Complete (q 2 + q + 8)/2-caps in the spaces P G(3, q), q ≡ 2 (mod 3) odd prime, and a complete 20-cap in P G(3, 5) Alexander A. Davydov · Stefano Marcugini · Fernanda Pambianco Abstract An infinite family of complete (q 2 + q + 8)/2-caps is constructed in P G(3, q) where q is an odd prime ≡ 2 ( mod 3), q ≥ 11. This yields a new lower bound on the second largest size of complete caps. A variant of our construction also produces one of the two previously known complete 20-caps in P G(3, 5). The associated code weight distribution and other combinatorial properties of the new (q 2 +q+8)/2-caps and the 20-cap in P G(3, 5) are investigated. The updated table of the known sizes of the complete caps in P G(3, q) is given. As a byproduct, we have found that the unique complete 14-arc in P G(2, 17) contains 10 points on a conic. Actually, this shows that an earlier general result dating back to the Seventies fails for q = 17. Keywords
Complete caps; Projective spaces of the dimension three; Projective planes
AMS Classification
51E21 · 51E22 · 94B05
A. A. Davydov (⊠) Institute for Information Transmission Problems, Russian Academy of Sciences, Bol’shoi Karetnyi per. 19, GSP-4, Moscow, 127994, Russian Federation e-mail:
[email protected] S. Marcugini
· F. Pambianco
Dipartimento di Matematica e Informatica, Universit`a degli Studi di Perugia, Via Vanvitelli 1, Perugia, 06123, Italy e-mail:
[email protected] F. Pambianco e-mail:
[email protected]
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1
Introduction
Let P G(n, q) be the projective space of dimension n over the Galois field Fq and let Fq∗ = Fq \ {0}. A k-cap in P G(n, q) is a set of k points, no three of which are collinear. A k-cap in P G(n, q) is called complete if it is not contained in a (k + 1)-cap of P G(n, q). If n = 2, then a k-cap is usually called a k-arc. A survey of results on caps and arcs is found in [16],[17], see also [18],[25],[27]. The main questions on caps in P G(n, q) are also of interest in Coding theory and are about the size of very large caps, especially about m2 (n, q), the size of the largest complete cap, and about m′2 (n, q), the size of the second largest complete cap. On the other hand, very small complete caps have also been investigated, especially the problem of determining t2 (n, q), the size of the smallest complete cap. The known constructions of caps mostly start with a given cap of size Tq = (q 2 +q+4)/2 containing all but just one point from an elliptic quadric of P G(3, q). As a matter of fact, Tq is an important number in the study of complete caps since any cap containing at least Tq points from an elliptic quadric is entirely contained in it. This was first pointed out by B. Segre in [25], see also [1],[17],[23], and gave a motivation for the study of Mq , the maximum size of complete k-caps other than elliptic quadrics containing as many as Tq − 1 points of an elliptic quadric. It is well known that m2 (3, q) = q 2 + 1 for all q ≥ 3, but the exact value of m′2 (3, q) has been determined so far only for q ≤ 7, namely m′2 (3, 3) = T3 , m′2 (3, 4) = T4 + 2, m′2 (3, 5) = T5 + 3, m′2 (3, 7) = T7 + 2, see [3],[11],[12],[18, Table 4.2(i)]. For q ≥ 8, the best known upper bound is approximately cq 2 with c = 1 for q > 2 even and for q odd but not prime, while c = 2641/2700 for q odd prime, see [1],[8],[13]-[15],[17],[18, Theorem 4.2,Table 4.2(ii)],[23]-[25],[27]. These and other results give a motivation for the construction of caps whose size is close to m′2 (3, q) which is the subject of the present paper. Our main result is the construction of an infinite family of complete caps of size Tq + 2 for all q ≥ 11 odd prime such that q ≡ 2 (mod 3). This gives an improvement on the best previous bound on m′2 (3, q): (q 2 + q + 8)/2 = Tq + 2 ≤ m′2 (3, q) for odd prime q ≡ 2 (mod 3), q ≥ 11.
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Furthermore, a variant of our construction produces one of the two complete m′2 (3, 5)caps in P G(3, 5). This is of interest, since m′2 (3, 5) = 20 and there exist only two projec2
tively non-equivalent complete 20-caps in P G(3, 5), see [3]. The points of both complete m′2 (3, 5)-caps in P G(3, 5) are listed in [3], and we give a synthetic construction for one of them. The (Tq + 2)-caps produced by our construction share Tq − 1 points with an elliptic quadric. Therefore, (q 2 + q + 8)/2 = Tq + 2 ≤ Mq for odd prime q ≡ 2 (mod 3).
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The known bounds for these q are Tq + 1 ≤ Mq ≤ Tq + q + 1 [25, Theorem V, p. 73]. For q = 5, from the above mentioned construction of 20-caps we obtain the exact value M5 = T5 + 3 = 20.
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For q odd, the best known lower bound on m′2 (3, q) is Tq +1, and is known to be attained for infinite values of q. From [23, Theorems IV,V], there exist complete k-caps for k = Tq + 1, q ≡ 1 (mod 4) or q = 2p − 1, p prime. and for Tq + 1 ≤ k, q odd, q 6= 2r − 1. Also, in [25, Theorem V, p. 73] complete k-caps in P G(3, q) are constructed for Tq + 1 ≤ k ≤ Tq + q + 1, q odd, q ≡ 2 (mod 3) or q = 3h ≥ 9. Subsets of our (Tq + 2)-caps can be adopted as a starter for the greedy algorithms often used for producing complete caps, see [8, Section 2]. This has allowed us to obtain many new entries in the spectrum of sizes of complete caps in P G(3, q), q ≤ 23, and to update the table of the known complete cap sizes in P G(3, q) as given in [8]. The essential idea of our construction of (Tq + 2)-caps, as many other known constructions, goes back to Segre, see [25, Theorem V, p. 73], see also [17, Section 18.2]. Given an elliptic quadric Q, in P G(3, q), our (Tq + 2)-cap comprises a point P 6∈ Q, the q + 1 common points of Q with the polar plane of P together with one point from each bisecant through P , chosen one of its two common points with Q. Segre proved that either the resulting Tq -cap K is complete or K can be completed by adding to it at most q + 1 points. Since the former case occurs with probability 1, a careful special choice of the points from the bisecants is needed in order to avoid that K be itself complete. Here we present such
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a special choice that allows us to add to K two more points obtaining a complete (Tq + 2)cap. Although our procedure for the choice is not of geometric nature, it gives a good geometrical result. The points of a k-cap K in P G(n, q) can be viewed as columns of the generating matrix of the associated code A(K) with length k and dimension n + 1. Taking into consideration the well known connection between secant hyperplanes of caps and the weight distribution of the associated code, [5, Theorem 16.1],[10],[15, Theorem 4.1],[20], our study of the code associated to the above (Tq + 2)-cap K is carried out using the combinatorial properties of K, the orbit structure of the collineation group preserving K, the plane sections of K, and the conics lying in K. Considering secant planes of our (T17 + 2)-cap we point out that the unique complete 14-arc of P G(2, 17) must contain 10 points from a conic. Actually, this disproves a result of [22, Theorem 1] where it is claimed that for q ≡ 1 (mod 4) in the plane P G(2, q) a complete arc which has (q + 3)/2 points on a conic can contain at most two points outside the conic. Some results of this work was presented without proofs in [9].
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An infinite family of (q 2 + q + 8)/2-caps in P G(3, q) with odd prime q ≡ 2 (mod 3)
Let q be odd prime, q ≡ 2 (mod 3). Since Fq is isomorphic to Zq , the elements of Fq are identified by the integers 0, 1, . . . , q − 1 and the operations in quantitative relations are carried out modulo q. Our aim is to construct an infinite family of (Tq + 2)-caps in P G(3, q) containing Tq − 1 points from an elliptic quadric. Let (x0 , x1 , x2 , x3 ) be a point of P G(3, q) with xi ∈ Fq , x0 ∈ {0, 1}. Choose a projective frame in P G(3, q) such that the elliptic quadric Q has equation x22 − αx21 − x0 x3 = 0, α ∈ Fq , α is a non square.
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Remark 1. In P G(3, q) through every point V 6∈ Q there are q + 1 tangents to Q and q(q−1)/2 bisecants of Q, see [17],[25]. We denote by πV the polar plane of the point V . The intersection of Q and πV is the conic CV = Q∩πV which is also the set of the tangency points of the tangents through V . For a point D = (d0 , d1, d2 , d3 ) with d0 ∈ {0, 1}, d1 , d2 , d3 ∈ Fq , 4
the equation of the polar plane πD can be written as Fx0 (D)x0 + Fx1 (D)x1 + Fx2 (D)x2 + Fx3 (D)x3 = 0 where Fxi is the corresponding partial derivative of the polynomial associated to the quadric. Our construction can be viewed as a variant of Segre’s construction recalled in Introduction. Construction A1. Let q be an odd prime, q ≡ 2 (mod 3). Then −3 is a non square element in Fq [16, Section 1.5]. In P G(3, q) a (Tq + 2)-set K having Tq − 1 points on the quadric Q of equation (4) is obtained as follows. Take the basic point P = (0, 1, x2 , x3 ) ∈ /Q √ where x2 = ± −3α and x3 is an arbitrary element of Fq . Every bisecant through P contains ′ ′ ′ ′ ′′ ′′ ′′ ′′ ′ ′′ two points from Q, namely A = (1, x1 , x2 , x3 ) and A = (1, x1 , x2 , x3 ). Here x1 6= x1 , and ′ ′′ ′ ′′ ′ ′′ we choose either A or A according as x1 < x1 or x1 > x1 . Following Segre’s construction, the set of size Tq comprising P , the polar conic πP and the above chosen points on the bisecants is a Tq -cap, let be K1 . Such a cap turns out to be incomplete as both points P1 = (0, 1, 0, 0), and P2 = (0, 1, −x2 , 4α − x3 ) can be added to obtain a (Tq + 2)-cap. The latter claim is our Theorem 1. It may be noted that the points P and P2 are dual to each other in the sense that P2 can be used as the basic point and then P plays the role of the point P2 . Although P is not uniquely taken, the (Tq + 2)-sets produced by Construction A1 seem to be projectively equivalent. This has been proven so far when 5 ≤ q ≤ 29 by a computer aided exhaustive search. Proposition 1. Let 5 ≤ q ≤ 29. For all non square elements α in (4) and for all x3 , the resulting (Tq + 2)-sets by Construction A1 are projectively equivalent. This suggests to specify the condition of the choice of P in Construction A1 by setting α = −3 in the quadric equation (4), and choosing P as follows. Condition 1. Let q be an odd prime and let q ≡ 2 (mod 3). Let the quadric Q be given by the equation 3x21 + x22 = x0 x3 . (5) √ Take x2 = −3α = 3 and x3 = 0 in the coordinates of P , so that the points outside Q in Construction A1 are P = (0, 1, 3, 0), P1 = (0, 1, 0, 0), and P2 = (0, 1, −3, −12). For the quadric of equation (5), the polar plane πD of a point D has equation πD :
−d3 x0 + 6d1 x1 + 2d2 x2 − d0 x3 = 0, 5
D = (d0 , d1 , d2 , d3 ).
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A straightforward computation proves the following result. Lemma 1. Under Condition 1, the polar planes πP , πP1 , and πP2 intersect the quadric Q of (5) in the conics CP , CP1 , and CP2 , respectively, CP = {(1, v, −v, 4v 2)|v ∈ Fq } ∪ {(0, 0, 0, 1)};
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2
CP1 = {(1, 0, d, d )|d ∈ Fq } ∪ {(0, 0, 0, 1)};
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2
CP2 = {(1, v, v + 2, 4v + 4v + 4)|v ∈ Fq } ∪ {(0, 0, 0, 1)}.
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Lemma 2. Under Condition 1, the Tq -cap K1 in Construction A1 is as follows: K1 = {(1, v, d, 3v 2 + d2 )| v ∈ Fq , d = v − 2i, i ∈ {v, v + 1, . . . , q − 1}} ∪ {A∞ , P } = {(1, v, d, 3v 2 + d2 )| v ∈ Fq , d ∈ {−v, −v − 2, . . . , v + 4, v + 2}} ∪ {A∞ , P }
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where A∞ = (0, 0, 0, 1), P = (0, 1, 3, 0), and the list of d values contains q − v terms. Proof. If the points A′ = (1, f, g, 3f 2 + g 2) ∈ Q and A′′ = (1, v, d, 3v 2 + d2 ) ∈ Q are on a bisecant through P = (0, 1, 3, 0), then u−1 (1, f, g, 3f 2 + g 2) − u−1 (1, v, d, 3v 2 + d2 ) = (0, 1, 3, 0), u ∈ Fq∗ , whence f − v = u;
g − d = 3u;
3f 2 + g 2 = 3v 2 + d2 .
From this v − d = 2f. If f > v then the point A′′ = (1, v, d, 3v 2 + d2 ) is in the cap K1 . So, by our procedure the points A∗ of K1 on the bisecants are A∗ = (1, v, d, 3v 2 + d2 ), v − d = 2i, i = v + 1, v + 2, . . . , q − 1. By (7), for points of CP we have v − d = 2v. Together with (11) it gives (10). Example 1. Let q = 5. By (10), the points of the T5 -cap K1 have the form v 0 0 0 0 0 1
i 0 1 2 3 4 1
d 0 3 1 4 2 4
point type v i d point 1000 CP 1 2 2 1122 1034 A∗ 1 3 0 1103 1011 A∗ 1 4 3 1132 1041 A∗ 2 2 3 1231 1024 A∗ 2 3 1 1213 1144 CP 2 4 4 1243
type A∗ A∗ A∗ CP A∗ A∗
Corollary 1. Condition 1 implies CP ∪ CP1 ∪ CP2 ⊂ K1 . 6
v 3 3 4
i 3 4 4
d 2 0 1
point type 1321 CP 1302 A∗ 1414 CP 0001 CP (A∞ ) 0130 P
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Proof. This follows from the definition of Construction A1 and from equations (8)-(10). Lemma 3. Condition 1 implies the point P1 = (0, 1, 0, 0) does not lie on any bisecant of K1 . Proof. If the points D ′ = (1, v, d, 3v 2 + d2 ) ∈ Q and D ′′ = (1, m, n, 3m2 + n2 ) ∈ Q lie on a bisecant of Q through P1 = (0, 1, 0, 0), then u−1 (1, v, d, 3v 2 + d2 ) − u−1 (1, m, n, 3m2 + n2 ) = (0, 1, 0, 0), v, d, m, n ∈ Fq , u ∈ Fq∗ , whence v − m = u;
3v 2 + d2 = 3m2 + n2 .
d = n;
From this v = −m, u = 2v, v 6= 0. So, P1 lies on the bisecant through the points D ′ and D ′′ where D ′ = (1, v, d, 3v 2 + d2 ) ∈ Q,
D ′′ = (1, −v, d, 3v 2 + d2 ) ∈ Q,
v 6= 0.
The points D ′ and D ′′ do not simultaneously belong to K1 . Let D ′ ∈ K1 . Then, by (10), d ∈ {−v, −v − 2, . . . , v + 4, v + 2}. If D ′′ ∈ K1 then d ∈ {v, v − 2, . . . , −v + 4, −v + 2}. But it is a contradiction since they are complementary subsets of Fq . Example 2. Let q = 11 and let D ′ = (1, 6, d, 9 + d2 ) ∈ K1 . Then d ∈ {5, 3, 1, −1, −3}. If D ′′ = (1, −6, d, 9 + d2 ) ∈ K1 then d ∈ {6, 4, 2, 0, −2, −4}. It is a contradiction. Lemma 4. Condition 1 implies the point P2 = (0, 1, −3, −12) does not lie on any bisecant of K1 . Proof. If the points V ′ = (1, v, d, 3v 2 + d2 ) ∈ Q and V ′′ = (1, m, n, 3m2 + n2 ) ∈ Q lie on a bisecant of Q through P2 = (0, 1, −3, −12), then u−1 (1, v, d, 3v 2 + d2 ) − u−1 (1, m, n, 3m2 + n2 ) = (0, 1, −3, −12), v, d, m, n ∈ Fq , u ∈ Fq∗ , whence v − m = u;
d − n = −3u;
3v 2 + d2 − 3m2 − n2 = −12u.
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Taking u 6= 0 into account, from (12) we obtain v − d = 2u − 2 6= −2,
m − n = −2u − 2 6= −2.
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We show that the points V ′ and V ′′ do not simultaneously belong to K1 . Let {V ′ , V ′′ } ⊂ K1 . Then, by (10), v − d ∈ {2v, 2(v + 1), . . . , 2(q − 2)}, v 6= q − 1.
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The number 2(q − 1) is not included in (14) as v − d 6= −2. By the same reason, v 6= q − 1 otherwise 2v = −2 and v − d ∈ {−2, 0, . . .}. Similarly, m − n ∈ {2m, 2(m + 1), . . . , 2(q − 2)}, m 6= q − 1.
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By (12), m = v − u. As m 6= q − 1 we have u 6= v + 1. Now, by ( 13) and (14), u=
v−d + 1, 2
u ∈ {v + 2, v + 3, . . . , q − 1}.
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From (16), u = v + ∆,
2 ≤ ∆ ≤ q − 1 − v.
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If v = q − 2 then the set {2v, 2(v + 1), . . . , 2(q − 2)} of (14) consists of the only element 2(q − 2), whence v − d = 2(q − 2). It implies u = v−d + 1 = (q − 2) + 1 = v + 1 that 2 contradicts to the second relation of (16). So, v 6= q − 2. Henceforth we represent v as a nonnegative integer, 0 ≤ v ≤ q − 3. From (12) and (17), m = v − u = −∆. Now, by (15), (m − n)/2 + ∆ ∈ {0, 1, . . . , ∆ − 2}. In the other hand, (m − n)/2 + ∆ = −u − 1 + ∆ = −v − 1, see (13),(17). Therefore −v−1 ∈ {0, 1, . . . , ∆−2}. As v ∈ / {q−1, q−2}, it holds that v ∈ {q−3, q−4, . . . , q+1−∆}. So, v ≥ q + 1 − ∆. But ∆ ≤ q − 1 − v, see (17). So, v ≥ v + 2; a contradiction. Example 3. Let q = 17, v = 4. By (14), v − d ∈ {8, 10, 12, . . . , 30}. By (16), u ∈ {6, . . . , 16}. By (17), u = v + ∆, 2 ≤ ∆ ≤ 12. As v ∈ {q − 3, q − 4, . . . , q + 1 − ∆}, we have v ∈ {14, 13, . . . , 18 − ∆}. So, v ≥ 18 − ∆ ≥ 6. It is a contradiction to v = 4. Theorem 1. Under Condition 1, where q is odd prime and q ≡ 2 (mod 3), the set K in Construction A1 is a (q 2 + q + 8)/2-cap in P G(3, q). Proof. By Lemmas 3 and 4, the points P1 and P2 are uncovered by bisecants of the cap K1 . In addition, the line P1 P2 is external to the quadric Q and to the cap K1 . Corollary 2. For odd prime q ≡ 2 (mod 3) the value (q 2 + q + 8)/2 is a lower bound on the second greatest size for the complete caps in P G(3, q).
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Remark 2. In Construction A1 the points P1 and P2, as well as the point P, can be regarded as a basic point for the (Tq + 2)-cap K, since every point off the elliptic quadric Q lies on some q + 1 tangents to Q and on q(q − 1)/2 bisecants of Q, see Remark 1. Moreover, it can be shown that, under Condition 1, using the same procedure for choice of the points A∗ in bisecants of Q with respect to the basic point P2 we obtain the same cap K1 as from the basic point P. We remind on the duality of the points P and P2 pointed out in the description of Construction A1. However, for the basic point P1 we need another procedure for the choosing of the points in the bisecants.
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Completeness of the (q 2 + q + 8)/2-caps in P G(3, q)
We start with the following result. Lemma 5. Let Π∞ be the tangent plane to the quadric Q with equation x0 = 0. Under Condition 1, all points of Π∞ outside the (Tq +2)-cap K of Construction A1 lie on bisecants of K. Proof. We relay on [25, Theorem V, p. 73, Proof],[17, Theorem 18.2.4, Proof ], and use Corollary 1. Assume that on Π∞ there are points P3 , P4,..., P∆ ∈ / Q which are not covered by bisecants of K. The polar plane of every such point, as well as the polar planes of P, P1, and S P2 , meet the quadric Q in a conic entirely contained in the cap K1 , i.e., CP ∪ ∆ i=1 CPi ⊂ K1 . By Construction A1, |Q ∩ K1 | = (q + 1) + q(q − 1)/2. If the conic CPi is not entirely contained in K1 then at least one bisecant of Q through Pi must contain two points on K1 , see Remark 1. From [25] and [17], every point of K1 belongs to at most three conics mentioned above. In particular, this yields that any tangent plane Π contains at most three points of the complete cap K which are not on Q, see [1, p. 7], as the tangency point of Π is contained in each conics cut out by the polar planes of other points of Π. We have {P, P1 , P2 } ⊂ Π∞ and the tangency point is A∞ = (0, 0, 0, 1) ∈ CP ∩ CP1 ∩ CP2 , see Lemma 1. Now consider points in the affine space AG(3, q) = P G(3, q) \ Π∞ with x0 = 1. Let Q = Q \ (K ∩ Q) be the set of points of the quadric Q not contained in the cap K. Let S ⊂ P G(3, q) be a cap. We say that a point H ∈ / S is covered by S, if H lies on a bisecant of S. Interestingly, if a point H ∈ / K ∪ Q is covered by Q then this point is covered by K too. This will be shown later in Lemma 6. Condition A plays a central role in the 9
proof of completeness of K. Condition B relays on Condition A and in turn Conditions C, D, and E are based on B. Hereinafter the following facts [16],[17],[25] are useful. In P G(3, q), q odd, every plane Φ which is not a tangent plane meets an elliptic quadric in some conic C(Φ). Every point of Φ \ C(Φ) is covered by any subset of (q + 5)/2 points of C(Φ). Lemma 6. Let R = {(q − 1)/2, (q + 1)/2, (q + 3)/2}. Under Condition 1, a point H = (1, b, c, d) ∈ / K ∪ Q is covered by bisecants of the cap K if any of the following sufficient conditions holds: A) The point H lies on a bisecant H 1 H 2 through the points H 1 , H 2 ∈ Q. B) The point H lies on a plane Φ meeting the quadric Q in a conic C(Φ) with |C(Φ) ∩ K1 | ∈ / R. C) b ∈ / R. D) For q ≥ 11, c − b ∈ / {1, 3, 5}. E) c + b ∈ / {1, q − 3, q − 1}. Proof. A) The point H lies on q + 1 tangents and q(q − 1)/2 bisecants of Q, see Remark 1. By Construction A1, |Q ∩ K1 | = (q + 1) + q(q − 1)/2. Hence, we prove that if Condition A holds, H lies on at least one bisecant of Q containing two points of K1 . Let ti and si denote, respectively, the number of tangents and bisecants to Q through H containing i points of K1 , i ≥ 0. Obviously, t0 + t1 = q + 1, s0 + s1 + s2 = q(q − 1)/2, t1 + s1 + 2s2 = |Q ∩ K1 | = (q + 1) + q(q − 1)/2. If s2 > 0 the point H is covered by K. Condition A implies s0 > 0. Hence, s1 < q(q − 1)/2, t1 + 2s2 > q + 1, and s2 > 0. B) As C(Φ) ⊂ Q ⊂ K ∪ Q we have H ∈ Φ \ C(Φ). Condition |C(Φ) ∩ K1 | ∈ / R means either |C(Φ) ∩ K1 | ≥ (q + 5)/2 or |C(Φ) ∩ K1 | ≤ (q − 3)/2. If |C(Φ) ∩ K1 | ≥ (q + 5)/2 all points of Φ \ C(Φ) are covered by C(Φ) ∩ K1 . If |C(Φ) ∩ K1 | ≤ (q − 3)/2 then |C(Φ) ∩ Q| ≥ (q + 5)/2 and all points of Φ \ C(Φ) are covered by C(Φ) ∩ Q. So, we may use Condition A. C) We consider the plane Φb with equation x1 = bx0 . Obviously, H ∈ Φb . The plane Φb meets Q in the conic C(Φb ) = {(1, b, d, 3b2 + d2 )| d ∈ Fq } ∪ A∞ with A∞ = (0, 0, 0, 1). By (10), |C(Φb ) ∩ K1 | = q − b + 1. If b ∈ / R then q − b + 1 ∈ / R and we may use Condition B. ∆ D) Let Φ− be the plane with equation x2 − x1 = ∆x0 , ∆ = c − b. Clearly, H ∈ Φ∆ − . The ∆ ∆ 2 2 plane Φ− intersects Q in the conic C(Φ− ) = {(1, v, d, 3v + d )|v ∈ Fq , d − v = ∆} ∪ A∞ . Let q ≥ 11. We show that |C(Φ∆ / R if ∆ ∈ / {1, 3, 5}. − ) ∩ K1 | ∈ The relation (10) can be rewritten as follows. K1 = {(1, 0, d, d2)|d ∈ Fq } ∪ {(1, v, d, 3v 2 + d2 )|v ∈ Fq∗ , d − v ∈ Mv } ∪ {A∞ , P }, 10
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Mv =
(
{2, 4, 6, . . . , q − 1} ∪ {1, 3, 5, . . . , q − 2v} if 1 ≤ v ≤ (q − 1)/2 . {2, 4, 6, . . . , 2(q − v)} if (q + 1)/2 ≤ v ≤ q − 1
Really, by (10), d − v = −2i ∈ {−2v, −2v − 2, −2v − 4, . . . , 4, 2}. If v = 0 then d ∈ {0} ∪ {q − 2, q − 4, . . . , 1} ∪ {q − 1, q − 3, . . . , 2} = Fq . If 1 ≤ v ≤ (q − 1)/2 then 2v < q, q − 2v is odd, and d − v ∈ {q − 2v, q − 2v − 2, . . . , 1} ∪ {q − 1, q − 3, . . . , 2}. Finally, if (q +1)/2 ≤ v ≤ q −1 then 2v > q, 2q −2v is even, and d−v ∈ {2q −2v, 2q −2v −2, . . . , 4, 2}. Let q−1 ≥ d−v = 2t ≥ 2. Then d−v ∈ Mv if 1 ≤ v ≤ (q−1)/2 or 2(q−v) ≥ 2t. The last inequality gives v ≤ q − t. As q − 1 ≥ 2t, we have (q + 1)/2 ≤ q − t. Therefore d − v ∈ Mv if and only if 1 ≤ v ≤ q −t. Also, by (18), (1, 0, 2t, 4t2) ∈ K1 . So, the cap K1 contains q −t+ 1 points (1, v, d, 3v 2 + d2) with d − v = 2t ≥ 2. Hence, |C(Φ2t − ) ∩ K1 | = (q − t + 1) + 1 where the last summand ”+1” comes from A∞ . As 2t ≤ q−1, it follows that |C(Φ2t − )∩ K1 | ≥ (q+5)/2. Let q−2 ≥ d−v = 2u−1. Then d−v ∈ Mv if and only if 1 ≤ v ≤ (q+1)/2−u that implies q − 2v ≥ 2u − 1. Also, (1, 0, 2u − 1, (2u − 1)2) ∈ K1 . So, K1 contains (q + 1)/2 − u + 1 points 2u−1 (1, v, d, 3v 2 + d2 ) with d −v = 2u −1. Hence, |C(Φ− ) ∩K1 | = (q + 3)/2 −u + 1 where ”+1” is connected with A∞ . If d − v ∈ / {1, 3, 5} then 2u − 1 ≥ 7 and |C(Φ2u−1 ) ∩ K1 | ≤ (q − 3)/2. − Finally, d − v = 0 only for the point (1, 0, 0, 0). So, |C(Φ0− ) ∩ K1 | = 1 + 1 < (q − 3)/2. We have shown that |C(Φ∆ / R if ∆ ∈ / {1, 3, 5}. This allows us to use Condition B. − )∩K1 | ∈ E) The proof is similar to that for Condition D. Let Φδ+ be the plane with equation x2 + x1 = δx0 where δ = c + b. We have H ∈ Φδ+ , C(Φδ+ ) = {(1, v, d, 3v 2 + d2 )|v ∈ Fq , d + v = δ} ∪ A∞ . The relation (10) can be rewritten as follows. K1 = {(1, 0, d, d2)|d ∈ Fq } ∪ {(1, v, d, 3v 2 + d2 )|v ∈ Fq∗ , d + v ∈ Tv } ∪ {A∞ , P }, (19) {0} ∪ {2v + 2, 2v + 4, . . . , q − 1} ∪ {1, 3, 5, . . . , q − 2} if 1 ≤ v ≤ (q − 3)/2 Tv = {0} ∪ {2v + 2 − q, 2v + 4 − q, 2v + 6 − q, . . . , q − 2} if (q − 1)/2 ≤ v ≤ q − 2 . {0} if v = q − 1
Really, by the second row of (10), d + v ∈ {0, −2, −4, . . . , 2v + 4, 2v + 2}. If v = 0 then d ∈ {0} ∪ {q − 2, q − 4, . . . , 1} ∪ {q − 1, q − 3, . . . , 2} = Fq . If 1 ≤ v ≤ (q − 3)/2 then 4 ≤ 2v+2 ≤ q−1, 2v+2 is even, and d+v ∈ {0}∪{q−2, q−4, . . . , 1}∪{q−1, q−3, . . . , 2v+2}. If (q − 1)/2 ≤ v ≤ q − 2 then q + 1 ≤ 2v + 2 ≤ 2q − 2, 2v + 2 is odd modulo q, and d + v ∈ {0} ∪ {2v + 2 − q, 2v + 4 − q, . . . , q − 2}. Finally, if v = q − 1 then 2v + 2 = 0. Let d + v = 2t ≥ 2. Then d + v ∈ Tv if and only if 1 ≤ v ≤ t − 1 that implies 2v + 2 ≤ 2t. Taking into account the points (1, 0, 2t, 4t2) and A∞ , we have |C(Φ2t + ) ∩ K1 | = t + 1. By 2t condition E, d + v ∈ / {q − 3, q − 1}. So, 2t ≤ q − 5 and |C(Φ+ ) ∩ K1 | ≤ (q − 3)/2. Let q − 2 ≥ d + v = 2u − 1. Then d + v ∈ Tv if 1 ≤ v ≤ (q − 3)/2 or 2v + 2 − q ≤ 2u − 1. 11
The last inequality implies v ≤ (q−3)/2+u. So, d+v ∈ Tv if and only if 1 ≤ v ≤ (q−3)/2+u. Due to the points (1, 0, 2u − 1, (2u − 1)2) and A∞ , we have |C(Φ2u−1 ) ∩ K1 | = (q + 1)/2 + u. + 2u−1 By condition E, d + v ∈ / {1}. So, 2u − 1 ≥ 3, u ≥ 2, and |C(Φ+ ) ∩ K1 | ≥ (q + 5)/2. At last, |C(Φ0+ ) ∩ K1 | = q + 1 as there is d + v = 0 for every v. We have shown that |C(Φδ+ ) ∩ K1 | ∈ / R if δ ∈ / {1, q − 3, q − 1}. Finally, Condition B is applied. Lemma 7. Under Condition 1, for q ≥ 11 all points of the form H = (1, b, c, d) ∈ / K∪Q are covered by bisecants of the (Tq + 2)-cap K in Construction A1. Proof. If either b ∈ / R or c−b ∈ / {1, 3, 5} the point H is covered due to Sufficient Condition C or D. Let b ∈ R and c − b ∈ {1, 3, 5}. Then the pair (b, c) ∈ {((q − 1)/2, (q + 1)/2), ((q − 1)/2, (q + 5)/2), ((q − 1)/2, (q + 9)/2), ((q + 1)/2, (q + 3)/2), ((q + 1)/2, (q + 7)/2), ((q + 1)/2, (q + 11)/2), ((q + 3)/2, (q + 5)/2), ((q + 3)/2, (q + 9)/2), ((q + 3)/2, (q + 13)/2)}. So, the sum b + c ∈ {0, 2, 4, 6, 8}. If q > 11 then {0, 2, 4, 6, 8} ∩ {1, q − 3, q − 1} = ∅ and the point H is covered due to Sufficient Condition E. Let q = 11. Then {0, 2, 4, 6, 8} ∩ {1, q − 3, q − 1} = q − 3 = 8 and we should consider the points H = (1, (q + 3)/2, (q + 13)/2, d) = (1, 7, 1, d). By (10), (1, b, 1, d) ∈ K1 if d = 1, 2, 4, 5, 6, 10; for these values of d the point (1, 7, 1, d) is collinear with (1, ¯b, 1, d) and A∞ since we have (1, ¯b, 1, d) + (7 − ¯b)(0, 1, 0, 0) = (1, 7, 1, d) . On the other hand, for the remaining values of d we have (1, 7, 1, 0) = 3(1, 6, 3, 7)−2(1, 0, 4, 5), (1, 7, 1, 3) = (1, 3, 8, 3)+ (1, 0, 5, 3), (1, 7, 1, 7) = 2(1, 1, 2, 7) − (1, 6, 3, 7), (1, 7, 1, 8) = 3(1, 3, 5, 8) − 2(1, 1, 7, 8), and (1, 7, 1, 9) = (1, 4, 7, 9) + 2(1, 3, 9, 9), where, by (10), all points of the linear combinations belong to K1 . Theorem 2. For q ≥ 11, under Condition 1, the (q 2 + q + 8)/2-cap K in Construction A1 is complete. Proof. By Construction A1, all points H ∈ Q are covered by bisecants of K through the point P. Points of the form H = (0, b, c, d) ∈ Π∞ and H = (1, b, c, d) ∈ / K ∪ Q are covered by Lemmas 5 and 7, respectively.
4
The associated code weight distribution and other combinatorial properties of the (q 2 + q + 8)/2-caps
For any cap K in P G(n, q), wi stands for the number of codewords with weight i in the associated code A(K), dmin (K) is the minimum distance of this code, and hj denotes the 12
number of hyperplanes in P G(n, q) meeting K in j ≥ 0 points. From previous works, see [15, Theorem 4.1],[10],[5, Theorem 16.1],[20], see also the references therein, we have the following result. Theorem 3. Let K be a k-cap in P G(n, q). Then w0 = 1,
wk−j = (q − 1)hj for j < k,
dmin (K) ≥ k − m2 (n − 1, q),
(20)
where hj ≥ 0 for m2 (n − 1, q) ≥ j ≥ 0 and hj = 0 for j > m2 (n − 1, q). We have dmin(K) = k − m2 (n − 1, q) if and only if the cap K contains an m2 (n − 1, q)-cap. Corollary 3. In P G(3, q), under Condition 1, the (Tq + 2)-cap K in Construction A1 has dmin(K) = Tq + 1 − q. In the weight distribution of the code A(K) we have wi > 0 for all values i in the region Tq + 2 ≥ i ≥ dmin (K) and wdmin (K) ≥ 3(q − 1). Proof. By Corollary 1, the cap K contains at least three conics. This shows the relations about dmin (K) and wdmin (K) . By the proof of Lemma 6 involving Sufficient Condition C, the plain Φb with x1 = bx0 , b ∈ Fq , contains q − b + 1 points of K1 and the same holds true for K. So, hj > 0 for q + 1 ≥ j ≥ 2. By (6),(10), the plane ϕ with 6x1 = 3x0 + x3 is the polar plane of the point A = (1, 1, 0, 3) ∈ K1 \{P } ⊂ Q. Hence ϕ is tangent to Q and to K1 at the tangency point A. As {P, P1 , P2 } ∩ ϕ = ∅, the plane ϕ is tangent to K as well and h1 > 0. Similarly, the polar plane ψ with equation 12x1 = 12x0 + x3 of the point (1, 2, 0, 12) ∈ / K1 is tangent to Q but external to K as {P, P1 , P2 } ∩ ψ = ∅. Hence h0 > 0. By computer aided arguments valid for q = 11, 17, 23, 29, 41 we established some properties of stabilizer group, that is the collineation group preserving the (Tq + 2)-cap. For groups we use the notation of [16, Table 2.3]. Proposition 2. Under Condition 1, for q = 11, 17, 23, 29, 41 the (q 2 + q + 8)/2-cap K obtained by Construction A1 has the following properties: (i) the stabilizer group of the cap is the symmetric group S3 of order 6; (ii) the stabilizer group partitions the cap into one 1-orbit {(0, 0, 0, 1)}, (q + 3)/2 3-orbits, and (q 2 − 2q − 3)/12 6-orbits; (iii) points of K outside Q form the 3-orbit {P, P1, P2 }. 13
Also, by computer aided arguments, for the (Tq + 2)-cap K under Condition 1 we have computed the numbers hj of planes intersecting K in j points, see Table 1. By (20), these values define the weight distribution of the associated code A(K). The number of conics entirely contained in K is equal to hq+1 . By Corollary 1, hq+1 ≥ 3. The conics CP , CP1 , and CP2 form a flat tetrahedral system [17, Lemma 18.4.4]. Each point of K1 lies on at most three conics. Any two conics are either of disjoint or have two common points. There is not a triple of disjoint conics. INSERT Table 1 HERE Remark 3. For q = 17 in P G(2, q) there is the unique complete (q + 11)/2-arc [13, Table 2.2]. Considering the secant planes through the line P P1 for the new (T17 + 2)-cap constructed in Section 2 we represented this arc in the form {(1, 0, 6), (1, 0, 11), (1, 1, 4), (1, 1, 13), (1, 6, 8), (1, 6, 9), (1, 10, 5), (1, 10, 12), (1, 14, 3), (1, 3, 14), (0, 1, 3), (0, 1, 0), (1, 5, 1), (1, 14, 10)}. The first (q + 3)/2 = 10 points lie on the conic 3x21 + x22 = 2x20 and the last four points are placed outside the conic. This is a contradiction to [22, Theorem 1] where it is claimed that for q ≡ 1 (mod 4) in the plane P G(2, q) a complete arc which has (q + 3)/2 points on an irreducible conic can contain at most two points outside the conic. Note that for q = 13 the similar contradiction was found in [4]. Note also that for q = 2p − 1 with odd prime p plane arcs sharing (q + 3)/2 points with a conic are studied in [19].
5
Construction of a complete 20-cap in P G(3, 5)
There exist only two projectively non-equivalent complete 20-caps in P G(3, 5) where 20 = T5 + 3 = m′2 (3, 5), see [3]. These caps are studied in [2],[3] where it is noted that each is preserved by a collineation group acting sharply transitively on it. The points of the caps are listed in [3] although no geometric construction is given. We show that our construction may be slightly modified in such a way that the resulting (T5 + 3)-cap is one of these two 20-caps, namely K2 with the notation used in [2] and [3]. We show that this cap contains 16 points from Q and that these points can be chosen by the same procedure as for our (Tq + 2)-caps. However one more point off Q can be added to the cap K for q = 5, which actually is the unique value of q uncovered in Lemma 7. Construction A2. Applying Construction A1 for q = 5 produces a (T5 + 2)-set K; a variant of Construction A1, under Condition 1, produces a (T5 + 3)-set K∗ = K ∪ {P3 } joining K with the further point P3 = (1, 3, 1, 0). 14
Theorem 4. The (T5 + 3)-set K∗ in Construction A2 is a complete 20-cap in P G(3, 5) projectively equivalent to the 20-cap K2 given in [3, Section 3]. Proof. Construction A1 and Theorem 1 provide a 19-cap K. Since it is known that m′2 (3, 5) = 20 and there are no complete 19-caps in P G(3, 5), see [3],[12],[13, Table 3.1], the cap K is incomplete and we can add to it just one point. Actually such a point must be P3 ; this can be shown by direct calculation. The two planes of equations x3 = ±2x0 are 5 -secant planes for the cap K∗ . By [3, Tables 1,2], the cap K2 is the unique 20-cap in P G(3, 5) provided with 5-secant planes. By Theorem 3 and [3, Table 2], the weight distribution of the code A(K2 ) is as follows: w0 = 1, w14 = 80, w15 = 240, w16 = 140, w18 = 120, w20 = 44. This code is optimal as a 5-ary linear code with length 20 and dimension 4 has the minimal distance at most 14 [7].
6
On the spectrum of complete caps sizes in P G(3, q)
The known sizes of complete caps in P G(3, q) are given in Table 2. We take the sizes of complete caps from the papers [3],[6],[8],[11]-[13],[17],[21],[23]-[25], see also the references therein. New entries obtained from our Theorem 2 and by a computer search are marked by asterisk ⋆. INSERT Table 2 HERE The following new lower bounds come from Construction A1 and Theorem 2: 70 = T11 + 2 ≤ m′2 (3, 11), 157 = T17 + 2 ≤ m′2 (3, 17), 280 = T23 + 2 ≤ m′2 (3, 23). New upper bounds t2 (3, 13) ≤ 36, t2 (3, 19) ≤ 58, t2 (3, 23) ≤ 72, are obtained in this paper by the randomized greedy algorithms [8, Section 2]. As the starter set for q = 19 we have used the first 20 points of the 40-cap of [26, Section 4.2]. For 7 ≤ q ≤ 19 the sizes and bounds already known from the literature are collected in [8, Table 1]. The results t2 (3, 7) = 17 [6] and m′2 (3, 16) ≤ 242 [14] are recent. In this paper we have obtained the following new sizes of the complete k -caps in P G(3, q): k = 90, 91, 92 for q = 13, k = 154 for q = 17, and k = 188, 190, 191 for q = 19. In P G(3, 23) the sizes k = 73 [8, Table 3], k = 120-122 [24], k = 186 [13, p. 90], k = 246 [24], k = 268 [21], and k = 278 [25], have already been known. The remaining sizes for q = 23 given in Table 2 are new entries. The bounds on t2 (3, 23) and m′2 (3, 23) come from [12, Lemma 2.5] and [17, Theorem 18.4.1].
15
For q = 13, 17, 19, 23, new sizes have been obtained using the greedy algorithms. To get starter caps for the algorithms, we have remove some points from the Tq -cap K1 in Construction A1. The completeness of the spectrum of complete caps sizes is marked in Table 2 by the dot after the value of q, see [3],[12],[13, Table 3.1]. Open problem. To complete the spectrum of sizes of complete caps in P G(3, 7), it only remains to solve the problem: ”Is there a complete 31-cap in P G(3, 7)?”
Acknowledgment The authors thank the anonymous referees for their useful comments.
16
Affiliation of authors Alexander A. Davydov Institute for Information Transmission Problems, Russian Academy of Sciences, Bol’shoi Karetnyi per. 19, GSP-4, Moscow, 127994, Russian Federation Stefano Marcugini and Fernanda Pambianco Dipartimento di Matematica e Informatica, Universit`a degli Studi di Perugia, Via Vanvitelli 1, Perugia, 06123, Italy
17
References [1] L. M. Abatangelo, On the number of points of caps obtained from an elliptic quadric of P G(3, q), European J. Combin. 3 (1982) 1-7. [2] V. Abatangelo, Classification of transitive caps in P G(3, 5), Pure Math. Appl. 11 (2000) 1-11. [3] V. Abatangelo, G. Korchm´aros and B. Larato, Classification of maximal caps in P G(3, 5) different from elliptic quadric, J. Geom. 57 (1996) 9-19. [4] A. H. Ali, J. W. P. Hirschfeld and H. Kaneta, The automorphism group of a complete (q − 1)-arc in P G(2, q), J. Combin. Des. 2 (1994) 131-145. [5] J. Bierbrauer, Introduction to Coding Theory, Chapman and Hall/CRC, Boca Raton, 2005. [6] J. Bierbrauer, S. Marcugini and F. Pambianco, The smallest size of a complete cap in P G(3, 7), Discrete Math. 306 (2006) 1257-1263. [7] A. E. Brouwer, Bounds on the minimum distance of linear codes: http://www.win. tue.nl/˜aeb/voorlincod.html of February 1, 2006. [8] A. A. Davydov, S. Marcugini and F. Pambianco, Complete caps in projective spaces P G(n, q), J. Geom. 80 (2004) 23-30. [9] A. A. Davydov, S. Marcugini and F. Pambianco, Complete (q 2 + q + 8)/2-caps in the projective space P G(3, q) with odd prime q ≡ 2 (mod 3), in: Proc. X Int. Workshop on Algebraic and Combin. Coding Theory, ACCT-X, Zvenigorod, Russia, Sept. 9-16, 2006, pp. 76-80. [10] S. Dodunekov and J. Simonis, Codes and Projective Multisets, Electron. J. Combin. 5 (1998) Research Paper 37, 23 pp. (electronic). [11] Y. Edel, L. Storme and P. Sziklai, New upper bounds on the sizes of caps in P G(N, 5) and P G(N, 7), J. Combin. Math. Combin. Comput. 60 (2007) 7-32. [12] G. Faina, S. Marcugini, A. Milani and F. Pambianco, The sizes k of complete k-caps in P G(n, q), for small q and 3 ≤ n ≤ 5, Ars Combin. 50 (1998) 235-243. 18
[13] G. Faina and F. Pambianco, On the spectrum of the values k for which a complete k-cap in P G(n, q) exists, J. Geom. 62 (1998) 84-98. [14] S. Ferret and L. Storme, On the size of complete caps in P G(3, 2h ), Finite Fields Appl. 10 (2004), 306-314. [15] R. Hill, Caps and codes, Discrete Math. 22 (1978) 111-137. [16] J. W. P. Hirschfeld, Projective Geometries over Finite Fields, second edition, Oxford University Press, Oxford, 1998. [17] J. W. P. Hirschfeld, Finite Projective Spaces of Three Dimensions, Oxford University Press, Oxford, 1985. [18] J. W. P. Hirschfeld and L. Storme, The packing problem in statistics, coding theory, and finite projective spaces: update 2001, in: Finite Geometries, Proceedings of the Fourth Isle of Thorns Conference, A. Blokhuis, J. W. P. Hirschfeld, D. Jungnickel and J. A. Thas, Eds., Developments in Mathematics, vol. 3, Kluwer Academic Publishers, Dordrecht, 2001, pp. 201-246. [19] G. Korchm´aros and A. Sonnino, Complete arcs arising from conics, Discrete Math. 267 (2003) 181-187. [20] I. Landjev and A. Rousseva, An extension theorem for arcs and linear codes, Problems of Information Transmission, 42 (2006) 319-329. [21] F. Pambianco and E. Ughi, A class of k-caps having k − 2 points in common with an elliptic quadric and two points on an external line, Australas. J. Combin. 21 (2000) 299-310. [22] G. Pellegrino, Un’osservazione sul problema dei k-archi completi in S2,q , con q ≡ 1 (mod 4), Atti Accad. Naz. Lincei Rend. 63 (1977) 33-44. [23] G. Pellegrino, Una rilettura di alcune proposizioni relative a calotte ed archi completi dello spazio P G(3, q) (r = 2, 3; q dispari), Giornate di Geometrie Combinatorie, Universit`a di Perugia, Perugia (1993) 269-287. [24] G. Pellegrino, Sulle calotte complete, non ovaloidi, dello spazio PG(3, q), q dispari, Rend. Circ. Mat. Palermo 47 (1998) 141-168. 19
[25] B. Segre, Le geometrie di Galois, Ann. Mat. Pura Appl. 48 (1959) 1-97. [26] L. Storme and H. Van Maldeghem, Cyclic caps in P G(3, q), Geom. Dedicata 56 (1995) 271-284. [27] T. Sz˝onyi, Arcs, caps, codes and 3-independent subsets, Giornate di Geometrie Combinatorie, Universit`a di Perugia, Perugia (1993) 57-80.
20
Table 1 The number hj of planes intersecting the (Tq +2)-cap K ⊂ P G(3, q) in j points q 11 17 23 29 41
h0 56 137 253 406 820
h1 36 105 210 351 741
h2 51 57 69 84 120
h3 19 16 7 6 6
h4 103 61 22 4 4
h5 102 39 12 6 3
h6 254 264 154 39 4
h7 297 153 57 24 3
h8 417 825 477 189 21
h9 h10 58 57 410 1509 222 1200 46 756 4 75
h11 h12 6 8 855 518 984 2214 336 1800 18 334
q h13 h14 h15 h16 h17 h18 h19 h20 h21 h22 h23 h24 h25 17 66 153 18 18 9 7 23 2217 2820 525 753 93 328 24 57 6 9 3 4 29 1083 3687 1981 6057 3222 2569 651 1242 171 399 18 103 12 41 93 1131 508 2556 1239 5043 3000 9951 5691 15132 7605 7185 1905 q h26 29 6 41 3687
h27 h28 3 3 827 1737
h29 3 255
q 41
h40 3
h42 3
h39 3
h41 3
h30 3 681
h31
h32
h33
h34
h35
h36
h37
h38
33
159
12
33
3
7
3
3
21
Table 2 The known sizes of the complete k-caps in P G(3, q), Tq = (q 2 + q + 4)/2 q
t2 (3, q)
The known sizes k of the complete k-caps with
m′2 (3, q) m2 (3, q) References
t2 (3, q) ≤ k ≤ m′2 (3, q) 3
8
8 = Tq
8
10 [12],[13],[25]
4
10
10, 12, 13, 14 = Tq + 2
14
17 [12],[13],[25]
5
12
12 ≤ k ≤ 18 and
20
26 [2],[3],[12],[13]
7
17
17 ≤ k ≤ 30 and
32
50 [6],[8],[11],[13],[17],[23]-[25]
8
≥ 14
20 ≤ k ≤ 41 = Tq + 3
≤ 60
65 [8],[12],[13],[17]
9
≥ 15
24 ≤ k ≤ 48 = Tq + 1
≤ 78
82 [8],[12],[13],[23]-[25]
11
≥ 18
30 ≤ k ≤ 70 = Tq + 2
≤ 116
122 ⋆, [8],[12],[13],[17],[23]-[25]
13
≥ 21
36 ≤ k ≤ 94 = Tq + 1
≤ 162
170 ⋆, [8],[12],[13],[23],[24]
16
≥ 25
41 ≤ k ≤ 138 = Tq
≤ 242
257 [8],[12]-[14],[17],[18]
17
≥ 26
51 ≤ k ≤ 157 = Tq + 2
≤ 278
290 ⋆, [8],[12],[13],[23]-[25]
19
≥ 29
58 ≤ k ≤ 192 = Tq
≤ 348
362 ⋆, [8],[12],[13],[17],[23]-[25]
23
≥ 35
72 ≤ k ≤ 280 = Tq + 2
≤ 512
530 ⋆, [8],[12],[13],[17],[21],[23]-[25]
k = 20 = Tq + 3 k = 32 = Tq + 2
22
