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A255191
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Number of solutions to the equation n = x^2 + y*(y+1)/2 + z*(z+1)/2 (x,y,z=0,1,...) with x == Floor[sqrt(n)] (mod 2) and y <= z.
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1
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1, 1, 1, 1, 2, 1, 3, 2, 1, 1, 2, 2, 2, 2, 1, 2, 4, 2, 2, 2, 4, 2, 4, 1, 2, 4, 2, 2, 2, 3, 3, 4, 3, 1, 2, 2, 4, 4, 4, 2, 4, 2, 5, 4, 1, 3, 6, 4, 2, 3, 3, 3, 5, 2, 2, 5, 5, 3, 3, 3, 3, 4, 3, 1, 5, 5, 4, 6, 2, 2, 6, 4, 6, 5, 4, 2, 6, 3, 3, 3, 5, 5, 6, 3, 2, 7, 2, 4, 4, 2, 5, 6, 6, 3, 4, 5, 2, 6, 3, 2, 6
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OFFSET
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0,5
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COMMENTS
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By Dickson's book in the reference, E. Lionnet observed in 1872 that any nonnegative integer can be written as the sum of a square and two triangular numbers.
We have a(n) > 0 for all n. This follows from Theorem 1(i) and Lemma 2 of Sun's reference in 2007. Note that if n = m*(m+1) with m a nonnegative integer then floor(sqrt(n)) = m.
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REFERENCES
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L. E. Dickson, History of the Theory of Numbers, vol. II, AMS Chelsea Publ., 1999, p. 24.
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LINKS
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EXAMPLE
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a(14) = 1 since 14 = 1^2 + 2*3/2 + 4*5/2 with floor(sqrt(14)) == 1 (mod 2).
a(44) = 1 since 44 = 4^2 + 0*1/2 + 7*8/2 with floor(sqrt(44)) == 4 (mod 2).
a(63) = 1 since 63 = 5^2 + 4*5/2 + 7*8/2 with floor(sqrt(63)) == 5 (mod 2).
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MATHEMATICA
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TQ[n_]:=IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[Mod[Floor[Sqrt[n]]-x, 2]==0&&TQ[n-x^2-y*(y+1)/2], r=r+1], {x, 0, Sqrt[n]}, {y, 0, (Sqrt[4(n-x^2)+1]-1)/2}];
Print[n, " ", r]; Label[aa]; 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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