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A290342
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Number of ways to write n as x^2 + 2*y^2 + z*(z+1)/2, where x is a nonnegative integer, and y and z are positive integers.
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5
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0, 0, 0, 1, 1, 1, 1, 1, 1, 3, 1, 1, 4, 2, 2, 2, 1, 2, 4, 3, 2, 4, 2, 4, 4, 3, 1, 4, 5, 2, 5, 1, 3, 6, 5, 2, 3, 6, 3, 9, 3, 1, 6, 3, 5, 4, 4, 6, 7, 3, 2, 5, 3, 6, 9, 6, 3, 7, 6, 2, 8, 5, 4, 8, 6, 3, 4, 6, 3, 12, 2
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OFFSET
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0,10
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COMMENTS
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Conjecture: a(n) > 0 for all n > 2. In other words, each n = 0,1,2,... can be written as x^2 + 2y*(y+2) + z*(z+3)/2 with x,y,z nonnegative integers.
As pointed out by Sun in his 2007 paper in Acta Arith., a result of Jones and Pall implies that every n = 0,1,2,... can be written as x^2 + 2*(2y)^2 + z*(z+1)/2 with x,y,z nonnegative integers.
Let a,c,e be positive integers, and let b,d,f be nonnegative integers with a-b, c-d, e-f all even. Suppose that a|b, c|d and e|f. The author studied in arXiv:1502.03056 when each nonnegative integer can be written as x*(a*x+b)/2 + y*(c*y+d)/2 + z*(e*z+f)/2 with x,y,z nonnegative integers, and conjectured that the answer is positive if (a,b,c,d,e,f) is among the following ten tuples (4,0,2,0,1,3), (4,0,2,0,1,5), (4,0,2,6,1,1), (4,0,2,6,2,0), (4,4,2,0,1,3), (4,8,2,0,1,1), (4,8,2,0,1,3), (4,12,2,0,1,1), (6,0,2,0,1,3), (6,6,2,0,1,3).
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LINKS
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EXAMPLE
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a(10) = 1 since 10 = 1^2 + 2*2^2 + 1*2/2.
a(11) = 1 since 11 = 0^2 + 2*2^2 + 2*3/2.
a(16) = 1 since 16 = 2^2 + 2*1^2 + 4*5/2.
a(26) = 1 since 26 = 3^2 + 2*1^2 + 5*6/2.
a(31) = 1 since 31 = 1^2 + 2*1^2 + 7*8/2.
a(41) = 1 since 41 = 6^2 + 2*1^2 + 2*3/2.
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MATHEMATICA
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TQ[n_]:=TQ[n]=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[TQ[n-x^2-2y^2], r=r+1], {x, 0, Sqrt[n]}, {y, 1, Sqrt[(n-x^2)/2]}]; Print[n, " ", 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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