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A254629
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Number of ways to write n as x^2 + y*(y+1) + z*(4*z+1) with x,y,z nonnegative integers.
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1
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1, 1, 1, 1, 1, 1, 3, 2, 1, 2, 1, 3, 2, 1, 1, 2, 3, 1, 3, 1, 3, 5, 2, 1, 3, 3, 2, 3, 2, 3, 3, 3, 1, 2, 4, 1, 5, 1, 2, 5, 2, 3, 5, 4, 1, 4, 4, 3, 4, 4, 2, 5, 2, 1, 4, 5, 5, 3, 1, 1, 7, 5, 1, 3, 4, 2, 5, 3, 2, 6, 5, 3, 4, 4, 5, 5, 4, 4, 5, 3, 1, 8, 2, 4, 7, 3, 4, 3, 5, 3, 6, 3, 3, 3, 6, 4, 5, 5, 2, 9, 2
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OFFSET
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0,7
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n, and a(n) > 1 for all n > 338.
(ii) If f(x) is one of the polynomials 3x^2+x, (7x^2+3x)/2, (9x^2-x)/2, (11x^2-7x)/2, (15x^2-7x)/2, (15x^2-11x)/2, then any nonnegative integer n can be written as x^2 + y*(y+1) + f(z) with x,y,z nonnegative integers.
We have proved that for each n = 0,1,... there are integers x,y,z such that n = x^2 + y*(y+1) + z*(4z+1).
It is known that {x^2+y*(y+1): x,y=0,1,...} = {x*(x+1)/2+y*(y+1)/2: x,y=0,1,...}.
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LINKS
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EXAMPLE
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a(103) = 1 since 103 = 8^2 + 0*1 + 3*(4*3+1).
a(122) = 1 since 122 = 9^2 + 1*2 +3*(4*3+1).
a(143) = 1 since 143 = 6^2 + 1*2 + 5*(4*5+1).
a(167) = 1 since 167 = 3^2 + 9*10 + 4*(4*4+1).
a(248) = 1 since 248 = 5^2 + 4*5 + 7*(4*7+1).
a(338) = 1 since 338 = 5^2 + 10*11 + 7*(4*7+1).
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MATHEMATICA
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SQ[n_]:=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-y(y+1)-z*(4z+1)], r=r+1], {y, 0, (Sqrt[4n+1]-1)/2}, {z, 0, (Sqrt[16(n-y(y+1))+1]-1)/8}];
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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