A276919 Number of solutions to x^3 + y^3 + z^3 + t^3 == 1 (mod n) for 1 <= x, y, z, t <= n.
1, 8, 27, 64, 125, 216, 336, 512, 1296, 1000, 1331, 1728, 1794, 2688, 3375, 4096, 4913, 10368, 7410, 8000, 9072, 10648, 12167, 13824, 15625, 14352, 34992, 21504, 24389, 27000, 30225, 32768, 35937, 39304, 42000, 82944, 48396, 59280, 48438, 64000, 68921, 72576, 77529, 85184, 162000, 97336
Offset: 1
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
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}] -
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
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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)} 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 -
Python
def A276919(n): ndict = {} for i in range(n): i3 = pow(i,3,n) for j in range(i+1): j3 = pow(j,3,n) m = (i3+j3) % n 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 = (1-i) % n if j in ndict: count += ndict[i]*ndict[j] return count # Chai Wah Wu, Jun 06 2017
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