Divisor matrices and magic sequences

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1 Discrete Mathematics 250 (2002) Divisor matrices and magic sequences R.H. Jeurissen Mathematical Institute, University of Nijmegen, Toernooiveld, 6525 ED Nijmegen, Netherlands Received 6 July 1999; revised 11 September 2000; accepted 29 January 2001 Abstract Yuster (Arithmetic progressions with constant weight, Discrete Math. 224 (2000) ) denes divisor matrices and uses them to derive results on magic sequences, i.e. nite sequences a 1;a 2;:::;a n with the property that for a certain all sums j=1 ai j with i 1;i 2;:::;i an arithmetic subsequence of 1; 2;:::; n, are equal. An important condition is the (conjectured) non-singularity of the elementary divisor matrices A, that could only be proved for with at most two prime divisors. We present a proof for general, thereby generalizing the results in Yuster [1] (Arithmetic progressions with constant weight, Discrete Math., to appear.). Our exploration of A also leads to new proofs, and enables us to add other results, in particular we give the dimension of the space of -magic sequences of length n for every and n and over every eld. c 2002 Elsevier Science B.V. All rights reserved. Keywords: Magic sequences; Divisor matrices 1. Divisor matrices In the following letters apparently denoting numbers will always denote positive integers. For d let M ;d be the d matrix composed by putting =d copies of the identity matrix I d in a row, so M ;d (i; j)=1 if d i j and M ;d (i; j) = 0 otherwise. Let M be the matrix obtained by putting all M ;d with d in a column, so M is a () matrix where ()= d d;weputm ;r above M ;s if r s. Let denote the Euler function. Let N ;d be the d matrix we get from M ;d by replacing the last d (d) rows by zero rows, and let A ;d be the matrix we get from N ;d by deleting these zero rows. Let N and A be composed from the N ;d and the A ;d, respectively, d, in the same way as M is composed from the M ;d. d (d)= so A is a matrix. On the next page we show A 20. address: jeuris@sci.un.nl (R.H. Jeurissen) X/02/$ - see front matter c 2002 Elsevier Science B.V. All rights reserved. PII: S X(01)

2 126 R.H. Jeurissen / Discrete Mathematics 250 (2002) Fig. 1. A divisor matrix. We shall prove: Theorem 1. For all and over all elds the matrix A is non-singular. The proof will be preceded by some remars serving to get the picture ; it will also require a lemma of which the proof is postponed so as not to disturb the line of thought. Note that since A is a matrix of integers, its non-singularity for all characteristics is equivalent to its determinant over Z being ±1. In [1] (Theorem 1:5) the non-singularity is proved for with at most two prime divisors (Fig. 1). Also it is conjectured (and veried for 180) that the determinant of A is ( 1) +1. We extended the verication up to = 287. Clearly, the ran of M is : it contains I as a submatrix. We shall prove the theorem by showing that N is equivalent to M, by which we mean that we can get N from M by repeating operations of the following type: adding a row to a lower row (i.e. one with higher number) subtracting a row from a lower row. We shall call this operating on M ; it is the same as multiplying from the left by an integer lower triangular matrix with 1 s on the diagonal and indeed denes an equivalence relation. Remar 1. Those operations not only preserve the ran, but also leave invariant the linear span (even over Z!) of the rst t rows, for any t.

3 R.H. Jeurissen / Discrete Mathematics 250 (2002) (This is the reason why we use N as an intermedium between M and A and only in the end delete the zero rows to get A ; thus the numbering of rows is not aected.) Suppose d. WegetM r;rd from M ;d by replacing in the latter every 1 by an r r identity matrix and every 0 by an r r zero matrix, a procedure which we call: blowing up by r. Operating with r r identity or zero blocs instead of 1 s and 0 s we see: Remar 2. Ifa0; 1-matrix N is equivalent to a 0; 1-matrix M and r M and r N result from M and N, respectively, by blowing up by r, then r N is equivalent to r M. Remar 3. Suppose by operating on M we can transform its submatrix M ;d into N ;d, then by operating on M r we can transform its submatrix M r;rd into the matrix we get by replacing its lowest r(d (d)) rows by zero rows, i.e. by blowing up N ;d by r. This results in N r;rd if and only if r(d)=(rd). Since, for a prime p with pam, we have (p j+1 m)=p(p j m) if and only if j 0, we have r(d)=(rd) if (and only if) every prime factor of r also occurs in d. Proof of Theorem 1. We prove by induction on that N is equivalent to M. For = 1 (and in fact also if is a prime) this is trivial. Suppose that 1 and that N j is equivalent to M j for all j. Let d, d. The submatrix of M composed from all its submatrices M ;s, s d, is a row consisting of =d copies of M d. By induction we can, operating on M, transform this part into a row of =d copies of N d, i.e. into N ;d. We can do this for every d with d ; that a submatrix M ;s may already have been transformed to N ;s because s is also a divisor of a divisor d of is harmless by remar 1. (Since M d can be transformed such that M d;d is transformed into N d;d while the other M d;s stay unaected, alternatively one could also transform each M ;d, d, separately into N ;d, starting with the largest d ). In the end we have replaced in M every submatrix M ;d by N ;d, except for the lowest matrix M ;.Ifnow = p 2 m, p a prime, we can, see Remars 1 and 3, operate further such that its nal p(pm (pm)) = p 2 m (p 2 m) rows are replaced by zero rows, and we have got N. We have left the case that is a product of dierent primes. Then we have got a matrix N consisting of the N ;d, d, d and M ;. Let N be this matrix with M ; omitted. In the subsequent lemma we shall prove that the ran of N is (). Now the row space of every M ;d is invariant under cyclic shifts of the coordinates, and since this property is preserved by elementary row operations, the same holds for the row spaces of N and N. Let r 1;r 2 ;:::;r denote the rows of M ;. Since N has ran () we must have in M ; a row r m+1, m 6 (), that is a linear combination of the rows of N and r 1 ;r 2 ;:::;r m. Tae m minimal. Then r m+2, being the cyclic shift of row m + 1, is a linear combination of the rows of N and r 2 ;r 3 ;:::;r m+1, so also from the rows of N and r 1 ;r 2 ;:::;r m ; liewise for r m+3 ;:::;r. So preserving the ran of N we can replace all rows below r m by zero rows; apparently we must have m (). So m = (). We now now that A is non-singular, but we still have to prove that r ()+1 ;:::;r are linear combinations of the rows of N and r 1 ;r 2 ;:::;r () with integer

4 128 R.H. Jeurissen / Discrete Mathematics 250 (2002) coecients, or, equivalently, of the rows of A (its upper () rows are the non-zero rows of N ) with integer coecients. Now the right upper ( ()) ( ()) submatrix T of A is non-singular, since the right lower () ( ()) submatrix of A is a zero matrix and its left lower () () submatrix is an identity matrix. This being true over every eld we have det(t )=±1, so T denes an automorphism of the Z-module Z (), implying that, for every row r j, j (), there exists a linear combination of the rst () rows of A with integer coecients of which the dierence with r j has only non-zero (integer) components in the rst () places. This dierence is of course a linear combination with integer coecients of r 1 ;r 2 ;:::;r (). Remars. Small examples suggested that the coecients with which r ()+1 is expressed as a linear combination of the rows of A are all 0 or ±1. A chec, using MAPLE, showed that this is true for =2; 3;:::;200 except for {60; 105; 120; 140; 165; 180; 195} but even then no other coecients than 0, ±1, ±2 and ±3 occur. For up to 100 the coecients for all r j ;() j6, are0or±1, except for =60 or = 84. If A is the matrix we get from A by replacing every submatrix A ;d by da ;d and the characteristic does not divide then every row r j, () j6, is again a linear combination of the rows of A with integer coecients as is easily seen. In fact, the coecient for a row of A ;d will be a multiple of =d. Now the sum of the coecients is always 1, since the inner product of every row with the all-one vector is. Lemma 2. Let = p 1 p 2 p r ; where the p i are dierent primes. Then the submatrix of A consisting of all A ;d ;d ; d ; has ran (); over any eld. Proof. Let D be that submatrix. As we noted above the row space of D is invariant under cyclic shifts of the coordinates. We apply an argument often used in the theory of cyclic codes. Consider the set S of all polynomials c 0 +c 1 x+ +c 1 x 1 for which (c 0 ;c 1 ;:::;c 1 ) is in the row space of D, over some eld F. View these polynomials as (representatives of) elements of the ring F[x]=(x 1), i.e. operate with them mod x 1. The invariance for cyclic shifts means that S is invariant under multiplication by x, and so by any polynomial: S is an ideal in F[x]=(x 1). Then it is the image, under the canonical map F[x] F[x]=(x 1), of a principal ideal J in F[x], generated by a monic divisor g(x) ofx 1, and the dimension of S, so of the row space of D, is deg g(x). This row space is the sum of the row spaces of the M ;d, d, d, and each of them corresponds in the same way to a divisor of x 1; in fact M ;d corresponds to f d (x):=(x 1)=(x d 1), as is easily seen. So J is generated by these f d (x), and since f d (x) f h (x) ifh d, J is already generated by the f Qi, where Q i = p 1 p i 1 ˆp i p i+1 p r. Thus the generator g(x) ofj is gcd(f Q1 ;f Q2 ;:::;f Qr ). If the characteristic of F does not divide, then g(x) has, in a suitable extension of F, as zeros (with multiplicity 1) those th roots of unity that are not a Q i th root of unity for any i. So the dimension deg g(x) of the row space is the number of th

5 R.H. Jeurissen / Discrete Mathematics 250 (2002) roots of unity that are Q i th roots of unity for at least one i. By inclusion and exclusion we nd for this number p 1 ˆp i p r p 1 ˆp i ˆp j p r 16i6r + 16i j t6r 16i j6r p 1 ˆp i ˆp j ˆp t p r = p 1 p r (p 1 1)(p 2 1) (p r 1) = (): If the characteristic of F divides, say it is p 1, then for i 1 we have f Qi (x)=((x Q1 1)=(x Si 1)) p1, where S i = Q i =p 1 ; their common zeros are the primitive Q 1 th roots of unity, each with multiplicity p 1. These are also zeros of f Q1 (x)=(x Q1 1) p1 =(x Q1 1)=(x Q1 1) p1 1, with multiplicity p 1 1. So the greatest common divisor has degree (p 1 1)(Q 1 )=(). As a corollary of Theorem 1 and the cyclic properties of the row spaces we also have: Proposition 3. For every and over every eld the ( ()) submatrix of A consisting of all A ;d ;d ; d ; has ran (). Each of its submatrices consisting of () consecutive (in cyclic sense) columns is non-singular. 2. Consequences for periodic sequences Note: when we say a sequence has period d or is d-periodic this does not imply that d is its smallest period. Let d ; d, and let p be the smallest prime divisor of. Then () (1 1=p)==p d. So a sequence of period d is certainly completely determined by its rst () terms and the space of d-periodic sequences of length r, r (), is isomorphic to that of d-periodic sequences of length (). (This may seem a bit overdone; however, () is not larger than d if is a prime power p e and d = p e 1.) The rows of M ;d form a base of the linear space of sequences of length with period d. So by considering the rst () rows of A we see: Theorem 4. Let m (). Over any eld the sequences of length m with a period dividing but smaller than generate a space of dimension (). Remar 4. In the theorem one can replace eld by Z or Z m or any other commutative ring with unit element (and space and dimension by free module and ran, of course.)

6 130 R.H. Jeurissen / Discrete Mathematics 250 (2002) Corollary 5. Every sequence of length () is the sum of sequences with periods and dividing. The theorem is trivial if is a prime power. Also if is the product p r1 1 pr2 2 of two prime powers: we then have the sum of two spaces of dimension =p 1 and =p 2, respectively, with an intersection of dimension =p 1 p 2, and =p 1 + =p 2 =p 1 p 2 = (). As a further example consider the case = 30, m =30 (30) = 22. The corollary tells us that every sequence of length 22 can be written as a sum of sequences with periods 6, 10 and 15, respectively. Let S i ;i=6; 10; 15, be the subspace of sequences of period i. Then 22 = dim(s 6 + S 10 + S 15 ) = dim(s 6 + S 10 ) + dim S 15 dim((s 6 + S 10 ) S 15 ) 6 dim(s 6 + S 10 ) + dim S 15 dim((s 6 S 15 )+(S 10 S 15 )) = dim S 6 +dim S 10 +dim S 15 dim(s 6 S 10 ) dim(s 6 S 15 ) dim(s 10 S 15 ) + dim(s 6 S 10 S 15 ) = =22: Apparently (S 6 + S 10 ) S 15 =(S 6 S 15 )+(S 10 S 15 ), in other words: if the sum of two sequences of length m 22 and periods 6 and 10, respectively, has period 15, it is also the sum of two sequences of period 3 and 5, respectively. Since S 3 S 5 contains only the constant sequences, those sequences of periods 3 and 5 are uniquely determined as soon as the rst element of one of them is prescribed. The argument is easily generalized to prove: Proposition 6. Let be the product of powers of three primes p 1 ;p 2 ;p 3. If a sequence of length m () is the sum of a sequence with period =p 1 and one with period =p 2 ; and has itself period =p 3 ; then it is also the sum of a sequence with period =p 1 p 3 and one with period =p 2 p 3. These sequences are uniquely determined by prescribing the rst =p 1 p 2 p 3 elements of one of them. 3. Magic sequences (For an elucidation on the concept magic and examples of its use we refer to [1] and its references.) A nite sequence a 1 ;a 2 ;:::;a n with elements from some eld F is called -magic ( 6 n) if all subsequences a i ;a i+r ;a i+2r ;:::;a i+( 1)r, with 1 6 i and i +( 1)r 6 n, have the same sum. It is convenient to say that such a subsequence starts in i with steps r. With subsequences taing steps 1 and starting in i and i +1, respectively, we see that a i = a i+ :a-magic sequence is -periodic. For given F and n the -magic sequences form a linear space V (n; ; F) with a subspace V 0 (n; ; F)

7 R.H. Jeurissen / Discrete Mathematics 250 (2002) consisting of the sequences for which the above sums are 0. Clearly, the dimensions of these spaces are nite, those of V (n; ; F) and V 0 (n; ; F) dier by at most 1, and they are non-increasing functions of n. In what follows a sequence a 1 ;a 2 ;:::;a n will always be understood to be -periodic; (a 1 ;a 2 ;:::;a ) is its (associated) vector. We may as well admit -periodic innite sequences ( n = ). Example 1. We use the associated vector (0; 0; 0; 1; 0; 1; 0; 0; 0; 1; 0; 1) to mae a 12-periodic sequence. It is magic (sums 0) if the length is 135, but not if it is 136, unless the characteristic is 2 or 3 (see the subsequence starting at 4 with steps 12). Let p = char(f) (p = 0 admitted). If pa, then the all-one sequence belongs to V (n; ; F) but not to V 0 (n; ; F), so their dimensions dier by 1. If p and n 1+ ( 1)p, then the sequences are long enough to contain a subsequence starting at 1 with steps p. Since, by -periodicity, its sum is p(a 1 +a 1+p +a 1+2p + +a 1+ p )=0, we then have V (n; ; F)=V 0 (n; ; F). The case p and n 1+( 1)p will be dealt with later. Consider the following problems: given F and, (1) What is the minimum dimension of V (n; ; F)? (2) What is the smallest n for which V (n; ; F) has that dimension? These were raised by Yuster, who solved (1) [1, Theorem 1:1] and also (2) [1, Theorem 1:6], but the latter under the condition that A is non-singular. We shall use the information we derived on A to answer the following extension of (1) and thus also give the answer to (2) for every F: (3) How does the dimension of V (n; ; F) depend on n? Lemma 7. Let 0 d f6. Then ( 1)d + (d) 6 ( 1)f; with equality only if =2. Proof. ( 1)(f d) 1 f 1 d (d). We have equality i f = d +1;= f and d = (d), so d =1. From now on we shall use the following notations: p denotes the characteristic of the base eld F, 1=d 1 ;d 2 ;:::;d s = are the divisors of, in ascending order, n is the length of the (always -periodic) sequences a 1 ;a 2 ;:::;a n under consideration. Subsequence means subsequence of length. By the lemma ( 1)d j + (d j ) belongs to the interval [( 1)d j +1; ( 1)d j+1 ], so for every n there is a unique j, 1 6 j 6 s, such that or n =( 1)d j + u with 1 6 u (d j )

8 132 R.H. Jeurissen / Discrete Mathematics 250 (2002) n =( 1)d j + u with { (dj ) 6 u 6 ( 1)(d j+1 d j ) if j s; (d j ) 6 u if j = s (the last line is meant to include n = for an innite sequence). The next lemma is also the main tool in [1]. Lemma 8. Let S be a subsequence starting at y with steps r. Let d = gcd(r; ) and let z y (mod d); 1 6 z 6 d. Then the sum of S is d(a z + a z+d + + a z+ d ). Proof. Let x y (mod ); 1 6 x 6. Then z x (mod d) and 1 i=0 a y+ir = 1 i=0 a x+ir = 1 i=0 a z+id = d(a z + a z+d + + a z+ d ). Note that the right-hand side is d times the inner product of the associated vector (a 1 ;a 2 ;:::;a ) and row number z of N ;d. Theorem 9. Let pa. Let n =( 1)d j + u with 1 6 u 6 ( 1)(d j+1 d j ). (A) If 1 6 u (d j ); then dim V 0 (n; ; F)= j 1 i=1 (d i) u. (B) If (d j ) 6 u 6 ( 1)(d j+1 d j ); then dim V 0 (n; ; F)= j i=1 (d i). Proof. Case A: For d = d 1 ;d 2 ;:::;d j 1 and every z with 1 6 z 6 (d) there are subsequences starting at z with steps d, since then z +( 1)d 6 ( 1)d + (d) 6 ( 1)d j n, by Lemma 7. So to belong to V 0 (n; ; F) our sequence must satisfy (see Lemma 8 and note that pad): a z + a z+d + + a z+ d = 0 for 1 6 z 6 (d) and d = d 1 ;d 2 ;:::;d j 1 : (1) Moreover, if 1 6 z 6 u then z +( 1)d j 6 n, so we must also have a z + a z+dj + + a z+ dj = 0 for 1 6 z 6 u: (2) Now (1) and (2) precisely mean that the associated vector (a 1 ;a 2 ;:::;a )isinthe ernel of the matrix consisting of the rst j 1 i=1 (d i)+u rows of the non-singular matrix A. Conversely, if that is the case, it also is in the ernel of the N ;di, i =1; 2;:::;j 1 so (1) also holds with 1 6 z 6 d. But this means that all subsequences with steps d j have sum 0, by Lemma 8. Subsequences with steps d j do not exist, since 1+( 1)(d j +1)=( 1)d j + ( 1)d j + (d j ) n (In case A we have 1). For a subsequence with steps d j starting in z we must have 1 6 z 6 u. Its sum is 0 by (2). So the sequence is magic indeed and V 0 (n; ; F) is the ernel of the above matrix with columns and ran j 1 i=1 (d i)+u. Case B: As in case A we nd, since now (d j )+( 1)d j 6 n, as necessary conditions: a z + a z+d + + a z+ d = 0 for 1 6 z 6 (d) and d = d 1 ;d 2 ;:::;d j : (3)

9 R.H. Jeurissen / Discrete Mathematics 250 (2002) This means that the associated vector (a 1 ;a 2 ;:::;a ) is in the ernel of the matrix consisting of the rst j i=1 (d i) rows of A. Conversely, if that is the case, it also is in the ernel of the N ;di ;i=1; 2;:::;j, so all subsequences with steps 6 d j have sum 0. Here subsequences with steps r d j may exist, but j s and gcd(r; ) d j+1 is impossible, since even step d j+1 is too large: 1 + ( 1)d j+1 ( 1)d j+1 n. So gcd(r; )=d i with 1 6 i 6 j if j s and of course also if j = s (i.e. d j = ). So every subsequence has the same sum as some subsequence with steps d i ;i=1; 2;:::;j, so sum 0. V 0 (n; ; F) is the ernel of the above matrix with columns and ran j i=1 (d i). Corollary 10. If char(f)a; then the minimum value of dim V (n; ; F) is 1. It is attained if and only if n ( 1) + (). This complies with Theorems 1:1 and 1:6 in [1]. Example 2. Let p 2; 3 and = 6. The following table gives dim V 0 (n; 6; F) for all possible lengths n. n =6;:::;10 n =11;:::;15 n =16 n =17;:::;30 n =31 n For 17 6 n 6 30 a basis for V (n; 6; F) is given by the associated vectors (1; 1; 1; 1; 1; 1), (1; 0; 1; 1; 0; 1) and (0; 1; 1; 0; 1; 1). Example 3. Let p 2; 3; 5 and = 30. If n = 291; 292; 293 then V 0 (n; 30; F) has dimension 19; 18; 17, respectively. It has dimension 16 for n = 294; 295;:::;435 and dimension 1ifn 878. Theorem 11. Let p. Let n =( 1)d j + u with 1 6 u 6 ( 1)(d j+1 d j ). (A) If p d j ; then dim V 0 (n; ; F)= i j;pad i (d i ). (B) If pad j and 1 6 u (d j ); then dim V 0 (n; ; F)= i j;pad i (d i ) u. (C) If pad j and (d j ) 6 u 6 ( 1)(d j+1 d j ), then dim V 0 (n; ; F)= i6j;pad i (d i ). Proof. We write p r to indicate that p r but p r+1 A. Let p r and = p r t. The proof is analogous to that of Theorem 9. The dierence is that, if p d, the expression d(a z + a z+d + + a z+ d ) in Lemma 8 is always 0, so the (necessary) conditions (as in (1) and (2), or (3)) are relaxed and restricted to those d for which pad, so that are divisors of t. Now for these d the submatrices M ;d in fact consist of p r matrices M t;d in a row, so in M, lie in M t, they can be transformed to the N ;d without using the rows of submatrices M ;d with p d. As in the proof of Theorem 9 it follows that the

10 134 R.H. Jeurissen / Discrete Mathematics 250 (2002) conditions are also sucient. This explains the rans lie i j;pad i (d i ) occurring in the theorem. Corollary 12. Let p = char(f); p ; p r and = p r t. Then the minimum value of dim V 0 (n; ; F) is t. It is attained if and only if n ( 1)t + (t). Proof. From the theorem we see, in (C), that the lowest dimension of V 0 (n; ; F) is reached as soon as n =( 1)d r + (d r ), where d r is the largest divisor of not divisible by p so d r = t. The dimension then is i6r;pad i (d i )= d t (d)= t. The smallest n for which a subsequence with steps p and thus sum 0, see Lemma 8, is possible, is ( 1)p+1.Soifn ( 1)p we have V (n; ; F)=V 0 (n; ; F). Now let n =( 1)p. Consider, as in [1], the sequence with the associated vector (a 1 ;a 2 ;:::;a ) where a i = 1 for 1 6 i 6 t, a i = 0 for i t. The sums of its subsequences are the d(a z + a z+d + + a z+ d ) with 1 6 z 6 d and d ; d p, by Lemma 8. But then d t and in a z + a z+d + + a z+ d precisely the rst t=d indices are 6 t. So all sums are equal to t and that sequence is in V (n; ; F) V 0 (n; ; F). One could also argue as follows. The submatrices A ;d of A with d, d p, together form a matrix with an independent set of rows; this is still the case if we replace every A ;d by da ;d. Let r be the number of rows. The matrix represents a surjective linear map from F onto F r, so there certainly is a vector (a 1 ;a 2 ;:::;a ) that has the all-one vector as its image. By Lemma 8 all subsequences of the corresponding sequence with length n 6 ( 1)p have sum 1, so that sequence is in V (n; ; F) V 0 (n; ; F). We have: Lemma 13. V (n; ; F)=V 0 (n; ; F) if and only if n ( 1)p. Comparing with the bound ( 1)t + (t) of Corollary 12 and noting that ( 1)p ( 1)t + (t) if p t and ( 1)p ( 1)t + (t) if p t (Lemma 7) we conclude: Theorem 14. Let p = char(f); p ; p r and = p r t. Then the minimum dimension of V (n; ; F) is t. If p t it is attained for n =( 1)t + (t). If p t it is attained for n =( 1)p +1. This also complies with Theorems 1:1 and 1:6 in [1]. Example 4. Let =6;p=2, so t = 3 and p t. V (n; 6; F 2 )=V 0 (n; 6; F 2 )in 11. The minimum of dim V (n; 6; F 2 ) is 3, it is attained for n = = 17. A basis (of associated vectors) then is {(1; 0; 0; 1; 0; 0); (0; 1; 0; 0; 1; 0); (0; 0; 1; 0; 0; 1)}. If we let =6, p =3, so t = 2 and p t the smallest n for which V (n; 6; F 3 )= V 0 (n; 6; F 3 ) and dim V (n; 6; F 3 ) taes its minimum value 4 is = 16. A basis then is {(1; 0; 0; 0; 1; 0); (0; 1; 0; 0; 0; 1); (0; 0; 1; 0; 1; 0); (0; 0; 0; 1; 0; 1)}.

11 R.H. Jeurissen / Discrete Mathematics 250 (2002) Example 5. If = 20;p= 2, then t = 5 and p t. The minimum dimension 15 of V (n; 20; F 2 ) is attained for n = 99, and V = V 0 from n = 39 on. If = 20, p = 5, then t = 4 and p t. The minimum dimension 16 of V 0 (n; 20; F 5 ) is attained for n = 78, but dim V (n; 20; F 5 ) = 17 for 78 6 n Example 6. Let = 30. Let p = 2; 3; 5. Then V (n; 30; F) has minimum dimension 15; 20; 24, respectively, and it is reached for n = 443; 294; 176, respectively. If p 2; 3; 5 then the minimum dimension 1 is reached for n = 878. References [1] R. Yuster, Arithmetic progressions with constant weight, Discrete Math. 224 (2000)

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