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A262813
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Number of ordered ways to write n as x^3 + y^2 + z*(z+1)/2 with x >= 0, y >=0 and z > 0.
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45
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1, 2, 2, 2, 2, 2, 3, 2, 1, 4, 5, 3, 2, 2, 5, 3, 2, 4, 4, 4, 1, 4, 4, 2, 3, 3, 5, 3, 5, 5, 4, 5, 3, 4, 1, 4, 9, 6, 4, 4, 3, 3, 3, 3, 7, 8, 4, 3, 3, 3, 3, 5, 7, 5, 5, 4, 4, 4, 4, 4, 3, 4, 3, 8, 6, 4, 8, 3, 4, 5, 8, 7, 5, 5, 5, 3, 2, 8, 8, 6, 4, 7, 8, 2, 5, 7, 4, 6, 2, 5, 7, 10, 6, 5, 7, 3, 5, 1, 6, 5
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OFFSET
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1,2
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COMMENTS
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Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 9, 21, 35, 98, 152, 306.
This has been verified for all n = 1..2*10^7.
If z >= 0, a(n) = 1 only for n = 21, 35, 98, 306. - Mauro Fiorentini, Jul 20 2023
In contrast with the conjecture, in 2015 the author refined a result of Euler by proving that any positive integer can be written as the sum of two squares and a positive triangular number.
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LINKS
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EXAMPLE
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a(1) = 1 since 1 = 0^3 + 0^2 + 1*2/2.
a(2) = 2 since 2 = 0^3 + 1^2 + 1*2/2 = 1^3 + 0^2 + 1*2/2.
a(6) = 2 since 6 = 0^3 + 0^2 + 3*4/2 = 1^3 + 2^2 + 1*2/2.
a(9) = 1 since 9 = 2^3 + 0^2 + 1*2/2.
a(21) = 1 since 21 = 0^3 + 0^2 + 6*7/2.
a(35) = 1 since 35 = 0^3 + 5^2 + 4*5/2.
a(98) = 1 since 98 = 3^3 + 4^2 + 10*11/2.
a(152) = 1 since 152 = 0^3 + 4^2 + 16*17/2.
a(306) = 1 since 306 = 1^3 + 13^2 + 16*17/2.
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MATHEMATICA
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TQ[n_]:=n>0&&IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[TQ[n-x^3-y^2], r=r+1], {x, 0, n^(1/3)}, {y, 0, Sqrt[n-x^3]}]; Print[n, " ", r]; Continue, {n, 1, 100}]
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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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