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A364421 For n >= 3, r >= 0, y an integer, a(n) is the number of integral solutions to the elliptic equation y^2 = n^3 + n^2 + 2*r*n + r^2.

%I A364421 #62 Sep 30 2023 21:42:36
%S A364421 2,2,2,3,2,4,3,4,2,8,2,4,8,5,2,7,2,9,8,4,2,15,3,4,5,10,2,15,2,7,8,4,8,
%T A364421 17,2,4,8,15,2,15,2,10,14,4,2,22,3,7,8,9,2,10,8,15,8,4,2,38,2,4,14,8,
%U A364421 8,15,2,9,7,16,2,27,2,4,13,9,8,15,2,22,6,4,2,39,8,4,7,16,2,27,8,10
%N A364421 For n >= 3, r >= 0, y an integer, a(n) is the number of integral solutions to the elliptic equation y^2 = n^3 + n^2 + 2*r*n + r^2.
%C A364421 The equation y^2 = n^3 + A*n^2 + B*n + C, where A = 1, B = 2*r, C = r^2 is a minimal model of an elliptic curve with integral coefficients, for details see the Links section.
%C A364421 For a prime number p >= 5, the equation y^2 = p^3 + (p + r)^2 has 2 solutions, r_1 = p*(p - 3)/2 and r_2 = (p + 1)*(p^2 - p - 1)/2.
%C A364421 Factoring the equation y^2 = n^3 + n^2 + 2*r*n + r^2 yields (y+n+r)*(y-n-r) = n^3, which implies y+n+r = d and y-n-r = n^3/d for some divisor d of n^3.  Thus a(n) is the number of divisors d of n^3 such that (d-n^3/d)/2 - n is a nonnegative integer.  This resolves some of Thomas Scheuerle's conjectures. - _Robin Visser_, Sep 30 2023
%H A364421 Robin Visser, <a href="/A364421/b364421.txt">Table of n, a(n) for n = 3..10000</a>
%H A364421 Joseph H. Silverman, <a href="https://www.math.ens.psl.eu/~benoist/refs/Silverman.pdf">The Arithmetic of Elliptic Curves</a>
%F A364421 a(p) = 2 for p prime >= 5, see Comments.
%F A364421 From _Thomas Scheuerle_, Sep 04 2023: (Start)
%F A364421 Conjecture: a(A190300(n)) = 3.
%F A364421 Conjecture: a(A196226(n)) = 4.
%F A364421 Conjecture: a(p^3) = 5 if p is an odd prime.
%F A364421 Conjecture: a(2*p^2) = 7 if p is an odd prime. But there exist other cases too, for example a(3*23) = 7.
%F A364421 Conjecture: a(prime(n)^prime(n)) = A245685(n - 1) - 1. (End)
%e A364421 n = 6: y^2 = 6^3 + (6 + r)^2 is valid for r = 9, 19, 47, thus a(6) = 3. The 3 solutions [y, n, n+r] are [21, 6, 15], [29, 6, 25], [55, 6, 53].
%o A364421 (PARI) a(n) = length(select((x) -> x[1] >= 0 && x[2] >= n, thue(thueinit(x^2-1,1),n^3),1)) \\ _Thomas Scheuerle_, Sep 03 2023
%o A364421 (Sage)
%o A364421 def a(n):
%o A364421     num_sols = 0
%o A364421     for d in Integer(n^3).divisors():
%o A364421         if ((d-n^3/d)%2 == 0) and ((d-n^3/d)/2 >= n): num_sols += 1
%o A364421     return num_sols  # _Robin Visser_, Sep 30 2023
%Y A364421 Cf. A081119, A134108, A190300, A196226, A245685, A364248.
%K A364421 nonn
%O A364421 3,1
%A A364421 _Ctibor O. Zizka_, Sep 01 2023
%E A364421 a(61)-a(92) from _Thomas Scheuerle_, Sep 01 2023