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A229294
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Number of solutions to x^2 + y^2 + z^2 + t^2 == n (mod 2*n) for x,y,z,t in [0, 2*n).
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4
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8, 96, 264, 384, 1160, 3168, 3080, 1536, 7560, 13920, 11528, 12672, 18824, 36960, 38280, 6144, 41480, 90720, 57608, 55680, 101640, 138336, 101384, 50688, 149000, 225888, 208008, 147840, 201608, 459360, 245768, 24576, 380424, 497760, 446600, 362880, 415880
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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)=8 and the sequence a(n)/8 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 + d^2, 2n] == n, 1, 0], {d, 0, 2n - 1}, {a, 0, 2n - 1}, {b, 0, 2n - 1}, {c, 0, 2n - 1}]; Table[Print[aaa = A[n]]; aaa, {n, 1, 40}]
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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)^4); polcoeff( lift(p), n)} \\ Andrew Howroyd, Aug 07 2018
(PARI) a(n)={my(f=factor(n)); 8*prod(i=1, #f~, my([p, e]=f[i, ]); if(p==2, 3*2^(2*e), p^(2*e-1)*(p^(e+1)+p^e-1)))} \\ Andrew Howroyd, Aug 07 2018
(Python)
ndict = {}
n2 = 2*n
for i in range(n2):
i3 = pow(i, 2, n2)
for j in range(i+1):
j3 = pow(j, 2, n2)
m = (i3+j3) % n2
if m in ndict:
if i == j:
ndict[m] += 1
else:
ndict[m] += 2
else:
if i == j:
ndict[m] = 1
else:
ndict[m] = 2
count = 0
for i in ndict:
j = (n-i) % n2
if j in ndict:
count += ndict[i]*ndict[j]
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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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