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A286885
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Number of ways to write 6*n+1 as x^2 + 3*y^2 + 54*z^2 with x,y,z nonnegative integers.
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4
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1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 1, 3, 2, 3, 1, 1, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 4, 4, 3, 2, 2, 4, 2, 3, 3, 3, 3, 3, 2, 2, 4, 3, 4, 1, 3, 2, 3, 4, 3, 3, 3, 3, 2, 3, 3, 2, 4, 3, 2, 3, 2
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OFFSET
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0,9
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COMMENTS
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Conjecture: a(n) > 0 for all n = 0,1,2,....
In the a-file, we list the tuples (m,r,a,b,c) with 30 >= m > max{2,r} >= 0, 100 >= a >= b >= c > 0, gcd(a,b,c) = 1, and the form a*x^2+b*y^2+c*z^2 irregular, such that all the numbers m*n+r (n = 0,1,2,...) should be representable by a*x^2+b*y^2+c*z^2 with x,y,z integers.
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LINKS
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Zhi-Wei Sun, Tuples (m,r,a,b,c) with 30 >= m > max{2,r} >= 0 and 100 >= a >= b >= c > 0, for which all the numbers m*n+r (n = 0,1,2,...) should be representable by a*x^2+b*y^2+c*z^2 with x,y,z integers.
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EXAMPLE
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a(9) = 1 since 6*9 + 1 = 1^2 + 3*0^2 + 54*1^2.
a(34) = 1 since 6*34 + 1 = 2^2 + 3*7^2 + 54*1^2.
a(125) = 1 since 6*125 + 1 = 26^2 + 3*5^2 + 54*0^2.
a(130) = 1 since 6*130 + 1 = 22^2 + 3*9^2 + 54*1^2.
a(133) = 1 since 6*133 + 1 = 11^2 + 3*8^2 + 54*3^2.
a(203) = 1 since 6*203 + 1 = 25^2 + 3*6^2 + 54*3^2.
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MATHEMATICA
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SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
table={}; Do[r=0; Do[If[SQ[6n+1-3y^2-54z^2], r=r+1], {y, 0, Sqrt[(6n+1)/3]}, {z, 0, Sqrt[(6n+1-3y^2)/54]}]; table=Append[table, r], {n, 0, 70}]
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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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