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A258928 a(n) = number of integral points on the elliptic curve y^2 = x^3 - (n^2)*x + 1, considering only nonnegative values of y.

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%I A258928 #16 Apr 23 2021 20:53:01
%S A258928 3,6,11,9,15,13,14,17,26,12,12,11,12,19,20,11,19,36,12,17,16,11,19,16,
%T A258928 15,27,17,17,18,16,12,15,17,11,12,11,28,16,12,11,15,24,27,11,17,12,26,
%U A258928 15,17,15,12,15,17,27,12,14,16,15,16,24,12,41,17,16,12,11,17,16,16,15,23,15,16,20,15
%N A258928 a(n) = number of integral points on the elliptic curve y^2 = x^3 - (n^2)*x + 1, considering only nonnegative values of y.
%C A258928 For n>3, the number of integral points on y = x^3 - (n^2)*x + 1 is at least 11. These 11 points correspond to the solutions x = {-1, 0, n, -n, n + 2, -n + 2, n^2 - 1, n^2 - 2n + 2, n^2 + 2n + 2, n^4 + 2n, n^4 - 2n}.
%H A258928 Robert Israel, <a href="/A258928/b258928.txt">Table of n, a(n) for n = 0..209</a>
%e A258928 a(0) = 3 because the integer points on y^2 = x^3 + 1 are (-1, 0), (0, 1), and (2, 3).
%o A258928 (Sage)
%o A258928 def f(n):
%o A258928   R.<x,y> = QQ[]
%o A258928   E = EllipticCurve(y^2 - x^3 + n^2*x - 1)
%o A258928   return len(E.integral_points(both_signs=false))
%o A258928 [f(x) for x in range(40)]  # _Robert Israel_, Apr 23 2021
%Y A258928 Cf. A081119, A081120, A259191.
%K A258928 nonn
%O A258928 0,1
%A A258928 _Morris Neene_, Jun 14 2015
%E A258928 More terms from _Robert Israel_, Apr 23 2021