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A229295
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Number of solutions to x^2 + y^2 + z^2 == n (mod 2n) for x,y,z in [0, 2*n).
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4
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4, 24, 36, 32, 100, 216, 196, 192, 396, 600, 484, 288, 676, 1176, 900, 256, 1156, 2376, 1444, 800, 1764, 2904, 2116, 1728, 2900, 4056, 3564, 1568, 3364, 5400, 3844, 1536, 4356, 6936, 4900, 3168, 5476, 8664, 6084, 4800, 6724, 10584, 7396, 3872, 9900, 12696
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OFFSET
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1,1
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COMMENTS
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All values are divisible by a(1)=4 and the sequence a(n)/4 is multiplicative. - Andrew Howroyd, Aug 07 2018
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LINKS
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FORMULA
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MATHEMATICA
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A[n_] := Sum[If[Mod[a^2 + b^2 + c^2, 2*n] == n, 1, 0], {a, 0, 2*n - 1}, {b, 0, 2*n - 1}, {c, 0, 2*n - 1}]; Array[A, 100]
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PROG
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(PARI) a(n)={my(m=2*n); my(p=Mod(sum(i=0, m-1, x^(i^2%m)), x^m-1)^3); polcoeff( lift(p), n)} \\ Andrew Howroyd, Aug 06 2018
(PARI) a(n)={my(f=factor(n)); 4*prod(i=1, #f~, my([p, e]=f[i, ]); if(p==2, if(e%2, 3, 1)*2^(e+e\2), p^(e+(e-1)\2)*(p^(e\2)*(p+1) - 1)))} \\ Andrew Howroyd, Aug 07 2018
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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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