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A338696
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Number of ways to write n as x^3 + y^2 + z*(3*z+2), where x and y are nonnegative integers, and z is an integer.
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2
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3, 3, 1, 1, 3, 3, 1, 2, 6, 5, 1, 2, 3, 2, 1, 3, 8, 4, 0, 2, 3, 4, 1, 3, 7, 4, 2, 3, 3, 3, 3, 4, 7, 4, 2, 4, 5, 5, 1, 2, 7, 5, 3, 6, 5, 1, 2, 3, 7, 5, 2, 6, 2, 2, 1, 2, 10, 5, 2, 4, 2, 1, 1, 7, 11, 8, 2, 5, 6, 5, 3, 4, 11, 3, 1, 5, 5, 2, 1, 5, 8, 6, 4, 5, 5, 5, 3, 2, 9, 7, 2, 6, 4, 5, 1, 5, 10, 5, 2, 4
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OFFSET
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1,1
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COMMENTS
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Conjecture: a(n) > 0 except for n = 19.
We have verified this for n up to 5*10^6.
As z*(3*z+2) = floor((3*z+1)^2/3) and 19 = 0^3 + 4^2 + floor(3^2/3), the conjecture implies that each n = 0,1,... can be written as x^3 + y^2 + floor(z^2/3) with x,y,z nonnegative integers.
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LINKS
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EXAMPLE
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a(63) = 1 with 63 = 3^3 + 6^2 + 0*(3*0+2).
a(327) = 1 with 327 = 5^3 + 13^2 + 3*(3*3+2).
a(478) = 1 with 478 = 6^3 + 1^2 + 9*(3*9+2).
a(847) = 1 with 847 = 1^3 + 29^2 + 1*(3*1+2).
a(1043) = 1 with 1043 = 3^3 + 20^2 + 14*(3*14+2).
a(3175) = 1 with 3175 = 5^3 + 35^2 + (-25)*(3*(-25)+2).
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MATHEMATICA
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OctQ[n_]:=OctQ[n]=IntegerQ[Sqrt[3n+1]];
tab={}; Do[r=0; Do[If[OctQ[n-x^3-y^2], r=r+1], {x, 0, n^(1/3)}, {y, 0, Sqrt[n-x^3]}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]
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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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