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Number of solutions to x^3 + y^3 + z^3 + t^3 == 1 (mod n) for 1 <= x, y, z, t <= n.
2

%I #26 Jun 07 2017 00:35:36

%S 1,8,27,64,125,216,336,512,1296,1000,1331,1728,1794,2688,3375,4096,

%T 4913,10368,7410,8000,9072,10648,12167,13824,15625,14352,34992,21504,

%U 24389,27000,30225,32768,35937,39304,42000,82944,48396,59280,48438,64000,68921,72576,77529,85184,162000,97336

%N Number of solutions to x^3 + y^3 + z^3 + t^3 == 1 (mod n) for 1 <= x, y, z, t <= n.

%C It appears that a(n) = n^3 for n in A088232. See also A066498. - _Michel Marcus_, Oct 11 2016

%H Chai Wah Wu, <a href="/A276919/b276919.txt">Table of n, a(n) for n = 1..10000</a>

%t JJJ[4, n, lam] = Sum[If[Mod[a^3 + b^3 + c^3 + d^3, n] == Mod[lam, n], 1, 0], {d, 0, n - 1}, {a, 0, n - 1}, {b, 0, n - 1}, {c, 0 , n - 1}]; Table[JJJ[4, n, 1], {n, 1, 50}]

%o (PARI) a(n) = sum(x=1, n, sum(y=1, n, sum(z=1, n, sum(t=1, n, Mod(x,n)^3 + Mod(y,n)^3 + Mod(z,n)^3 + Mod(t,n)^3 == 1)))); \\ _Michel Marcus_, Oct 11 2016

%o (PARI) qperms(v) = {my(r=1,t); v = vecsort(v); for(i=1,#v-1, if(v[i]==v[i+1], t++, r*=binomial(i,t+1);t=0));r*=binomial(#v,t+1)}

%o a(n) = {my(t=0); forvec(v=vector(4,i,[1,n]), if(sum(i=1, 4, Mod(v[i], n)^3)==1, print1(v", "); t+=qperms(v)),1);t} \\ _David A. Corneth_, Oct 11 2016

%o (Python)

%o def A276919(n):

%o ndict = {}

%o for i in range(n):

%o i3 = pow(i,3,n)

%o for j in range(i+1):

%o j3 = pow(j,3,n)

%o m = (i3+j3) % n

%o if m in ndict:

%o if i == j:

%o ndict[m] += 1

%o else:

%o ndict[m] += 2

%o else:

%o if i == j:

%o ndict[m] = 1

%o else:

%o ndict[m] = 2

%o count = 0

%o for i in ndict:

%o j = (1-i) % n

%o if j in ndict:

%o count += ndict[i]*ndict[j]

%o return count # _Chai Wah Wu_, Jun 06 2017

%Y Cf. A000189, A047726, A060839, A063454, A087412, A087786, A254073, A276920.

%K nonn,mult

%O 1,2

%A _José María Grau Ribas_, Sep 22 2016