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A349992
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Number of ways to write n as x^4 + y^2 + (z^2 + 2*4^w)/3, where x, y, z are nonnegative integers, and w is 0 or 1.
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3
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1, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 2, 2, 2, 4, 8, 7, 7, 6, 5, 6, 6, 6, 8, 7, 8, 6, 1, 4, 2, 6, 8, 6, 7, 5, 7, 6, 6, 6, 7, 7, 8, 7, 3, 5, 3, 4, 6, 6, 6, 7, 5, 3, 5, 4, 9, 8, 9, 8, 2, 4, 1, 2, 9, 8, 10, 8, 4, 6, 4, 9, 6, 6, 6, 4, 2, 2, 1, 2, 10, 10, 13, 8, 9, 7, 9, 9, 7, 10, 6, 10, 4, 3, 4, 3, 11, 10, 9
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OFFSET
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1,2
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COMMENTS
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Conjecture 1: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 30, 64, 80, 302, 350, 472, 480, 847, 3497, 13582, 25630, 38064.
This has been verified for n up to 10^6.
Conjecture 2: If (a,b,c,m) is one of the ordered tuples (1,1,11,12), (1,1,11,60), (1,1,14,15), (1,1,23,24), (1,1,23,32), (1,1,23,48), (1,2,23,96), (2,1,11,60), (2,1,23,24), (2,1,23,48), (4,1,23,48), then each n = 1 2,3,... can be written as a*x^4 + b*y^2 + (z^2 + c*4^w)/m, where x,y,z are nonnegative integers, and w is 0 or 1.
We have verified Conjecture 2 for n up to 2*10^5.
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LINKS
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EXAMPLE
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a(30) = 1 with 30 = 1^4 + 5^2 + (2^2 + 2*4)/3.
a(480) = 1 with 480 = 1^4 + 14^2 + (29^2 + 2*4)/3.
a(847) = 1 with 847 = 0^4 + 29^2 + (4^2 + 2*4^0)/3.
a(3497) = 1 with 3497 = 4^4 + 48^2 + (53^2 + 2*4^0)/3.
a(13582) = 1 with 13582 = 9^4 + 28^2 + (53^2 + 2*4^0)/3.
a(25630) = 1 with 25630 = 5^4 + 158^2 + (11^2 + 2*4^0)/3.
a(38064) = 1 with 38064 = 3^4 + 157^2 + (200^2 + 2*4^0)/3.
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MATHEMATICA
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SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[3(n-x^4-y^2)-2*4^z], r=r+1], {x, 0, (n-1)^(1/4)}, {y, 0, Sqrt[n-1-x^4]}, {z, 0, 1}]; 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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