Weight Distributions of Hamming Codes (II)

Moisio discovered a handful of new power moments of Kloosterman sums over F q , when the characteristic of F q is 2 and 3 ([3], [4], [6], [7]). The idea is, via Pless power moment identity, to connect moments of Kloosterman sums and frequencies of weights in the binary Zetterberg code of length q +1, or those in the ternary Melas code of length q -1.

In [1], we adopted his idea of utilizing Pless power moment identity and exponential sum techniques so that we were able to derive Theorem 1 below under the restriction that (m, q -1) = 1. This restriction was needed to assume that H(m, q) is cyclic (cf. Theorem 3). It is somewhat surprising that there has been no such recursive formulas giving the weight distributions of the Hamming codes in the nonbinary cases, whereas there has been one in the binary case(cf. Theorem 2).

In this correspondence, we will give an elementary proof showing that the restriction (m, q -1) = 1 can be removed.

Theorem 1: Let {C h } n h=0 (n = (q m -1)/(q -1)) denote the weight distribution of the q-ary Hamming code H(m, q). Then, for h with

where S(h, t) denotes the Stirling number of the second kind defined by

Theorem 2 (p.129 in [2]): The author is with the Department of Mathematics, Sogang University, Seoul 121-742, Korea(e-mail: dskim@sogang.ac.kr).

H(m, 2). Then the weight distribution satisfies the following recurrence relation:

. Theorem 3 ([5]): Let n = (q m -1)/(q -1), where (m, q -1) = 1. Let γ be a primitive element of F q m . Then the cyclic code of length n with the defining zero γ q-1 is equivalent to the q-ary Hamming code H(m, q).

We know that the formula (1) holds for (m, q -1) = 1([1, Theorem 1]). By the recursive formula in (1), we see that all C i (i = 0, 1, 2, • • • , n = (q m -1)/(q -1)) are formally polynomials in q with rational coefficients, which depend on m (cf. Corollary 2 in [1] for the explicit expressions of C i for i ≤ 10). Put C i = P i (q; m), for i = 0, 1, 2, • • • , n = (q m -1)/(q -1). Then (1) can be rewritten as

(1 ≤ h ≤ n = (q m -1)/(q -1)).

Let m, h be fixed positive integers. Then the LHS and the RHS of (3) are formally polynomials in q and (3) is valid whenever q is replaced by prime powers p r satisfying (m, p r -1) = 1 and h ≤ (p rm -1)/(p r -1).

So it is enough to show that there are infinitely many prime powers p r such that (m, p r -1) = 1, since then (3) is really a polynomial identity in q, so that the restriction of our concern can be removed. There are three cases to be considered.

Case 1) 2 does not divide m. Let m = p e1 1 p e2 2 • • • p er r , where p 1 , p 2 , • • • , p r are distinct odd primes and e j ’s are positive integers. Then, by Dirichlet’s theorem on arithmetic progressions, there are infinitely many prime numbers p such that p ≡ 2(mod m). For each such an p, p ≡ 2(mod p j ), for j = 1, • • • , r. Then p j does not divide p -1, for all j, so that all p j is relatively prime to p -1. So (m, p -1) = 1, for all such primes p.

Case 2) 2 is the only prime divisor of m. In this case, 2 l -1(l = 1, 2, • • • ) are all relatively prime to m.

Case 3) 2 and some odd prime divide m.

, where e, e 1 , • • • , e r , r are positive integers and p 1 , p 2 , • • • , p r are distinct odd primes. Noting that (2, m 1 ) = 1, we let f = ord m1 2 be the order of 2 modulo m 1 . Then 2 lf ≡ 1(mod m 1 ), for all positive integers l. So 2 lf ≡ 1(mod p j ), for all j = 1, • • • , r. Thus 2 lf +1 ≡ 2(mod p j ), for all j, and hence p j does not divide