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A262815
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Number of ordered ways to write n as x^3 + y*(y+1)/2 + z*(3*z+1)/2, where x, y and z are nonnegative integers.
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25
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1, 2, 2, 3, 2, 1, 2, 2, 4, 3, 3, 4, 1, 3, 2, 3, 5, 3, 5, 1, 1, 3, 3, 4, 2, 3, 3, 3, 4, 6, 6, 3, 2, 3, 2, 4, 6, 6, 3, 2, 3, 3, 4, 5, 8, 2, 3, 3, 5, 3, 2, 5, 3, 3, 3, 7, 3, 4, 4, 3, 3, 3, 5, 8, 2, 4, 3, 5, 7, 4, 7, 4, 5, 3, 6, 1, 4, 4, 6, 7, 4, 8, 5, 1, 4, 7, 7, 4, 4, 5, 2, 3, 5, 10, 6, 4, 2, 1, 3, 5, 7
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OFFSET
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0,2
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COMMENTS
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Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 5, 12, 19, 20, 75, 83, 97, 117.
By Theorem 1.7(i) in the linked paper, each natural number can be written as the sum of a triangular number, an even square and a generalized pentagonal number.
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LINKS
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EXAMPLE
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a(0) = 1 since 0 = 0^3 + 0*1/2 + 0*(3*0+1)/2.
a(5) = 1 since 5 = 0^3 + 2*3/2 + 1*(3*1+1)/2.
a(12) = 1 since 12 = 0^3 + 4*5/2 + 1*(3*1+1)/2.
a(19) = 1 since 19 = 1^3 + 2*3/2 + 3*(3*3+1)/2.
a(20) = 1 since 20 = 2^3 + 4*5/2 + 1*(3*1+1)/2.
a(75) = 1 since 75 = 2^3 + 4*5/2 + 6*(3*6+1)/2.
a(83) = 1 since 83 = 0^3 + 3*4/2 + 7*(3*7+1)/2.
a(97) = 1 since 97 = 3^3 + 10*11/2 + 3*(3*3+1)/2.
a(117) = 1 since 117 = 0^3 + 13*14/2 + 4*(3*4+1)/2.
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MATHEMATICA
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TQ[n_]:=IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[TQ[n-x^3-z(3z+1)/2], r=r+1], {x, 0, n^(1/3)}, {z, 0, (Sqrt[24(n-x^3)+1]-1)/6}]; Print[n, " ", r]; Continue, {n, 0, 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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