W= -b'e + af + cg + dh, X= a'e + bf + dg' - ch', Y = -d'e - cf+ ag' - bh, Z = e'e - df + bg + ah',

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1 PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 88, Number 4. August 1983 LAGRANGE IDENTITY FOR POLYNOMIALS AND Ô-CODES OF LENGTHS It AND 13/ C. H. YANG Abstract. It is known that application of the Lagrange identity for polynomials (see [1]) is the key to composing four-symbol S-codes of length (2.V + I); fors = 2"10''26' and odd t < 59 or t = 2J\Qe2ti^ + 1, where a, b, c. d, e and / are nonnegative integers. Applications of the Lagrange identity also lead to constructions of foursymbol S-codes of length u for u = It or 13;. Consequently, new families of Hadamard matrices of orders 4uw and 20»»' can be constructed, where w is the order of Williamson matrices. Related topics on zero correlation codes are also discussed. 1. Introduction. A sequence V = (vk)s = (vx, v2.cs) of vectors vk, where vk is one of m orthonormal vectors ix,i2,...,im or their negatives, is said to be an m-symbol 8-code of length s if v(j) = 0 for j = 0, where v(j) = 5lsk~ J\Vk vk+j is the nonperiodic autocorrelation function of V. Another characterization of V being an m-symbol 5-code of length 5 is that the associated polynomial a- m V(z)=2 vkzk-x= 2Pj(*)ij k=\ 7=1 (where /z) = Tk=lPjkzk-x, l"/=x\pjk\=l and all pm G {0,1,-1}) satisfy S"L 1 I Pj(z) 2 = s, for any z in the unit circle K = (z G C: z = 1}. When m = 4 it is convenient to set/, = (1,1,0,0), i2 = (1,-1,0,0), i3 = (0,0, 1, 1) and z4 = (0,0,1,-1) as four column vectors. All distinct i are obviously orthogonal to each other and have L \= - 2 for each Such a four-symbol ô-code of length s is called a regular 8-code of length s (abbreviated RD(s)). A sequence V = (vk)s of normalized vectors vk is called a zero correlation code of length s if v(j) Q fox j =0. A regular zero correlation code of length z (abbreviated RZC(s)) is a zero correlation code of length 5 with vectors vk = (qk, rk, sk, tk) such that \qk\ + \ rk \ + j sk. I +1 tk I = 2 for each k, where all qk, rk, sk, and tk G {0,1, -1}. We note here that I vk I = ]/2 for each k, and a regular ô-code is a regular zero correlation code in which all distinct vectors (excluding negatives) are orthogonal to each other. The following theorem is proved in [1]. Theorem 1 (Lagrange Identity for Polynomials). Let a, b, c, d, e, f, g, h be polynomials of z in K with real coefficients. Also let W= -b'e + af + cg + dh, X= a'e + bf + dg' - ch', Y = -d'e - cf+ ag' - bh, Z = e'e - df + bg + ah', Received by the editors October 21, Mathematics Subject Classification. Primary 05B20, 05A19, 62K American Mathematical Society /82/ /S01.75

2 LAGRANGE IDENTITY FOR POLYNOMIALS 747 wherep' p(z x)forp p(z). Then \W\2 +\X\2 +\Y\2 +\Z\2 = {\a\2+\b\2+\c\2 +M2)(k 2 + / 2 + g 2 + «2). Turyn base sequences for length t = 2m + p (abbreviated TBS(O) are f ur (1,-1)- sequences (A, B; CD), respectively, of lengths m + p and m, pairwise, such that A = (ak)m+p, B = (bk)m+p, C = (ck)m and D = (dk)m satisfy a(j) + b(j) + c(j) + d(j) = Ofor/^O. TBS(r) exist for odd / = 59 or t = 2"10''26'; + 1, where a, b and c are nonnegative integers (see [4, 5]). We note here that, from given TBS(r), RD(r) can be found as follows: (A,Q), (5,0), (0, C) and (0, D). 2. General results. We shall use = - = (-pk) and ' = (pn_k+x)n to represent the negative and the reverse of a sequence = (pk). It should be noted that, as polynomials, for P = P(z) = l"xpkzk-x. P' = P(z-x)^Ípkzx-k = zx"'2p^k+xzk-x 1 I Theorem 2. Let (A, B; C, D) be given TBS(r). Then the following (Q, R; S, T) is RD(7r). ô = {qk) = (Â,C;0,0;A, D; 0,0; A, C; 0,0; B',0), R = (rk) = (B,D;0,0;B,C;0,0;B,D;0,0;A',0), S= (sk) = (0,0;A,C;0,0;B,C;0,0;A,C;0,D'), = (tk) = (0,0; B, D;0,0;A, D; 0,0; B, D; 0, C). Proof. In (2), each column vector (qk, rk, sk, tk) is obviously one of /,, i2, i3, i4, or their negatives. Consequently, it suffices to prove that 2] )(z) 2 = It for any z in, where,=«? + R)/2, P2 = (ß - )/2, 3 = (5 + )/2 and 4 = (S )/2; or, equivalently, that (2') \Q\2 +\R\2 + S 2+ 2 = 14? for any z in K. First, we put a = A, b = ', ç = -C, d=-d', e = 1, /=z"2' + z2', g = (z"2' + 1 z2')zw and «= z4'-1, where M = t m = m + p, in (1). We can easily prove (2') from Theorem 1 by observing that Q = Yz2,+M, R = Z'z2,+M, S = IFz3' and T = A"z3';? ~) ~) ,1 \a\ + \b\ + M + \d\ =\ \ + \B\ +\C\ + \D\ = 2t, \e\2 =\h\2 = 1 and l/p + g 2 = 5 for any z in K. The above construction of RD(7?) is based on the fact that there are nice sequences, e = (1), f = (1,0,1), g = (1,1,-1) and h = (1), satisfying e 2 + f 2 + g 2 + h 2 = 7 for any z in K; e(z) = e(z2') and/(z) = i(z2')z~2' are symmetric, i.e., e' = e(z~]) = e and /' =/; also e + f and g = g(z2')zm"2' fit well to each other. Similarly Theorem 3 below is based on the fact that there are nice sequences,

3 748 C. H. YANG e = (1,0,1,0, 1), f = (-1,0,1,0,-1, 1) and g = (-1,1, 1, 1, 1,-1), satisfying e 2 + f 2 +1 g 2 = 13 for any z in K; g g(z')z"5'/2 is symmetric; also, e, f (defined below) and g fit well. Theorem 3. Let (A, B;C,D)bea RD(13i). (3) given TBS(r). Then the following (Q, R; S, T) is Q = (A, D';Ä,C;Ä,»'; Ä, C; Ä, D'\ A, C; 0, C; 0,0; 0,0; 0,0; 0,0; 0,0; 0,0), R = (B,C; B, D; B,C; B, D; B.C; B, D; 0, D; 0,0; 0,0; 0,0; 0,0; 0,0; 0,0), 5 = (0,0; 0,0; 0,0; 0,0; 0,0; 0,0; J,0; A, C; B',C;I,C; B',C;A,C; B',C), T= (0,0; 0,0; 0,0; 0,0; 0,0; 0,0; 73,0; B, D; A', D; B, D; A', D; B, D; A'', D). Proof. In (3) it is obvious that each (qk, rk, sk, tk) is one of the i,'s or their negatives. Therefore it is sufficient to prove that (3') ô 2 +\R\2 + S = 26? for any z in K. We put a = A, b = B, c = C, d= D, e = (z~2' + 1 +z2')z'/2~' = e(z')z"3'/2~', /= f(z')z'/2_m and g = g(z')z~5,/2 in (1). We can prove (3') without difficulty from Theorem 1 by observing that Q = -Yzs,/2, R = Zz5,/2, S= -WzX9t/2+M and = XzX9,/2 + M, where M=t-m = m+p. Golay complementary sequences of length s (abbreviated GCL(í)) are two (1,-1)- sequences of length s, f = (fk)s and g = (gk)s, having f(y') + g(y') = 0 for y = 0 or, equivalently, the associated polynomials satisfy f 2 + g 2 = 2s for any z in K (see [3]). The following theorem is somewhat similar to Theorem 2 of [1], Theorem 4. Let (A, B; C, D) be TBS(t) and ( fk)s and (gk)s GCL(i), where s>l. Then the following (Q, R, S, T) is RZC((2s + l)t). Q = (Afs,Cgi;Afs_x,Cg2;... ; Af,Cgs; 0, D; 0,0; 0,0;... ; 0,0), R = (Ags,Cfx;Ags_x,Cf2;...;Agx,Cfs;B,0;0,0;0,0;...;0.0), S = (0,0; 0,0;...; 0,0; 0, C; Bfs, Dgs; Bf_x, 7)gs_,;...; /,, Dgx), T= (0,0; 0,0;...;0,0; A,0; Bgx, Dfx; Bg2, Df2;...; Bgs, Df). Proof. In (4) each (qk, rk, sk, tk) has y/2 as its norm. First we put a = A, b = B, c= C,d= D,e = 0, r-\ r-\ f= J V l f 7(2k+\)i/2 lr + k+\z + M ' - h V J (2k+\)i/2 Kr + k+ Z ' k = -r k = -r where r = s/2, and h = zu+v),/2 in (1). By observing that Q= Wzu-\)t/i+M^ R = yz<f~"'/2, S = Az(3j+1"/2+wand = Zz(3s+1),/2, and by noting that A \2 + B 2 + C 2 + D 2 = 2t, I/)2 + g 2 = 2s and «2 = 1 for any z in K, we obtain, from Theorem 1, Iß) = {\A\ C 2 + 7) 2)( /12 + g 2 + «2) = 2í(2j + 1).

4 Let LAGRANGE IDENTITY FOR POLYNOMIALS 749 (4') Q = (A,0),R = (B,0),S = (0,C)andT= (0, D) be RD(u), where A, B, C and D are (0, ± l)-sequences such that (U, V) is a (1, -l)-sequence for U = A or B, and V- C or D.1 We note here that in Theorem 4, by replacing TBS(r): (A, B; C, D) with RD(w) of (4') and t with u in RZC, our argument is still valid. Consequently we have Theorem 4'. Let ((A,0),(B,0); (0, C),(0. D)) be RD(w) as defined in (4'). Then (4) is RZC((2s + l)u). Four (1, -1 )-sequences (,, G, 77) of length «are said to be complementary if e(j) +f(j) + g(j) + h(j) = 0 for/ = 0 or, equivalently, if j G = 4«for any z in K. We obtain such complementary sequences (,, G, 77) from Theorem 4', where E=Q+T, F=Q-T, G = R + S and H-R-S. Corollary 1. There exist four complementary (1,-1 )-sequences of length «= (2 s + l)(2g+ 1)?, where s = I^IO^ZÓ', g = 3,6 or 2í/10í'26/, and t < 59 or? = 2'Ï0>26*"+ 1; a, ft, c, a\ e, f, i, j, and k are nonnegative integers. Corollary 2. There exist Goethals-Seidel ( Hadamard ) matrices of order Ah, where h is defined in Corollary 1. Although we have not found nice (0, ± l)-sequences for composing RD(11?), RZC(lli) can be constructed as follows. Theorem 5. Let (A, B; C, D) be TBS(f). Then the following (Q, R, S, T) is RZC(llr). Q = (A,C;A,C; B', C; I, C; 0, D; 0, D; 0, D; 0,0; 0,0; 0,0; 0,0), R = (Ä,C;A,D';A,C;Ä,C;B,0;B,0;B,0;0,0;0,0;0,0;0,0), S= (0,0; 0,0; 0,0; 0,0; A,0; A,0; A,0; B, D; B,C; B, D; B, D), T= (0,0; 0,0; 0,0; 0,0; 0, C; 0, C; 0, C; B, D; B, D;A', D;B,D). Proof. In (5), each (qk, rk, sk,tk) has the same norm JÏ. It is known that the four sequences e = (1), f = (-1,0,-1,1), g = (1,-1,-1,1) and h = (1,1,1) satisfy e 2 + f 2 + g 2 + h 2 = 11 for any z in K. In (1), we let e = z'/2_l, / = f(z')z"3,/2 + M, g = g(z')z~3,/2 and «= h(z')z5'/2. Also let p = for p = a,b,c and d. By observing that Q = IFz3,/2+w, R = -Yz3,/2, S = Zzxl,/2 and = -äv7'/2 + w, we obtain from Theorem l,\q\2 + \R\2+ \S\2 + \T\2 =22t for any z in. From Theorem 5 we also obtain four complementary sequences of length 11?: E= Q + S, F= Q- S, G=7x + and77 = -. 'Eg. in RD(3rUA,0) = (A, C; 0,0; fl',0), (B,0) = (B, D; 0,0; A',0), (0,C) = (0,0; A, C; 0, D') and (0,D) = (0,0; B, D; 0,0); thus (A,C) = (A.C: A,C; B', D'), etc.

5 750 C. H. YANG Corollary 3. There exist four complementary (l,~l)-sequences of length lit. Acknowledgment. I wish to thank the referee for comments and suggestions. References 1. C H. Yang, A composition theorem for 8-codes (to appear). 2._Hadamard matrices and 8-codes of length 3n, Proc. Amer. Math. Soc. 85 (1982), _, Haáamará matrices, finite sequences, and polynomials defined on the unit circle. Math. Comp. 33(1979), R. J. Turyn, Hadamard matrices, Baumert-Hall units, four symbol sequences, pulse compression, and surface wave encodings, J. Combin. Theory Ser. A 16 (1974), _Compulation of certain Hadamard matrices. Notices Amer. Math. Soc. 20 (1973), A-l. Department of Mathematical Sciences, State University of New York, Oneonta, New York 13820

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