It is found that four-symbol 6-codes of length t = 3n can be composed for odd n 0. Consequently new families of Hadamard matrices of orders 4tw and 20tw can be constructed, where w is the order of Williamson matrices. Introduction. An Hadamard matrix Hn = (hij) of order n is an n X n matrix with entries 1 or -1 such that HnHT = nIn, where In is the n X n identity matrix and T indicates the transposed matrix. In Hn, row vectors vi = (hil, hi2, ... , hin) are orthogonal, i.e. vi vj = En 1 hikhjk = 0, i $4 j. Hn exists only if n = 1, 2, or 4k. A sequence of vectors V = (vk)n (v1, v2,..., vn)4 where vk is one of m orthonormal vectors il, i2, ...i, or their negatives, is said to be an m-symbol 6-code of length n, if (I) v(j) = 0 for j $4 0, where v(j) Vk . Vk+j is the nonperiodic auto-correlation function of V. Another characterization of V = (vk)n being an m-symbol 6-code is that its associated polynomial V(z) _ Ekv=1 VkZZ,m 1 Pj(z)ij, where Pj(z) = EZn pjkzk-1, 1 0) will be called Turyn base sequences for length 2m + p (abbreviated as TBS(2m + p)) if they satisfy (3) u(j) + w(j) + x(j) + y(j) = 0 for j 7$ 0. Condition (3) is also equivalent to 1U12 + IW12 + IX12 + lYl2 = 2(2m + p) for any z E K. If TBS(2m + p): U, W; X and Y exist, then TS(2m + p) can be formed (cf. [1]) as follows: I (U' + W, 0), I (U W, 0), I (O', X + Y), and I (O', X Y), where o = 0m (the sequence of zeros of length m) and O' = 0m+p. THEOREM. Let U = (Uk)m+p, W = (Wk)m+p; X = (Xk)m and Y (Yk)m be TBS(n) for n = 2m + p. Then the following are TS(3n): 2 Q = (U+W,X+Y;O/,O;(U W)*,O), 2 R -(U W, X =Y; O' O; (U + W) O), (4) 2 1 S = -(0', 0; U + W, -(X + Y); 0', (X -Y)*) 2 1 T = (?' 0; U WI (X Y); O', (X + Y)*) 2 or Q = ((U W)*,O;U+W,X+Y;O',O), 2 R = -(-(U+ W)*,;U -W,X -Y;O',O), (5) 2 1 S =2(0', (X Y)*;O',O ; U+W w (X+ Y)) 2 1 T = (?'' (X + Y)*; O', O; U W (X Y)), 2 where A* (aN, aN-1, .. ., a,) is the reverse of A = (a,, a2, .. ., aN). LEMMA. Let a, b, c and d be polynomials with real coefficients in z E K. And lete =a+b+c, f =a -b+d,g =a -c -d, andh =b -c+d. Then IeI2 + If 12 + 1g12 + Ihi2 = 3(aI2 + IbI2 + Ic12 + IdI2) for any z E K. The Lemma can be proved easily by straightforward computations and by observing that IpI2 = pp', where p' = p(z-1) for any z E K. PROOF OF THEOREM. Let a = U, b = -zn-mX, C = z2n-my* , and d = -z2nW* in the Lemma. Then as sequences, e = (U, -X; O', -Y*), f = (U,X;O',0; W*), g = (U,0;O',Y*;W*) and h = (0, X;O',Y*; W*). Consequently g* = (W, Y; O', 0; U*) and h* = (-W, Y; O', -X*; 0'). In case (4), we have Q = (f +g*)/2, R = (f g*)/2, S = zn(e-h*)/2, and T = zn(e+h*)/2. By noting that IzI = 1 and Ip*12 p12 since Ip*(z)I = Ip(z-')I, we obtain IQI2 + 2This neat form of case (4), which contains less *'s than my original one, was suggested by R. J. Turyn. This content downloaded from 207.46.13.57 on Mon, 08 Aug 2016 04:56:54 UTC All use subject to http://about.jstor.org/terms