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A244819
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Positive numbers primitively represented by the binary quadratic form (1, 0, 3).
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5
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1, 3, 4, 7, 12, 13, 19, 21, 28, 31, 37, 39, 43, 49, 52, 57, 61, 67, 73, 76, 79, 84, 91, 93, 97, 103, 109, 111, 124, 127, 129, 133, 139, 147, 148, 151, 156, 157, 163, 169, 172, 181, 183, 193, 196, 199, 201, 211, 217, 219, 223, 228, 229, 237, 241, 244, 247, 259
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OFFSET
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1,2
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COMMENTS
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Discriminant = -12.
x^2 + 3*y^2 represents positive k properly (gcd(x, y) = 1), with nonnegative x, and the following multiplicities m(k): m(1) = 1, m(3) = 1, m(4) = 2, and if k = 3^a*4^b*Product_{j=1..P1} p1(j)^e1(j), with p1(j) primes 1 (mod 6) (A002476), e1(j) nonnegative integer numbers, and a and b from {0, 1}, then m(k) = 2^(P1+b). Shown by the lifting theorem (e.g., Apostol) for prime powers. Note that for prime 2 there is one solution of j^2 + 3 == 0 (mod 2) this corresponds the imprimitive reduced form (2, 2, 2), not to the one reduced primitive form (1, 0, 3) for discriminant -12 (A000003(3) = 1). - Wolfdieter Lang, Mar 02 2021
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REFERENCES
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Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976 (1986), Theorem 5.30, pp. 121-122.
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LINKS
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EXAMPLE
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Proper solution to x^2 + 3*y^2 = a(n), with x nonnegative: a(12 = 3*4) with (x, y) = (3, pm 1), pm = +1 or -1, multiplicity m(12) = 2, (a, b, P1) = (1, 1, 0); a(21 = 3*7) with (3, pm 2), m(21) = 2, (a, b, P1) = (1, 0, 1); a(49 = 7^2) with (1, pm 4), m(49) = 2 (a, b, P1) = (0, 0, 1)). - Wolfdieter Lang, Mar 02 2021
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MAPLE
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MATHEMATICA
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Reap[For[n = 1, n < 1000, n++, r = Reduce[x^2 + 3 y^2 == n, {x, y}, Integers]; If[r =!= False, If[AnyTrue[{x, y} /. {ToRules[r /. C[1] -> 0]}, CoprimeQ @@ # &], Sow[n]]]]][[2, 1]] (* Jean-François Alcover, Oct 31 2016 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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