A229294 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).
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
Offset: 1
Keywords
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
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}] -
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
def A229294(n): 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] return count # Chai Wah Wu, Jun 07 2017
Formula
a(n) = 8*A240547(n) for odd n, a(2^k) = 24*2^(2*k). - Andrew Howroyd, Aug 07 2018
Comments