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A331605
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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.
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4
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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a(4) = 10 because 10 = ((-1)^2 + 2^2 + 5^2)/((-1)*2 + 2*5 + 5*(-1)).
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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