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A300396
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Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that 2*x or y is a power of 4 (including 4^0 = 1) and x + 63*y = 2^(2k+1) for some k = 0,1,2,....
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29
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0, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 1, 3, 1, 3, 4, 2, 2, 4, 1, 2, 1, 1, 2, 4, 3, 1, 1, 2, 1, 6, 2, 2, 2, 5, 1, 4, 1, 2, 6, 3, 3, 3, 1, 2, 3, 4, 3, 3, 2, 4, 2, 2, 1, 7, 3, 1, 4, 1, 2, 8, 1, 3, 7, 3, 4, 6, 3, 4, 4, 6, 4, 3, 2, 4, 3, 1, 2
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OFFSET
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1,3
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COMMENTS
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Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 5, 13, 25, 29, 59, 61, 79, 91, 95, 101, 103, 1315, 2^k (k = 1,2,3,...), 2^(2k+1)*m (k = 0,1,2,... and m = 3, 5, 7, 11, 15, 19, 23, 887).
This is stronger than the conjecture that A300360(n) > 0 for all n > 1. Note that a(387) = 3 < A300360(387) = 4 and a(1774) = 1 < A300360(1774) = 2.
We have verified that a(n) > 0 for all n = 2..10^7.
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LINKS
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EXAMPLE
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a(29) = 1 since 29^2 = 2^2 + 2^2 + 7^2 + 28^2 with 2*2 = 4^1 and 2 + 63*2 = 2^7.
a(86) = 2 since 65^2 + 1^2 + 19^2 + 53^2 = 65^2 + 1^2 + 31^2 + 47^2 with 1 = 4^0 and 65 + 63*1 = 2^7.
a(1774) = 1 since 1774^2 = 8^2 + 520^2 + 14^2 + 1696^2 with 2*8 = 4^2 and 8 + 63*520 = 2^15.
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MATHEMATICA
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SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Pow[n_]:=Pow[n]=IntegerQ[Log[4, n]];
tab={}; Do[r=0; Do[If[Pow[y]||Pow[(2*4^k-63y)/2], Do[If[SQ[n^2-y^2-(2*4^k-63y)^2-z^2], r=r+1], {z, 0, Sqrt[Max[0, (n^2-y^2-(2*4^k-63y)^2)/2]]}]], {k, 0, Log[4, Sqrt[63^2+1]*n/2]}, {y, 0, Min[n, 2*4^k/63]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
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CROSSREFS
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Cf. A000118, A000290, A000302, A271518, A279612, A281976, A299924, A299537, A299794, A300219, A300356, A300360, A300362.
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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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