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%I #9 Jun 25 2023 08:13:48
%S 76,526,6508,34228,324562,747004,2849576,4965266,12896050,41071144,
%T 57316082,138789292,231850328,294172382,458893426,836688844,
%U 1430252626,1689646684,2700843026,3609164734,4146921368,6155086706,7879211410,11169529016,17176506056,21022261804,23187646130
%N a(n) is the number of isomorphism classes of genus 3 hyperelliptic curves over the finite field of order prime(n).
%H E. Nart, <a href="https://doi.org/10.1016/j.aim.2009.01.001">Counting hyperelliptic curves</a>, Adv. Math. 221 (2009), no. 3, 774-787.
%H E. Nart and D. Sadornil, <a href="https://doi.org/10.1016/j.ffa.2003.08.002">Hyperelliptic curves of genus three over finite fields of even characteristic</a>, Finite Fields Appl. 10 (2004), no. 2, 198-220.
%F a(1) = 76, and for n > 1, a(n) = 2*prime(n)^5 + 2*prime(n)^3 - 2 - 2*(prime(n)^2 - prime(n))*[prime(n) == 3 (mod 4)] + 2*(prime(n)-1)*[prime(n) > 3] + 4*[prime(n) == 1 (mod 8)] + 12*[prime(n) == 1 (mod 7)] + 2*[prime(n) == 7] + 2*[prime(n) == 1 or 5 (mod 12)].
%e For n = 1, E. Nart and D. Sadornil showed that there are 76 genus 3 hyperelliptic curves over F_2, so a(1) = 76.
%o (Sage)
%o def a(n):
%o if n == 1: return 76
%o p = Primes()[n-1]
%o ans = 2*p^5 + 2*p^3 - 2
%o if p%4 == 3: ans -= 2*(p^2 - p)
%o if p > 3: ans += 2*(p - 1)
%o if p%8 == 1: ans += 4
%o if p%7 == 1: ans += 12
%o if p == 7: ans += 2
%o if p%12 in [1, 5]: ans += 2
%o return ans
%Y Cf. A362243, A363840.
%K nonn
%O 1,1
%A _Robin Visser_, Jun 23 2023
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