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

1,2

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

From Michael Shmoish, Jan 10 2026: (Start)

k is in the sequence iff the squarefree part of (k-1)*(k+2) contains no prime factors of the form 3*j+2.

a(n) = A392201(n) - 1 since (x^2 + y^2 + z^2)/(x*y - y*z + x*z) = k >= 1 for some integers x, y and z if and only if the squarefree part of (k-2)*(k+1) contains no prime factors of the form 3*j+2. (End)

Links

Table of n, a(n) for n=1..57.

Alon Amit, Quora answer about divisibility by 3

Jinyuan Wang, PARI program and details with k less than 100

Formula

a(n) = A392201(n) - 1. - Michael Shmoish, Jan 10 2026 [Corrected by Sean A. Irvine, Feb 04 2026]

Example

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

Prog

(Python)
from itertools import count, islice
from sympy import factorint
def A331605_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n: all(e&1 == 0 for p, e in factorint(n-1).items() if p % 3 == 2) and all(e&1 == 0 for p, e in factorint(n+2).items() if p % 3 == 2), count(max(startvalue, 1)))
A331605_list = list(islice(A331605_gen(), 57)) # Chai Wah Wu, Jan 16 2026

Crossrefs

Cf. A392201.

Sequence in context: A140411 A053353 A099792 * A115757 A159032 A082745

Adjacent sequences: A331602 A331603 A331604 * A331606 A331607 A331608

Keyword

nonn

Author

Jinyuan Wang, Jan 22 2020

Extensions

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

Status

approved