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%I #13 Oct 23 2023 11:47:24
%S 4,4,6,6,8,10,10,12,14,18,18,20,24,26,28,30,32,34,38,42,44,48,50,54,
%T 60,62,66,68,72,74,80,82,84,90,98,102,104,108,110,114,122,126,128,130,
%U 132,138,140,150,152,158,164,168,170,174,180,182,192,194,198,200,212,224,228,230,234
%N Maximal k such that there exists a k-arc on the projective plane over GF(q), where q = A246655(n) is the n-th prime power > 1.
%C A k-arc is a set of k points in PG(2,q) (the projective plane over GF(q)) such that no three are collinear.
%D J. W. P. Hirschfeld, Linear codes and algebraic curves, pp. 35-53 of F. C. Holroyd and R. J. Wilson, editors, Geometrical Combinatorics. Pitman, Boston, 1984.
%H Paolo Xausa, <a href="/A365216/b365216.txt">Table of n, a(n) for n = 1..10000</a>
%H J. W. P. Hirschfeld, G. Korchmáros, and F. Torres, <a href="https://doi.org/10.1515/9781400847419">Algebraic curves over a finite field</a>, Princeton Ser. Appl. Math., Princeton University Press, Princeton, NJ, 2008.
%F a(n) = q + 1 if q is odd, otherwise a(n) = q + 2, where q = A246655(n).
%e For n = 1, the four points (0:0:1), (1:0:1), (0:1:1), (1:1:1) form a 4-arc in PG(2,2); the projective plane over GF(2). Moreover, any five points in PG(2,2) contain three points which are collinear, thus a(1) = 4.
%e For n = 4, the six points (0:0:1), (1:0:1), (0:1:1), (1:1:1), (3:2:1), (3:4:1) form a 6-arc in PG(2,5); the projective plane over GF(5). Moreover, any seven points in PG(2,5) contain three points which are collinear, thus a(4) = 6.
%t Map[#+2-Mod[#,2]&,Select[Range[200],PrimePowerQ]] (* _Paolo Xausa_, Oct 23 2023 *)
%o (Sage)
%o for q in range(2, 1000):
%o if Integer(q).is_prime_power(): print(q + 2 - (q%2))
%Y Cf. A000509, A005524.
%K nonn
%O 1,1
%A _Robin Visser_, Aug 26 2023
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