A331605 Positive integers k such that k = (a^2 + b^2 + c^2)/(a*b + b*c + c*a) for some integers a, b and c.

1, 2, 5, 10, 14, 17, 26, 29, 37, 50, 62, 65, 74, 77, 82, 98, 101, 109, 110, 122, 125, 145, 149, 170, 173, 190, 194, 197, 209, 226, 242, 245, 257, 269, 290, 302, 305, 314, 325, 334, 362, 365, 398, 401, 410, 434, 437, 442, 469, 482, 485, 497, 509, 514, 530, 554, 557

Offset: 1

Author

Jinyuan Wang, Jan 22 2020

Keywords

• nonn

Comments

This sequence is infinite because k = x^2 + 1 is a term, where a = x + 1, b = x^2 + 1 and c = x^4 + x^3 + 3*x^2 + 2*x + 1. There are other forms of k:

k = (a^2-a+2)^2 - 2 when a + b = 1 and c = a^2 - a + 1.

k = x^4 + 2*x^3 + 5*x^2 + 4*x + 2 when a = k*(b+c) + x, b = x^2 + x + 1 and c = x + 1.

k = ((a^2+a*b+b^2)^2 - 2*a*b + 1)/(a + b)^2 when a = Fibonacci(2*m-1), b = Fibonacci(2*m) and c = ((a^2+a*b+b^2)^2 - a*b)/(a + b).

a(n) == 1 or 2 (mod 4). Proof: a^2 - k*(b+c)*a + (b^2+c^2-k*b*c) = 0, hence discriminant D = (k^2-4)*(b^2+c^2) + (2*k^2+4*k)*b*c is a square. Because (a/g, b/g, c/g) is also a set of solution if k = (a^2 + b^2 + c^2)/(a*b + b*c + c*a) and gcd(a, b, c) = g, we only need to consider the case of gcd(a,b,c) = 1.

Case (i). k = 4*r, then D/4 = (4*r^2-1)*(b^2+c^2) + (8*r^2+4*r)*b*c == 2 or 3 (mod 4), hence D is not a square, a contradiction.

Case (ii). k = 4*r - 1, then (a+b)^2 + (b+c)^2 + (c+a)^2 = 8*r*(a*b + b*c + c*a) is divisible by 4, hence a, b and c are odd numbers. Therefore, ((a+b)^2 + (b+c)^2 + (c+a)^2)/4 == 1 (mod 2), a contradiction.

a(n) is never divisible by 3. This follows from writing the equation as (a+b+c)^2 = (k+2)(ab+ac+bc) and then working mod a prime p==2 (mod 3) dividing k+2 an odd number of times. See linked Quora answer below. - Alon Amit, Aug 11 2024

k is in the sequence iff k-1 and k+2 are both Loeschian numbers (A003136). This can be proved with Legendre's theorem. - Yifan Xie, Feb 15 2025

Examples

			a(4) = 10 because 10 = ((-1)^2 + 2^2 + 5^2)/((-1)*2 + 2*5 + 5*(-1)).
		

Links

• Jinyuan Wang, PARI program and details with k less than 100
• Alon Amit, Quora answer about divisibility by 3

Extensions

a(22)-a(57) from Giovanni Resta, Jan 29 2020