. / Journal of Cobinatorial Theory, Series A 114 (2007) [4] P. Charpin, V.A. Zinoviev, On coset weight distributions of the 3-error-correcting BCH-codes, SIAM J. Discrete Math. 10 (1) (February 1997) [5] P. Charpin, T. Helleseth, V. Zinoviev, On cosets of weight 4 of binary BCH codes of length 2 ( odd), with inial distance 8, and exponential sus, Probl. Inf. Trans. 41 (4) (2005) [6] P. Charpin, T. Helleseth, V.A. Zinoviev, On binary BCH codes with inial distance 8 and Kloosteran sus, in: Proceedings of Ninth Int. Workshop on Algebraic and Cobinatorial Coding Theory, Kranevo, Bulgaria, June, 2004, pp [7] P. Charpin, T. Helleseth, V.A. Zinoviev, The coset distribution of triple-error-correcting binary priitive BCH codes, in: Proceedings 2005 IEEE International Syposiu on Inforation Theory, ISIT 05, Adelaide, Australie, Septeber [8] J.F. Dillon, Eleentary Hadaard difference sets, in: Proc. 6th S E Conf. Cobinatorics, Graph Theory, and Coputing, Congr. Nuer 14 (1975) [9] T. Helleseth, V.A. Zinoviev, On Z 4 -linear Goethals codes and Kloosteran sus, Des. Codes Cryptogr. 17 (1 3) (1999) [10] T. Helleseth, P.V. Kuar, Sequences with low correlation, in: V.S. Pless, W.C. Huffan (Eds.), Handbook of Coding Theory, Part 3: Applications, Elsevier, Asterda, 1998, R.A. Brualdi (assistant Ed.) (Chapter 21). [11] G. Lachaud, J. Wolfann, The weights of the orthogonals of the extended quadratic binary Goppa codes, IEEE Trans. Infor. Theory 36 (3) (May 1990) [12] F.J. MacWillias, N.J.A. Sloane, The Theory of Error Correcting Codes, North-Holland, [13] R.J. McEliece, Finite Fields for Coputer Scientists and Engineers, Kluwer, Boston, [14] J.C.C.M. Reijn, H.J. Tiersa, A duality theore for the weight distribution of soe cyclic codes, IEEE Trans. Infor. Theory 34 (5) (Septeber 1988)