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.

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, 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, 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

Offset: 3

Author

Ctibor O. Zizka, Sep 01 2023

Keywords

• nonn

Comments

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.

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.

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

Examples

			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].
		

Links

• Robin Visser, Table of n, a(n) for n = 3..10000
• Joseph H. Silverman, The Arithmetic of Elliptic Curves

Crossrefs

Cf. A081119, A134108, A190300, A196226, A245685, A364248.

Programs

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

Sage

def a(n):
    num_sols = 0
    for d in Integer(n^3).divisors():
        if ((d-n^3/d)%2 == 0) and ((d-n^3/d)/2 >= n): num_sols += 1
    return num_sols  # Robin Visser, Sep 30 2023

Formula

a(p) = 2 for p prime >= 5, see Comments.

From Thomas Scheuerle, Sep 04 2023: (Start)

Conjecture: a(A190300(n)) = 3.

Conjecture: a(A196226(n)) = 4.

Conjecture: a(p^3) = 5 if p is an odd prime.

Conjecture: a(2*p^2) = 7 if p is an odd prime. But there exist other cases too, for example a(3*23) = 7.

Conjecture: a(prime(n)^prime(n)) = A245685(n - 1) - 1. (End)

Extensions

a(61)-a(92) from Thomas Scheuerle, Sep 01 2023