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A301303
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Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= 0 <= z <= w such that x^2 + 23*y^2 = 2^k*m^3 for some k = 0,1,2 and m = 1,2,3,....
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4
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1, 1, 1, 1, 3, 3, 1, 1, 3, 3, 2, 2, 2, 4, 2, 1, 3, 5, 3, 5, 5, 4, 3, 3, 3, 5, 3, 3, 4, 6, 2, 1, 4, 3, 4, 5, 3, 4, 2, 3, 6, 5, 2, 4, 6, 4, 2, 2, 3, 6, 3, 3, 4, 6, 4, 4, 5, 4, 4, 5, 2, 4, 4, 1, 7, 8, 2, 7, 8, 5, 3, 5, 5, 7, 4, 5, 6, 5, 2, 5
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OFFSET
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1,5
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COMMENTS
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Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k (k = 0,1,2,...), 3, 7, 115, 151, 219, 267, 1151, 1367.
We have verified a(n) > 0 for all n = 1..10^8.
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LINKS
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EXAMPLE
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a(2) = 1 since 2 = 1^2 + 0^2 + 0^2 + 1^2 with 1 + 23*0 = 1^3.
a(4) = 1 since 4 = 2^2 + 0^2 + 0^2 + 0^2 with 2^2 + 23*0 = 2^2*1^3.
a(7) = 1 since 7 = 2^2 + 1^2 + 1^2 + 1^2 with 2^2 + 23*1^2 = 3^3.
a(48) = 2 since 48 = 4^2 + 0^2 + 4^2 + 4^2 = 6^2 + 2^2 + 2^2 + 2^2 with 4^2 + 23*0^2 = 2*2^3 and 6^2 + 23*2^2 = 2*4^3.
a(115) = 1 since 115 = 3^2 + 3^2 + 4^2 + 9^2 with 3^2 + 23*3^2 = 6^3.
a(267) = 1 since 267 = 3^2 + 1^2 + 1^2 + 16^2 with 3^2 + 23*1^2 = 2^2*2^3.
a(1151) = 1 since 1151 = 7^2 + 3^2 + 2^2 + 33^2 with 7^2 + 23*3^2 = 2^2*4^3.
a(1367) = 1 since 1367 = 17^2 + 5^2 + 18^2 + 27^2 with 17^2 + 23*5^2 = 2^2*6^3.
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MATHEMATICA
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SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
QQ[n_]:=n>0&&(CQ[n]||CQ[n/2]||CQ[n/4]);
tab={}; Do[r=0; Do[If[QQ[x^2+23y^2], Do[If[SQ[n-x^2-y^2-z^2], r=r+1], {z, 0, Sqrt[(n-x^2-y^2)/2]}]], {y, 0, Sqrt[n/2]}, {x, y, Sqrt[n-y^2]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
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CROSSREFS
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Cf. A000118, A000290, A271510, A271513, A271518, A281976, A300666, A300667, A300708, A300712, A300751, A300752, A300791, A300792, A300844, A300908, A301304, A301314.
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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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